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Transcript
TEXT FLY WITHIN
THE BOOK ONLY
DO
1620025m
OUP
-24-4-1-695,0(0
OSMANIA UNIVERSITY LIBRARY
Gall
No
Author
5 1 2* & $ & fi Accession No. Gt ** ^ ^
'Vr/-v<*^< ^^^.cfok^
'
.
This lx>
should be returned on or before the date
last
marked below.
Algebra and Trigonometry
Algebra
and
Trigonometry
ALVIN
Head
BETTINGER
K.
Department and
Professor of Mathematics
of
The Crcif/hton University
and
JOHN
A.
ENGLUND
Formerly Assistant Professor
of Mathematics
The Cr eight on University
INTERNATIONAL TEXTBOOK COMPANY
Scranton, Pennsylvania
INTERNATIONAL TEXTBOOKS IN MATHEMATICS
L
R.
W//COX
Professor of Mathematics
Illinois Institute of
Technology
CONSULTING EDITOR
Second Printing, ] armory 1963
>
Copyright
I960, by International Textbook Company. All rights reserved.
Printed in the United States of America by The Haddon Craftsmen, Inc.
at Scranton, Pennsylvania. Library of Congress Card Number: 60-9987.
t
Preface
The authors
believe that in this book the basic material of college
algebra and trigonometry has been presented with suflicient rigor
to provide a firm and coherent groundwork for subsequent courses
in mathematics. The material is presented in such a way that it
can be grasped by the student without undue assistance.
The first chapter consists of a number of introductory topics
which are intended to serve as a review of elementary algebra.
Actually, something more than a mere review is available in this
chapter. Not only are the review topics considered from a more
mature point of view than is usual, but the treatment is interwoven with concepts that are basic for an understanding of more
advanced mathematical topics. We begin with the algebra of the
real-number system. Axioms pertaining to fundamental operations
are given, and the various rules for the elementary operations of
algebra are derived and logically connected with the bcisic assumptions. We are led naturally to an ordering of the real-number
system and to the foundation for a later chapter on inequalities
that is easier to understand and more useful than the treatment
one customarily finds in textbooks.
The second chapter introduces the student to the function concept, which serves as a basis for much of the remaining work of
Certain aspects of the discussion become somewhat
abstract, but the student is reminded that a proper understanding
of the true nature of a function is important for virtually all later
courses in mathematics.
In line with modern demands, the trigonometric functions are
initially introduced in the third chapter as functions of real numbers. Following this presentation, the transition to functions of
the book.
angles
is relatively simple.
rest of the volume contains all the usual topics from college
algebra and trigonometry. In certain instances a particular devel-
The
differ somewhat from that usually found. In such
the
authors believe, the departure is to the advantage of
cases,
the student.
opment may
Preface
vi
We
are indebted to our colleague, Professor Morris Dansky, for
his valuable suggestions while the manuscript was in preparation.
wish particularly to express our deep appreciation to Pro-
We
fessor L. R. Wilcox for his thorough criticism of the manuscript
and his invaluable suggestions for improvement of the text. Finally,
a special
pany for
is due the International Textbook Comand
cooperation
patience.
word
its
of thanks
A. K. BETTINGER
J. A. ENGLUND
Omaha, Nebraska
August, 1960
Contents
1.
INTRODUCTORY TOPICS
1-1.
The Real-Number System
1-2.
Fundamental Assumptions
1
1
1
5
1-4.
Operations With Zero
Reciprocals
1-5.
The Real-Number Scale
6
1-3.
5
Rules of Signs
Fundamental Operations on Fractions
1-8.
Order Relations for Real Numbers
1-9.
Absolute Value
1-10. Inequalities Involving Absolute Values
1-6.
1-7.
1-11.
Positive Integral Exponents
1-12.
Algebraic Expressions
Equations and Identities
1-13.
1-14.
Symbols of Grouping
Order of Fundamental Operations
1-16. Addition and Subtraction of Algebraic Expressions
1-15.
1-17.
Multiplication of Algebraic Expressions
1-18.
Special Products
Division of Algebraic Expressions
1-19.
1-20.
1-21.
1-22.
1-23.
1-24.
1-25.
1-26.
1-27.
1-28.
1-29.
1-30.
2.
Factoring
Important Type Forms for Factoring
Greatest Common Divisor
Least Common Multiple
Reduction of Fractions
Signs Associated With Fractions
Addition and Subtraction of Fractions
Multiplication and Division of Fractions
Complex Fractions
Linear Equations
Linear Equations in One Unknown
THE FUNCTION CONCEPT
2-1.
2-2.
2-3.
2-4.
2-5.
2-6.
2-7.
Rectangular Coordinate Systems
Distance Between Two Points
Functions
Functional Notation
Some Special Functions
Variation
Classification of Functions
vii
in
a Plane
7
9
12
14
14
16
17
18
18
20
21
22
22
23
25
27
30
32
33
34
36
39
40
42
43
49
49
50
52
55
57
57
61
Confenfs
viii
3.
THE TRIGONOMETRIC FUNCTIONS
3-1.
Definitions of the Trigonometric Functions
Identities
68
3-4.
Tables of Trigonometric Functions
Positive and Negative Angles and Standard Position
71
75
76
77
78
3-7.
3-8.
3-9.
3-10.
4-2.
4-3.
4-4.
Trigonometric Functions of Angles
Tables of Natural Trigonometric Functions of Angles
OF EXPONENTS
81
82
86
86
88
DO
Positive Integral Exponents
Meaning of
Negative Exponents
Scientific Notation
1)2
1)2
4-6.
Rational Exponents
The Factorial Symbol
4-7.
The Binomial Theorem
1)7
48.
General
Term
1)9
4-5.
in the
J)7
Binomial Expansion
101
LOGARITHMS
5-1.
5-2.
5-3.
5-4.
101
Definition of a Logarithm
Laws of Logarithms
Systems of Logarithms
Common Logarithms
5-5.
Rules for Characteristic and Mantissa
5-6.
How
How
5-7.
5-8.
5-9.
to
to
Write Logarithms
of Mantissas
Use a Table
Logarithmic Computation
Change of Base
RIGHT TRIANGLES AND VECTORS
6-1.
Rounding Off Numbers
6-2.
6-3.
6-4.
Trigonometric Functions of Acute Angles
Procedures for Solving Right Triangles
Angles of Elevation and Depression
6-5.
Bearing
6-6.
Projections
Scalar and Vector Quantities
Logarithms of Trigonometric Functions
Logarithmic Solution of Right Triangles
6-7.
6-8.
6-9.
7.
Measurement of Angles
The Relation Between Radians and Degrees
Arc Length and Area of a Sector
THE LAWS
4-1.
6.
(54
3-2.
3-6.
5.
G3
63
3-3.
3-5.
4.
The Point Function P(t)
in
Navigation
arid
Surveying
TRIGONOMETRIC FUNCTIONS OF SUMS AND DIFFERENCES
7-1.
Derivation of the Addition Formulas
7-2.
The Double-Angle Formulas
7-3.
The Half-Angle Formulas
7-4.
Products of Two Functions Expressed as Sums, and Sums
Expressed as Products
102
105
105
106
108
108
110
113
115
115
116
117
120
121
122
125
131
133
135
135
139
140
143
Contents
8.
GRAPHS OF TRIGONOMETRIC FUNCTIONS; INVERSE FUNCTIONS
AND THEIR GRAPHS
8-1.
8-2.
8-3.
117
148
and Phase
8-4.
Periodicity, Amplitude,
Inverse Functions
Inverses of the Trigonometric Functions
14i)
ir>r>
156
LINEAR EQUATIONS AND GRAPHS
!)-!.
9-2.
1
163
Solutions of Simultaneous Equations
Algebraic Solution of Linear Equations in
9-4.
Linear Equations in Three Unknown*
Graphs of Linear Functions
9-5.
Intercepts
9-6.
Graphical Solution of Linear Equations in
9-3.
10.
146
146
Variation of the Trigonometric Functions
The Graph of the Sine Function
The Graphs of the Cosine and Tangent Functions
8-5.
8-6.
9.
ix
1(5*5
Two Unknowns...
Two Unknowns
.
.
.
173
173
175
177
)ETERMINANTS
10-1
.
10-2.
10-3.
10-4.
I
)eti'rminants of the Second Order
Determinants of the Third Order
Properties of Determinants
Solution of Three Simultaneous Linear Equations
in
Three
Unknowns
10-5.
10-7.
1 1
.
in
Three Unknowns
11-5.
183
()
184
185
189
t
^
189
191
192
105
106
Graphical Representation
Trigonometric Representation
11-6.
Multiplication and Division in Trigonometric
11-7.
DC Moivrc's Theorem
Roots of Complex Numbers
11-8.
When
Homogeneous Equations
Sum and Product of Determinants
COMPLEX NUMBERS
11-1.
The Complex Number System
The Standard Notation for Complex Numbers
11-2.
11-3.
Operations on Complex Numbers in Standard Form
11-4.
12.
181
Systems of Three Linear Equations
D=
10-6.
Form
198
199
200
EQUATIONS IN QUADRATIC FORM
12-1.
Quadratic Equations in One Unknown
12-9.
Solution of Quadratic Equations by Factoring
Completing the Square
Solution of Quadratic Equations by the Quadratic Formula
Equations Involving Radicals
Equations in Quadratic Form
The Discriminant
Sum and Product of the Roots
Graphs of Quadratic Functions
12-10.
Quadratic Equations
1
2-2.
12-3.
12-4.
12-5.
12-6.
12-7.
12-8.
166
167
169
170
171
in
Two Unknowns
204
204
20
1
206
.
.
.
209
212
214
215
217
218
221
Confenfs
x
12-11.
12-12.
12-13.
12-14.
13.
THEORY OF EQUATIONS
1 3-1.
Introductory Remarks
235
235
235
240
13-7.
The Remainder Theorem
241
The Fundamental Theorem of Algebra
243
Pairs of Complex Roots of an Equation
244
The Graph of a Polynomial for Large Values of aRoots Between a and b If /(a) and f(b) Have Opposite Signs 245
13-8.
Rational Roots
INEQUALITIES
14-3.
Introduction
Properties of Inequalities
Solution of Conditional Inequalities
14-4.
Absolute Inequalities
14-1.
14-2.
PROGRESSIONS
15-5.
Sequences and Series
Arithmetic Progressions
The General Term of an Arithmetic Progression
Sum of the First 71 Terms of an Arithmetic Progression
Arithmetic Means
15-6.
Harmonic Progressions
15-1
.
15-2.
15-3.
15-4.
1
5-7.
15-8.
15-9.
15-10.
15-11.
15-12.
15-13.
Geometric Progression
The General Term of a Geometric Progression
Sum of the First n Terms of a Geometric Progression
Geometric Means
Infinite Geometric Progression
Repeating Decimals
The Binomial Series
MATHEMATICAL INDUCTION
16-1.
16-2.
17.
233
Synthetic Division
13-6.
16.
231
13-3.
13-5.
15.
224
227
13-2.
13-4.
14.
Graphical Solutions of Systems of Equations Involving
Quadratics
Algebraic Solutions of Systems Involving Quadratics
Exponential and Logarithmic Equations
Graphs of Logarithmic and Exponential Functions
248
248
248
249
254
256
256
260
260
261
262
264
265
265
266
267
268
269
271
273
273
Method of Mathematical Induction
Proof of the Binomial Theorem for Positive Integral Exponents 275
PERMUTATIONS, COMBINATIONS, AND PROBABILITY
Fundamental Principle
17-2.
Permutations
17-3.
Permutations of n Things Not All Different
17-4.
Combinations
Binomial Coefficients
17-5.
Mathematical Probability
17-6.
Most Probable Number and Mathematical Expectation
17-7.
17-1.
17-8.
245
Statistical, or Empirical, Probability
278
278
279
280
281
282
283
284
284
Confenfs
17-9.
17-10.
17-11.
18.
xi
Mutually Exclusive Events
Dependent and Independent Events
Repeated Trials
285
286
287
SOLUTION OF THE GENERAL TRIANGLE
18-1.
Classes of Problems
18-2.
The Law of Sines
18-3.
Solution of Case
I
by the
Law
289
289
289
of Sines: Given
One Side and
Two Angles
290
18-4.
Solution of Case II by the Law of Sines: Given
the Angle Opposite One of Them
18-5.
The Law of Cosines
18-6.
Solution of Case III and Case IV by the Law
The Law of Tangents
The Half-Angle Formulas
Area of a Trjangle
18-7.
18-8.
18-9.
Two
Sides and
of Cosines
291
296
297
298
300
302
APPENDIX
A.
Tables
"B.
Answers
INDEX
to
Odd-Numbered Problems
307
337
353
1
1-1.
Introductory Topics
THE REAL-NUMBER SYSTEM
The real-number system that we
use in the early part of this
a development from the original counting numbers, or
positive integers, such as 1, 2, and 3. Almost simultaneously with
the invention of positive integers, practical problems of measurement gave rise to positive fractions, such as 1/2, 5/6, and 16/7.
Much later, in comparatively modern times, the concepts of negative
numbers and of other types of numbers were gradually developed.
Negative numbers were invented when the problem of subtracting
one number from a smaller one presented itself. Thus, the number
system was soon enlarged to include the negative integers and
course
is
fractions. These positive and negative numbers, together with zero,
are called the rational numbers. Hence, a rational number is
defined to be any number that can be expressed as the quotient, or
ratio, of two integers. For example,
2/3, 5 (which may be considered as 5/1), and
7 are rational numbers.
The number system was then extended to include also numbers
which cannot be expressed as the quotient of two integers, namely,
the irrational numbers; examples are ^/2 ami rr. The two classes of
numbers, rational and irrational, comprise the real numbers. These
numbers are so called in contrast to the imaginary or complex
numbers considered in Chapter 11.
1-2.
FUNDAMENTAL ASSUMPTIONS
We shall proceed to introduce the four fundamental operations
of addition, subtraction, multiplication, and division into the system
of real numbers. The reader has probably been performing these
operations in arithmetic and algebra without being conscious that
certain basic laws were being obeyed. We shall introduce the four
fundamental operations and state, without proof, the laws or
assumptions governing them.
i
2
Sec.
Introductory Topics
1-2
is assumed that there is a mode of combining any
numbers a and 6 so as to produce a definite real number
called their sum. This mode of combination is called addition. The
sum of a and b is denoted by a + b. In this sum a and b are called
Addition. It
two
real
terms.
Multiplication. It is assumed that there is a mode of combining
any two real numbers a and b to produce a definite real number
mode of combination is called multiplicaThe product of a and b is denoted by a b or by ab. The individual numbers a and b are called factors of the product.
called their product. This
tion.
Commutative Law for Addition.
If
a and b are any real numbers,
then
a
(1-1)
+
b
=
b
+
a.
Thus the sum of two mumbers is the same regardless
in which they are added. For example,
1
,
+ 3=3 + 2.
2
Associative
Law
of the order
for Addition. If a,
b, c
are any real numbers,
then
(1-2)
(a
+
6)
+c =a+
(6
+
c).
we obtain the same result whether we add the sum of a and
6 to c, or we add a to the sum of b and c. Since the way in which
we associate or group these numbers is immaterial, we may write
this common value as a + 6 4- c without fear of ambiguity. For
That
is,
example,
2
+3+4 =
Commutative Law for
(2
+
3)
=
+4
2
(3
+ 4).
and
b are
+
Multiplication. If a
any
real
num-
bers, then
ab
(1-3)
-
ba.
is, the product of two numbers is the same regardless of the
order in which they are multiplied. For example,
That
2-3=3-2.
Associative
Law
for Multiplication. If a, 6, c are
,
any real num-
bers, then
(1-4)
1
(o6)c
=
a(6c).
Illustrations of the laws are given here only for the
bers, the positive integers.
all real numbers.
It is understood,
most familiar num-
however, that the laws apply to
12
Sec.
That
is,
3
Introductory Topics
we
we
obtain the same result whether
multiply the product
of a and b by c, or we multiply a by the product of b and c. Since
the way in which we associate or group these numbers is immaterial, we may write the result as abc without fear of ambiguity.
Thus
=
2-:5-4
Distributive Law. If
(1
(2-:})
a(b
2- (:;-4).
are any real numbers, then 2
a, b, c
f>)
=
i
+
=
r)
+
ah
or.
This law, which is usually known as the distributive law for multiplication with respect to addition, effects a connection between
addition and multiplication. The distributive law forms the basis
for the factoring process in algebra, as will be seen.
simple example of the distributive law is
A
+
2 -(3
4)
=2-:5
+
2-
1.
This law can be extended to the case where the
three or
more terms, as
sum
consists of
in the following illustration:
For positive integers, multiplication may also be interpreted as
repeated addition. Thus, by the distributive law,
S
4
;>,.4
Zero.
=
=
=
(1
+ + 1)
+ + +
+ + +
1
.">(!
:$
It is
\
I
:!
;;
--
1)
(1
-4)
+
(1
4)
+
(1
4)
=
(3-
1)
+
Cl-
1)
+
(o-
0,
-
=
4
1)
+
4
+
C>-
1)
+
4,
:j.
assumed that there
and denoted by
(1
-
1
is
number
number
a special
such that, for every real
a
(i)
+
=
a.
i
=
called zero
,
For example,
i]
It
of
a
a
+
=
+
0'
From
2
+
o
j,
+
o
=
o.
can be easily shown that only one number with the property
can exist. For let 0' be another such number. Then, since
= a and b + 0' = b for any numbers r/, b, it follows, by taking
and
b
=
0,
that
()'
tion,
o
o=;j,
+
=
the commutative law,
The right
we agree
side of
0',
and
+
=
0.
0'.
(1-5) should read (ah)
to omit the parentheses
performed before any addition.
0'
when
-f
all
(ar).
However, by conven-
multiplications are to be
4
Sec.
Introductory Topics
12
Negative of a Number. It is assumed that for every real number
a there exists a corresponding number, called the negative of a and
designated by a, such that
+ (-
a
7)
(1
=
a)
0.
For example,
+
1
(-
=0,
1)
(-
+
2)
2
=0.
That each number has but one negative may be shown in the
0.
following way: Let x be another negative of a, so that a + x
Then
By
= (-
a
n)
+0
= (-
-
a
~ ((-
a
+
+
<i)
+
(a
x).
associativity,
o)
+
n)
+
x,
or
In particular, the negative of zero
.r.
is
- -
-
x
=
+
0.
The Unit. It is assumed that there is a special number called the
unit and denoted by 1, such that, for every real number a,
(1
a
8)
1
There cannot be a second unit
i
whence
=
1
r =
.
not
is
1'.
i,
a.
we
If there were,
r-
1
-
could say that
r,
1'.
Reciprocal of a Number. It
which
=
0,
there
is
is
assumed that for every number a
an associated number -
>
a
called the rccip-
rocal of a, such that
a
(1-9)
The reader may verify the
of each number.
a
x
1,
then x
~
Thus,
if
'7
= L
fact that there is only one reciprocal
another reciprocal of a, that is, if
is
,r
a
in the definition of
important to note the restriction a ^
the reciprocal. In the next section we shall see why this restriction
It is
is
needed.
Subtraction.
6, is
(1
The
difference a
6,
of
any
defined by
10)
o
-
b
=a +
(-
6).
real
numbers a and
Sec.
14
5
/nfroc/ucfory Topics
The operation indicated by the si#n minus which produces for any
two real numbers a and b the real number a b is called
subtraction.
The quotient
Division.
a
f
b or
or a
.
b of
-r-
real
any
numbers a
t)
and
&,
where b
^ 0,
is
defined by
'
The operation associating with
quotient
numbers a and
real
b (b
=/-
0) their
called division.
is
should be noted that subtraction and division are subordinate
and multiplication, in that they are defined in terms of
b is that number .r for which
these latter. The difference a
b + .r - a. Also, tho quotient a b is that number y for which
b y ~ a. It should be noted that
+ (-) = for every
number a, and that a 'a a (I/ a)
1 for every number a ~f 0.
It
to addition
OPERATIONS WITH ZERO
1-3.
It
erty
a
has already been noted that the special number
has the prop- a for every real number a. In particular, we may let
(t +
=
to obtain
=
+
It has already been noted that
for every real number a.
Next, we prove that for every real
a
(1-12)
a
Let x
0.
.r
()
0.
=
0,
so that
number
=
a
+
x.
=a
4-
a,
-r 0.
Then, by the distributive law,
=
a
=
a
(0
+
0)
=
a
-
+
-
a
we add -.r, we obtain
= + (- .r) = (.r + .r) + (- .r) = x + (x + (Since a: = a 0, (1-12) is established.
From this last result, it follows that, for b 0,
x
If
;r
x))
=
x
+
=
x.
-
013)
1-4.
It
H
RECIPROCALS
was noted that every non-zero number has a
reciprocal.
We
can now see why cannot have a reciprocal. If has a reciprocal x,
x = 1. Since it has been shown that
x = 0, we would have
then
6
Sec.
Introductory Topics
=
to conclude that
1.
However,
=
if
1 is allowed,
1-4
then for every
number a we have
would be the only number
Hence,
number system. This
in the
cannot have a
situation obviously should be ruled out. Therefore,
=
a
the
since
a/6
(1/6),
reciprocal. Moreover,
quotient a/6 is not
defined
when
6
=
The reciprocal
0.
way
-111
:
n
1^
(1-J4)'
^
r
b
a
provided that neither a nor 6
with
a
r
b
b
To prove
I
7
then multiply by
---,*
a
is 0.
1
11
-.-.&&=a
a b
We
two non-zero numbers can be
of the product of
expressed in another
'T-O*
a
I
we begin
=1.
i
,
this result,
,
1
b
r to obtain
L__
-1
!
/;
a
b
.
-.
,
a
a
b
a
b
In the preceding proof, free use has been
made
-!
i
i
a
,
b
of the commutative
and associative laws.
Finally, if a
^ 0,
the reciprocal of the reciprocal of a
is
a
itself.
Thus,
^=
(1-15)
Since
a.
=
^- (!/)
!,
multiplication by a gives
a
-
=
a
1
-T-
I/a
1-5.
(I/a)
a
=
-7I/a
1
=
.
I/a
THE REAL-NUMBER SCALE
Real numbers may be represented by points on a straight line.
such a line select an arbitrary point
as origin and lay off
equal unit distances in both directions, as shown in Fig. 1-1. (The
On
i
i
i
i
i
i
i
i
-7-6-5-4-3-2-10
i
i
i
i
i
i
i
I
2
3
4
5
6
7
Fiu. 1-1
unit segment may have any length whatsoever.) Label the points
thus far specified as indicated
is the origin, 1 is the first point to
:
Sec.
16
7
Introductory Topics
the right, 2 is the second point to the right,
1 is the first point to
the left, and so on. Rational numbers that are not integers correspond to certain other points in a natural way. For example, 1/2
corresponds to the midpoint of the segment joining points labeled
and 1; and
7/3 represents the point one-third of the distance from the point 2 to the point
3. It is a basic assumption
the
real
numbers
that every point corresponds to a
concerning
unique real number, and that every real number corresponds to
The full significance of this assumption cannot
be developed in an elementary text.
One observation of importance can be made at this time. The
non-zero real numbers are divided into two classes. One class consist of numbers representing points to the "right" of 0, and the
other consists of numbers representing points to the "left" of 0.
The first class consists of positive numbers, and the second of
negative numbers. The number may be considered as constituting
a third class. It is understood that no two of the three classes zero,
exactly one point.
positive numbers,
a
common. Thus
and negative numbers
number cannot be both
have any numbers in
and zero, both
positive
positive and negative, or both negative and zero. The specific desigwhich
nation of any negative number will include an explicit sign
,
prefixed. (This convention, however, does not exclude the possibility of allowing a general symbol, such as x, to stand for a negaPositive numbers do not require such a sign,
tive number.)
although frequently the sign f is used.
is
It is
to be
positive, as is
1-6.
assumed that the sum of two positive numbers
also the product of two positive numbers.
is
RULES OF SIGNS
To operate effectively with real numbers, a knowledge of the
rules of signs and of properties of negative numbers is essential.
In each of the following relationships, a and b are any two real
numbers, except that the denominator of a fraction
- (-
(1-16)
-
(1-17)
-
(1-18)
(1-19)
(1-20)
(1-21)
(-
a)b
= -
(a
(a
(6);
+
-
a}
=
a.
6)
= -
o
-
6.
6)
= -
a
+
6.
in particular,
(-)(- 6)
-L = -
=<*b.
J
may not be
(-
1)6
= -
6.
zero.
8
Sec.
Introductory Topics
a
a
(1-22)
-b
a
b
a
b
_
~
^T
1-6
a
6'
Proo/s o/ (i-i
(1-16)
(1-17)
(1-18)
- (- a) = - (- a) = a + (- a) + (- (- a)) = o + = a.
- (a + b) = + - (a + V) = - a + a + (- &) + b - (a + 6)
= - a - 6 + (a + 6) - (a + 6)
= -a-6 + = -a-6
by (1-6), (1-7).
- (a - 6) = - (a + (- 6)) = - a - (- 6) = - a + 6
by
(1-19) Since
(- a)
(1-20)
= (-
a)
0- 6 = (- o
b
a b - (a
(-
(-
6)
(b
,
a) -6
a)
+
+
+
= = ab
((-
-
= 0-6 =
-
=
b)
a)
6
a)
6)
(a
= - (-
= -
6)
-
(a
=
<"
a
1
(1-12),
b).
(1-19), (1-16).
by(1
-
20)
-
=
= ,( ~
N
a)
^| =
d-23)
by
(a
6))
by
/i oo\
(I-22 )
(1-17), (1-16).
1
= ~
-
/
(
'6
a-
1\
ft)
=
(-
|)
|
= ~
a
i
by
6
by
/i
(1
' in\
19)
*
(1-22), (1-16).
has already been observed that non-zero numbers are divided
classes, namely, positive and negative. It is assumed that
if # is positive, then -a is negative and that if a is negative, then
a is positive. All calculations involving negative numbers can be
made by performing calculations with positive numbers and applying one or more relationships just given. It follows from (1-17),
for example, that the sum of two negative numbers is negative, and
It
,
into
two
;
is
equal to the negative of the
sum
of the negatives of the given
from (1-20) it follows that the product of two
negative numbers is positive, and is equal to the product of the
numbers.
Also,
negatives of the given numbers. By (1-19) the product of a positive number and a negative number is negative.
Sec.
1-7.
1-7
9
Introductory Topics
FUNDAMENTAL OPERATIONS ON FRACTIONS
A further study of the algebra of real numbers leads us to the
consideration of the fundamental operations as applied to fractions.
By definition, a fraction is the quotient obtained by dividing
one number a by another number 6, where b is not zero. We
a the numerator and b the denominator; and we generally
write the fraction a/6, read "a over b" or "a divided by b."
call
We
shall list the following basic relationships for applying the
four operations to fractions. In them a, b, c, d are any real numno factor in the denominator of a fraction may
be zero.
bers, except that
ac
tt-24)
(1 24)
a
U*.
+
(1-25)
a
.b
c
d
(1-27)
a
+
ad
__
~~
6
_
c
ad
_
~~
d
c
a
b I
be
cd
c
c
(1-28)
A
_
bc~b
be
cd
ac
c
d
b
be
c
special case of (1-30) is
_
~
/ c
/ d
d
'
c
which states that the reciprocal of a fraction is found by inverting
the fraction. Also, by (1-30), dividing by a fraction is equivalent
to multiplying by its reciprocal.
Proofs of (1-24)
=
~
(1-29) Since
b
by
^
ac
7"
to (1-30)
*
~5
=
d
7"
1
*
C
b
*
"1
d
(&
1
,
,
""
*
w
*
1
*
bIT 3
d
=
a
C
lac
T"~5
==
bd
=
= J.1=J
6c6c66
2
U-24)
/^ rp\
(1-25)
(1-14),
1
a
:
a,6
+-
=
J.
l.,l
+ 6 .- =
a .-
a + 6
/"'i\l
+ 6) ,- = ^t-
(o
T~5
bd
by
^
/
(1-29).
\
v/ie\
.by (1-5).
10
/
rt
ra-^
f+l-ffz +
<'-26>
#
6
---
*x
(1-27)'
v
=
a
~
c
c
c
-
.
6)-
(
+-
=
1
-
a
c
c
+
,
f
,
(
v
6)'
x
-1 =
c
*->.
-
/
(a
v
=
t"\
t*
On\
/
,
(_
__
_ _ + (-
U/
Vx
a
6)
b)
fl(*
W/U/
Cfr
J.
d
=3
+9
11
c)
(2
+ 3) + 7 = (11 +
= 2 (3 5).
5
3)
+ 4) = 5 3 + 5 4.
+
d) 5(3
c
(1-22), (1-5).
be
(1-29), (1-24).
State which of the fundamental assumptions are employed in
each of the following equations:
+ (-5)
+ (-5)
o) 3
6)
1
-
ttCt
by
1-1.
*>
c
erf
Example
N
6)'
by
a
& _ a
-____
1-7
Sec.
fnfrocfucfory Topics
(6
+9.
+
6)
(3
+
7).
Solution:
d)
6)
c)
d)
The
The
The
The
and commutative laws
associative
associative law for multiplication.
distributive law.
Example
1-2.
In each of the following, perform the indicated operation:
a)
2+3.
b)
d)
(+5) -(+3).
e)
Solution:
6)
c)
d)
e}
/)
Each
+
(-2) -f (-3).
(-7) -(+6).
+
c)
5
/)
(-7) -(-6).
(-3).
case can be treated as an addition.
= 5.
(-2) +(-3) =
5 + (-3) =2.
(+5) -(+3) =
(- 7) - (+ 6) =
(-7) -(-6) =
a) 2
for addition.
associative law for addition.
3
-5.
+5 +
-7 +
-7 +
(-3)
(- 6)
<+6)
= +2.
= - 13.
= -1.
Example 1-3. By using the fundamental assumptions and rules for
and transforming the left side into the right side, justify the equation
(a
+ 6) -
(c
-
d)
=
(6
-
c)
+
(a
+ d),
operations^
Sec.
1-7
Introductory Topics
By
Solution:
By
(1-1)
and
Example
1 1
(1-18),
(a
+
6)
-
(a
+
6)
-
-
(c
=
d)
(a
+ 6) -
+A
c
(1-2),
1-4.
+
c
d
=
-
(b
c)
+
+
(a
d).
using the fundamental assumptions and rules of operations,
By
justify the equation
a
Solution:
By
b
(1-29) and (1-24),
6
ac
6
a
_
a
c
be
6
c
"~
a
_
"~
ac
b
c
(be)
(6
6c
__
~~
6
c)
c
EXERCISE 1-1
1.
Identify the fundamental law or laws that justify each of the following equations:
a.
x
c.
2(3
=
(2
e.
(a -f 2) (6
-
2.
Find the value
a.
j.
(-3)
(+7)
(- 5)
(-5)
3.
Evaluate each of the following:
a.
(+2) (-3).
(-7) (+5).
b.
c.
e.
2(-5).
f.
(-7)0.
g.
(-!)(-
h.
(-4) (-5) -(_2)(-l).
i.
(+2) (-3) -(+7) (-5).
4.
Determine the negative of each of the following
d.
g.
+
?/=?/
5)
+
=
of
+(+5).
-(+2).
- 0.
- (+5).
2)
d. 5(a
3)5.
3)
-
(6
3) (a
+ 2).
h.
k.
-
2-3.
J-
3#
5.
i
!
+ cb.
ca
(- 7) - (- 5).
(+32) -(-23) +(-45).
1.
(-3) (-5).
(-5) (-9).
&)]
3.
+
2.
:
d. 2x.
0.
c.
2a - 36.
k. x - 3.
g.
-
h.
a
1.
-
(x
c.
b.* 2/3.
1.02.
f.
_L_.
x + y
r
a
+6.
,_ 2?
&J
*
g.
-
i
|^
*
j^
*
t
k
A*
-
y).
+ 0.
*
Find the reciprocal of each of the following:
a. ],
e.
(-1) -(-2).
0-(-2).
i.
+(-3)(0).
f.
(-
c.
f.
d.
e. 2/3.
[(a)
(a
(-5) +(-3).
(-8) -(-9).
15 + (- 3).
+ (-3) -(+4).
b.
e.
b.
i-
f.
each of the following:
a. 5.
-
= sr.
+ b) - 5a + 56.
+ b)c = c(a + b) =
b. rs
x.
3
+|O
--
1 '
d. 2
t
h.
a
o
l
r
-
+
5
'
"
0.1
2
-
0.3*
12
6*
Sec.
Introductory Topics
Prove each of the following equations by using rules
and
for signs
1-7
for operations
with fractions:
a.
a
c.
-
o-
-
(
[6
6)
-
+a
(a
-
<J&C
(
-
c)]
=
= -
c)
(a
-
6)
a
-
(b
-t-
b,
c).
-
d. 6
c.
-
(ac
/ --
/
rt
~~*
jr/
<*-<
OC
ad)
h
= a[d + c = a.
(
c)].
C
-
f-
ORDER RELATIONS FOR REAL NUMBERS
1-8.
We
to express the fact that a is a
a>
and the notation a < to indicate that a is negative. The symbol > means is greater than, and < means is less
than. These symbols are called order symbols.
Assume that a and b are any two given numbers. If a - b > 0,
we shall write a > 6, or b < a, and shall read "a is greater than b,"
or "& is less than a." As can be easily seen, a > b means that a lies
shall use the notation
positive number,
to the right of 6 on the real-number line.
a
b
>
tive;
b
0,
then &
which equals
a,
and conversely.
- a < 0.
The student
is
Hence,
When
(a
a>
b,
that
is,
when
by (1-18), is negaif and only if
6)
a>&(or&<a)
familiar with the symbol
-
(for equality),
which
used to indicate that two quantities are the same. Thus a = b
means that the two symbols a and b represent the same matheis
matical object. For example, 6-3*2.
If a and 6 are two distinct numbers on the scale, we say "a is
different from &" or "a does not equal &," and we write symbolically
a^b. The symbol
means does not equal and
=
is
called the
\
inequality symbol.
In general, the oblique line or vertical line through any symbol
will form a new symbol which is the negation of the original one.
Thus, a < b means "a is not less than b." In other words, a = b or
a > b (by Property 1 below). For example, 5 < 3.
Sometimes we shall find it convenient to combine the symbols
< and = or > and =. We write ^ to mean is less than or equal to,
and we write ^ to mean is greater than or equal to.
We thus have order relations on pairs of real numbers, defined
by either of the following equivalent statements
:
The system
tion
>
(or
a
>
b (or b
<
a)
if
and only
if
a
b is positive;
a
>
b (or 6
<
a)
if
and only
if
b
a
is
negative.
numbers is then said to be ordered by the relathe relation <) Assertions of the type a < b or a > b are
of real
.
See.
1-8
13
Introductory Topics
The ordering of the
called inequalities.
real
numbers has the
fol-
lowing properties.
Property 1. For every pair of real numbers, a and
only one of the following relationships holds:
=
a
<
or a
6,
6,
or a
>
6,
one and
b.
Proof of Property 1: If a = 6, the statement is certainly true.
b = 0, so that
Saying that a = b is equivalent to saying that a
a
6 is either positive or negative. Thus, if a
6 is positive, we
have a > b. If, however, a
b is neither positive nor zero, then it
is negative, and a < b. If two of the three possibilities occurred
together, we should have, say, a = b and a > b, or a > b and a < b.
b would be both zero and positive, or both positive and
Thus, a
negative. Since no overlapping may occur among the three classes
we
of numbers,
Property
are thus led to a contradiction.
For any
2.
if
numbers
real
<
a
b
Proof of Property 2: If
and
a<
b
b
<
a, 6, c, it is
c,
then a
and 6 <
<
true that
c.
then both 6
c,
a and
a as (c
Let us write c
(b
6)
a). We
have assumed that the sum of two positive numbers is positive.
Since c
b and 6
a are positive by assumption, their sum, which
is c
a, is also positive. Hence, c > a, or a < c.
c
b are positive.
Property
3.
4-
For any
if
real
by
Proof of Property 3:
positive. But, by (1-17),
(a
It
a
+
c)
-
+
(b
a, b, c, it is
+c>b+
By definition, a> b
>
a
numbers
then a
true that
c.
means that a
+ (-b) + (-c) = a - 6.
- (6 + c) is positive, and
(a + c)
=
c)
therefore follows that
b + c.
a
+
6
is
c
that
+c>
Property
4.
For any
if
real
>
a
b
numbers
and
c
>
0,
a, 6, c, it is
then ac
>
true that
be.
We
have assumed that the product of two
6 and c are positive, it
positive. Since both a
Proof of Property 4:
positive numbers is
follows that their product
Therefore, ac
Property
5.
be
is also positive.
is positive,
For any
if
a
real
>
b
and ac
numbers
and
c
<
>
But (a
6)c
= ac
be.
a, 6, c, it is
0, then ac
<
be.
true that
be.
14
Sec.
Introductory Topics
Proof of Property 5:
We
numbers with unlike signs
is
1-8
have noted that the product of two
- b is positive, but
negative. Here a
= ac 6c, and (a 6) c is negative, it
c is negative. Since (a
6) c
follows that ac
be < 0, or that ac
<
be.
According to Property 3, the order symbol in an inequality is not
changed if the same number is added to or subtracted from both
sides. It therefore follows that a term on one side of an inequality
may be transposed to the other side with its sign changed. For
6 > c, then a > c + b.
example, if a
to
According
Property 5, the order symbol in an inequality is
reversed if both sides are multiplied or divided by the same negative number.
1-9.
ABSOLUTE VALUE
As a consequence
of the properties of the ordering of real numcan be associated with each number a certain nonnegative number called its absolute value. For any real number a,
we define the absolute value of a, denoted by |a|, as follows:
bers, there
|
Thus,
1-10.
3
|
=
|
a
=
|
3, since 3
a, if
>
a
^
0; also
0,
and
|- 3
|
|
a
=
a, if
\
= - (-
3)
=
<
a
0.
3, since
-
3
<
0.
INEQUALITIES INVOLVING ABSOLUTE VALUES
We
now
we let
consider some inequalities involving absolute
number x be represented by a point P on a
number scale, then \x\ is the numerical distance between P and the
origin. If we let a be a positive number, then \x\ < a means that
the point P is less than a units from the origin; that is, x lies
shall
If
values.
between
a and
a.
the
We can write this in the form
<
a
<
more briefly in the form
a < x < a mean exactly the same thing.
\x\ < a and
A more general inequality which often occurs
6 < a.
where a > 0. This is equivalent to a < x
a
or
to each term,
ments
\x
we may
b\<a
and
write 6
6
x
a<
a.
x
< x and x <
a,
Therefore, the statements
<
a<x<b + a
b
+
a.
mean
is
|#
If 6 is
6|
<
a,
added
Hence, the stateexactly the same
thing.
2<x 3<2 and
For example, \x~ 3|<2 may be written
means that the distance between x and 3 is less than 2. To solve
this inequality for x we add 3 to each term of the inequality,
t
obtaining 1< x < 5.
The following illustrative examples may help to give a better
understanding of the processes involved in the solution of the problems in Exercise 1-2.
Sec.
1-10
1-5.
Example
Arrange the following numbers in increasing order:
-
2,
Since
Solution:
TT
1-6.
Example
3.5, 0, 2, 3.14,
Since
or
|,
|-2
1-7.
Example
-
|
|
-2 =
> -2.
|
integers a
|-2
and since - 2 <
2,
6
and
such that a
Example
1-8.
we have the inequality - 2
2,
<
<
\/2
b.
be represented approximately by 1.414, the values
may
satisfy the inequalities.
6
|.
such that a
6=2
and
1
as follows:
is
|-5|.
TT,
2 and
Find integers a and
Solution: Since \/2
=
a
|.
Insert the proper inequality sign (order symbol) between the
following numbers:
Solution:
|-5
3.5, O,TT, 3.14,
approximately 3.1416, the desired order
is
-
< |-2
15
Introductory Topics
^
and
1
Express the inequality
Thus, 1 < \/2 < 2. Any other pair of
would also satisfy the inequalities,
6^2
|
<
x
\
3 without using the absolute-value
symbol.
Solution:
We know that the statements
the same thing.
Here a
point represented
by x
and
3.
The
inequality
is
x < a and
a < x < a mean exactly
the positive number 3, and x
< 3 means that the
3
less than 3 units from the origin; that is, x is between
|
\
is
|
be written
may
<
3
<
x
3.
Example 1-9. Explain the meaning of the inequality
without using the absolute- value symbol.
Solution:
The inequality x
|
x
<
be written
inequalities, we obtain 1 < x
2
|
|
may
1
\
x
|
2
<
|
1
and write
it
b
< a is equivalent to a < x b < a. Hence
- 1 < x - 2 < 1. If we add 2 to each term of the
< 3.
\
EXERCISE 1-2
1.
a.
c.
Arrange the numbers in each of the following
-
e. 3,
- 2, 5.
3, 0, 4,
- 1, - 1/3, 2, 10,
- 2, 1, V3, - 3/2.
2. Insert
sets in increasing order:
b.
d.
4.
-
-
6,
f . 1.4. 0,
1/2,
8, 2, 0,
10, 9, 4,
-
2,
-
3, 3/8,
-
V,
I
- 3/4.
- 6/5.
3
|.
the proper order symbol between the two numbers in each of the
following:
,,
and 1/3.
- 3 and -
a. 3
d.
3.
b.
2.
Examine each
e.
f
-
3 and
22/7 and
|
3
|.
w.
of the following inequalities,
c.
\/2 and 1.414.
t.
1/8 and 1/6.
and determine whether or not
it is
true.
a.
d.
TT
5
>
> 22/7.
3.
b.
e.
|
- 3 -f 2 < 0.
-2 < 2
|
|
|.
.
c.
-
3
3
-
|
f.
|
|
7
> >
|
3.
|
5
-
2
|,
16
4.
Find the value of each of the following:
b.|-3| + |+3|.
a.|+2|-|-2|.
-6
d. |-7| + |-5|-|+5|. e.
12 - 4 h.
3.
g. (-18)*
|-9|*|4|.
j. 0|4|-|-5|.
|
5.
1-10
Sac.
Introductory Topics
|
|
c.|+4|-|-4|.
f.
|5-3|+|3|-|2|.
|.
1.0*14.
Express each of the following inequalities without using the absolute-value
symbol:
a.
I
d.
|
x
<
I
2x
2
<
|
<
b.
1.
4.
e.
|
x
-
c.
1.
<
1
|
3.
f.
|
6. In each of the following, find a pair of integers, a
inequalities are satisfied:
b. a <
3 < b.
a. a < 5 < b.
and
d, a
<
TT
<
6.
7. If
a
^
3,
place the proper order symbol between a
8. If
a
*z 5,
a
what can be
<
<
\/3
e.
s
a
f. a
&.
said about the value of 3a
+
^
I
-
a.
1/2
<
|
3/2.
such that the given
6,
-
e.
\x
<
<
<
1
1
b.
-
2
|
<
6
7 and 10.
2?
POSITIVE INTEGRAL EXPONENTS
1-11.
If two or more equal quantities are multiplied by one another,
the product of the equal factors is called a power of the repeated
factor. Thus 5 2 read "5 squared," means 5 5 5 3 read "5 cubed/'
means 5-5-5. In general, an means the product of n factors each
;
,
,
equal to a. We call a the base and n the exponent of the power. It
follows from the associative law that
a2
Also,
if
a3
^
a
=
(a
a
a) (a
a)
=
a a a a a
=
=
a5
-
,,5-3
a 2 *3
.
0,
a
a3
a a
a
a
a
a
a
a
=
a
=
a
"2
a
These and similar results suggest the following laws of expo-
m
and n are positive integers. The proofs of these
them
laws are reserved for a later chapter.
nents. In
Law
of Multiplication.
To multiply two powers
of the
same
base,
add the exponents :
am
(1-31)
Law
of Division,
power of the same
To
an
divide one
=
am+n
.
power of a given base by another
base, subtract the exponents
:
'
(1-32)
^=
a"*-*,
if
a
^
0,
m>
n.
Sec.
1-12
Law
Power of a Power. To
for a
power, multiply the exponents
(1-33)
(a
For example,
Law
17
Introductory Topics
(a
for a
3 2
)
=
Power
a3
a3
=
m
a3
raise a
power of a given base to a
:
=
n
)
'
2
=
a mn
a6
.
.
To obtain a power of a product,
power
of a Product.
raise each factor of the product to the given
(1-34)
(aft)*
Thus, (3a
Law
3 2
)
=
3 2 (a 3 ) 2
=
3 2a6
=
=
9a 6
:
n
a b.
.
Power of a Quotient. To obtain a power of a quotient,
numerator and the denominator to the given power
for a
raise the
:
H
(n\
I)
Thus
if
inus, n
=,
nn
if
6*0.
6*0
* o,
o
ALGEBRAIC EXPRESSIONS
1-12.
An
means
algebraic expression is formed by combining numbers by
of the fundamental operations of algebra. The distinct parts
by plus and minus signs are called
The terms of the expression 3x 2 -5xy 2 +7z are 3# 2 -5xy2
Here the numbers 3, 5, and 7 are called numerical coffi-
of the expression connected
terms.
and
7z.
,
,
x 2 xy 2 and z are called the literal parts.
An expression containing one or more terms is called a multinomial. A multinomial consisting of one term is a monomial. A
binomial is a multinomial consisting of two terms, and a trinomial
is a multinomial with three terms. A polynomial is a multinomial
are
whose terms are of the form ax m y nz p
, where m, n, p,
cients, or just coefficients;
,
,
positive integers and a is a numerical coefficient, and where one or
more of the factors x m , y", z p ,
may be absent. Thus, 7, 5# 4 , and
3xy + 2 are polynomials, while x
+ -y
is not.
The degree of a term of a polynomial is the sum of all the exponents in its literal part. For example, the degree of 3# 2 is 2, the
degree of 5xy 2 is 3, and the degree of 7z is 1, because the sums of
the exponents are, respectively, 2, 3, and 1.
The degree of a polynomial is the degree of its highest-degree
term. Thus, in the trinomial 3x 2 -5xy* + 7z, the third-degree
2 2
5xy* + 7z
term, -5xy , is its highest-degree term? Therefore, 3#
is
a polynomial of the third degree.
18
Sec.
Introductory Topics
By
a polynomial in x of degree n
we mean an
112
expression of the
form
a xn
+
where the coefficients a
a T^ 0, and w is a positive
aix
n~ l
+
h an,
a n are numerical
*
&i,
,
,
n=
coefficients,
we
agree that the polynomial reduces to a number a which is not 0, and that the degree
is zero. The number
is regarded as a polynomial also, but as one
integer. If
0,
,
having no degree.
If the typical polynomial just given has degree n, the coefficient
a is called the leading coefficient. If its leading coefficient is 1, a
2x 2 + 5x + 1 is a monic polypolynomial is called monic. Thus, x*
nomial of degree 3.
EQUATIONS AND
1-13.
IDENTITIES
An
equation is a statement of equality between two numbers or
algebraic expressions. The two expressions are called members, or
sides, of the equation.
Equations are of two kinds, namely,
conditional equations and identities. A conditional equation, or
simply an equation, may be true only for certain values (possibly
none at all) of the literal quantities appearing. An identity is true
for all numerical values that can be substituted for the literal
quantities.
Illustrations of equations are
-
3x
5
=
x
+
1
5x
+
4
=
0.
and
x2
The
x
=
first
1 or
x
one
is
true only
-
if
x
= 3,
and the second
is
true only if
= 4.
Illustrations of identities are
-
2)
=
+4=
(x
3(s
3x
-
6
and
x2
-
Each of these equations
1-14.
5x
is
true for
all
1) (x
-
4).
values of x.
SYMBOLS OF GROUPING
Parentheses
(
and other symbols of grouping which have the
)
same meaning as parentheses, namely, brackets [ ], braces { }, and
the vinculum
are used to associate two or more terms which are
to be combined to form a single quantity. The word "parentheses"
,
Sec.
1-14
19
Introductory Topics
is often used to indicate any or all of these
symbols of grouping.
Removal of the symbols of grouping is accomplished by applying
the laws of algebra, such as the laws of signs and the distributive
law. The following examples illustrate the procedure.
Example
1-10.
Solution:
The
Remove
steps
-
may
(2x
parentheses from
3).
= - 1) (2x - 3)
= (-l)(2x) +(-!)(= - 2x + 3.
3)
(
3)
from 8x
of grouping
2[5y
+3
(x
y)]
basic rules for enclosing a group of terms in parentheses
may
x
[2y
Solution:
8x
-
3y} and collect terms.
One way
-
2[5y
of obtaining the desired result follows:
+ 3(z -
be stated as follows
-
y)]
=
=
=
The
-
(2x
be indicated as follows:
-
Remove symbols
Example 1-1 1.
-
Sx
Sx
8x
{2y
-
-
x
2[5y
2[2y
4y
-
-
Zy]
+3x - 3y] - {2y + 3x] - {5y - x}
6x
-
5y
x
+
3y}
+x
:
To write a given expression
sign, write the
and write
+
in parentheses preceded by a plus
terms as they are given, enclose them in parentheses,
in front of the parentheses.
Thus,
in parentheses preceded by a minus
change the sign of each term, write the resulting terms in
in front of the parentheses. Thus,
parentheses, and write
To write a given expression
sign,
a
The
-
first rule is
&
= -
(
-
+
6)
= -
(6
-
a).
obvious, and the second follows
from the rule of
signs (1-18).
Example 1-12.
a) Enclose the last
two terms
of 2 4- 3x
y within parentheses preceded by a
plus sign.
6)
Enclose these terms within parentheses preceded by a minus sign.
Solution:
+, we enclose 3x - y in its
given form within the parentheses preceded by a plus sign. In this case the ex*
- y).
pression becomes 2 -f (3x
a) Since the sign before the parentheses is to be
20
Sec.
Introductory Topics
1-14
Since the sign before the parentheses is to be
, we change the sign of each
- y and enclose 3x y within the parentheses preceded by a minus
6)
+
term of 3x
Then the
sign.
expression becomes 2
3#
(
+ y).
ORDER OF FUNDAMENTAL OPERATIONS
1-15.
Parentheses and other symbols of grouping are useful in indicating which operation is to be performed first. We have used them
in this way from the outset. In order to avoid using them unnecessarily, as has been already pointed out, the convention is adopted
to perform all multiplications first and then the additions (or
subtractions). If two or more of these symbols of grouping are
used in the same expression, we usually (though not necessarily)
remove the innermost pair of symbols first.
Illustrations in which symbols of grouping are removed follow
:
6)
- (6/2) =4-3 = 1.
- 6)/2 = (- 2)/2 = (4
c)
6
a)
d)
4
+ [15/(3
6 + (15/3)
5)]
'5
1.
+ (15/15) =6 + 1=7.
= 6 + 5-5 = 6 + 25 = 31.
=
6
EXERCISE 1-3
In each problem in the group from
1 to 36,
perform the indicated operation or
operations.
25
1.
2.
.
5. 10'.
6.
34
3.
.
(-
2)
4.
7.
((-
I)
6)
3
.
4.
53
3.
8.
-
.
10 6
.
4
10.
9. 7*.
(-4) 8
11.
.
()'
"'
*
17. i(4 3 )
J
18. f(3 4 )
19. a(5 8 ).
21. a 3 a 4 .
22. (a 3 ) 4
25.
.
2
(
6
o
3 2
.
26. a
)
34.
a8
a
-
.
2
&YV4/
a
3.
12
-
(|)
*>
(-!)'
23. (a&) 4
27. a
2
a
20.
24.
.
4
a 2 6(-
a7
28.
,
ttpV.
w/
36.
(W\a /
8
In each problem in the group from 37 to 48, remove symbols of grouping and
simplify.
37. 3 - (b
-
+
39. 4[a
(6
a)].
- 36).
41. a - t
(2a
+
43. (a
-
26)
-
(3a
+ (y - z).
- (3a (a + 26)
38. x
2).
-
+ 6).
40.
42. a
44.
-
[3
W-
+
(2a
-
[3x
+
(2y
b).
4)].
-
z 2 )].
Sec.
1-16
-
45. a* 2
46. 3n6
48.
+
2bxy
4ac
{
-
47. 2a
-
-
[(6y
-
+ ab] 3a6
- 6 + a~^~6 -
(3a
-[ -(a -b -c) -a +
{
In each problem from 49 to
+
52. x
+y
55. z 2
58. ax
+
b
-
-1.
2
?/
2
c.
-
z2 .
2ax?/
}
2
i/
)].
.
(3a 4- 2c)
}
4c
+ a].
-c)]J.
(6
two terms in parentheses. First
and then use a minus sign.
60, enclose the last
50. a 2
-
2ab
53. 2a
+
6
-
x2
-
59. 2z
-
56.
+ y*.
+
(axy
2ac
-f [ab
[36 -f 4c
-
2cx*)
use a plus sign before the parentheses,
49. a
21
Introductory Topics
+
-
-
2
-
3y
.
54. 3x
3c.
?/
- 62 + c2
- 4y + 20.
51. a 2
6*.
z2.
57.
60.
4z.
-
+
a36
x
2
-
ab
3
+
+
&2-
1.
In each of the following problems, evaluate the given expression.
61. 16
64. 4
-
67. (3
70. 3
-
- 2).
7).
(6
- (4 2).
3)
3 - (4 2).
16-6-2.
62.
(6
65. 4
6
68. 3
3
+ (4/2).
(8 + 4)/(2 + 3).
74. (8
76.
77. 8
ADDITION
63.
4
69. 3
2.
+ 4)/2.
+ (4/2) + 3.
2
3
Terms, such as 2x y and 5x y
3)
75. 8
78. 8
B
,
(-4) -
- 7.
6)
(3
4)
72. 8/(2
AND SUBTRACTION OF
2
(-
66. (4
7.
+ 4.
71. (8/2)
73. 8
1-16.
-
(4)
(-
2).
2.
+ 4).
+ 4(1/2) + 3.
+ ((4/2) + 3).
ALGEBRAIC EXPRESSIONS
which have the same literal parts,
may be added or subtracted by
are called similar or like terms and
adding or subtracting their
an example.
coefficients.
To
illustrate, let
us con-
sider
Example 1-13. Add 5zy 2
,
7x*y,
Solution: Collecting like terms
(5
-
2)zt/
2
+
(7
-
-
2zy
2
,
-
and adding
2
9)x y
+ 4z 2
3
i/
9x*y,
coefficients,
we have
= 3xy - 2x*y
The procedure used in the solution of Example 1-13 follows at
once from the distributive law. For example, the sum of the terms
2
5xy* and -2xy 2 is obtained as (5 - 2)xy* or Say
This leads at once to the rule for the addition (or subtraction) of
.
algebraic expressions. In practice, we usually arrange like terms
in vertical columns, and then we find the sum of each column by
sum of the numerical coefficients
procedure may be made clearer by means of the
prefixing the
in the column.
The
following examples.
22
Sec.
fnfroe/ucfory Topics
-
2
Example 1-14. Add 2x
3xy
+ z,
-
2z 2
x
-
x2
3z2/
-f 3z
2xy
-
36
xy
+c
2
z.
-
c2
-
2a
36 -f
c2
+
-
2
a2
-
from 3a 2
3a 2
Solution:
+z
-5z
2
3x 2
Example 1-15. Subtract 2a 2
+ 3z.
2xy
50,
1-16
36
2c 2
c2.
.
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS
1-17.
With the help of the distributive law for multiplication, the
product of two algebraic expressions is found by multiplying each
term of one by each term of the other and combining like terms.
Example 1-16. Multiply 3x 2
3z 2
Solution:
-
2xy
+y
2xy
2x
+ y2
-
2
by 2x -
3y
2xy
6z 3
-
I3x 2 y
3y.
H-
2
Sxy
2
-
3y
3
.
SPECIAL PRODUCTS
1-18.
The following
typical forms of multiplication occur so frequently
should learn to recognize them quickly and to obtain the
products without resorting to the general process of multiplication.
that
we
They should be learned thoroughly.
a(b
(1-5)
(1-36)
(1-37)
(1-38)
(1-39)
(1-40)
(1-41)
+ c) =
ab
+
ac.
+ b) (a - 6) = a 2 - b 2
2 ab + b 2 = a 3 + 6 3
(a + 6) (a
- 6) (a 2 + ab + b 2 = a 3 - 6 3
(a
2 =
a 2 + 2a6 + b 2
(a +
- b) 2 = a 2 - 2ab + b 2
(a
2
2
(ax + by) (ex + dy) = acx + (ad + bc)xy + bdy
(a
.
.
)
.
)
.
ft)
.
.
See.
1-19
1-19.
23
Introductory Topics
DIVISION OF ALGEBRAIC EXPRESSIONS
To divide a polynomial by a monomial, divide each term of the
polynomial by the monomial and add the results.
This rule follows immediately from the rule for fractions
expressed by (1-25).
Example 1-17. Divide 6a 2 z 2
12a 4 z
~ , ,.
6a 2 z 2
Solution:
- 12a% -
30a 6 z 6 by 15a 3 z 2
.
30a 6 x 5
15a 3 z 2
_
5a
To
5x
divide a polynomial
(the dividend) by a polynomial (the
both
divisor) , arrange
according to descending or ascending powers
of some common literal quantity. Then proceed as follows:
Divide the first term of the dividend by the first term of the
divisor to obtain the first term of the quotient.
Multiply the entire divisor by the first term of the quotient,
subtract this product from the dividend.
and
Use the remainder found by this process as a new dividend, and
repeat the process. Continue the work until you obtain a remainder
that is of lower degree in the common literal quantity than the
divisor.
Example 1-18. Divide 6* 3 - 5z 2
2x
Solution:
-
1
|
-
6z 3
6s 2
+
+
3x
+
3x
2x 2
+
3z
2x 2
4-
x
5x 2
3s
-
by 2x
1
+
1
|
-
3x 2
1.
x
+
1
2
2x
2x
+
1
-
I
2.
The division can be checked by finding the product of
and adding
2,
proving that 6z
Any problem
3
-
5x
2
-f
3z
+
in division, in general,
the relationship
dividend
1
=
(2x
(2s
may
-
-
1)
1)
and (3z 2
(3z
2
-
x
-
+
x
-f 1)
1) -f 2.
be checked by means df
?
This equation
literal
is
=
(divisor) (quotient)
an identity that
quantities.
;
is, it is
+ remainder.
true for
all
values of the
Indeed, this equation supplies the underlying
meaning of the process of
division,
24
Sec.
Introductory Topics
1-19
EXERCISE 1-4
In each problem in the group from
separated by commas.
-
1.
xy, 3xy.
+ 116,
6a
4.
-
+ 26.
3a
2.
4x 2 y 2
5.
2
4a
26
add the given expressions which are
2xy.
,
-
to 12,
1
2
,
-
a
2
46
2
.
3.
3x 2 y 2
6.
-
3a
f
26
2x 2 y*.
+ 4c, 4a + 36 -
6c.
-3a+6~c, -a-6+c.
7.
+ 2y - z, x + y - 3z, 4x - 3y + 20.
- 2xy + y 4xy, - y 2
3z 2 - 4j?/ 2 + 2?/ 3 4z 2 - 2x 2 - y 3 4xy 2 - 2y 8
2x* - 3z + 1, x 2 + 2x - 3, 2z 2 - x s + 4 - 2z.
4az + 3bxy - 4z 2 26z - bx 2 + 2azy, 3z - 2y.
3x
8.
x2
9.
10.
2
.
,
?/
,
11.
12.
.
,
,
In each problem from 13 to 24, subtract the second expression from the
- xy, 3xy.
14. 4x 2 y 2 - 2xy.
13.
first.
,
-
15. 3z 2 2/ 2 ,
-
17.
4a 2
19.
-
3a
21. x 2
+
26
2x 2 2/ 2
2
,
+6 2a;y 4-
-
22. 3
23.
.
-
a2
-
c,
2
2/
2x
46 2
>
+
.
-
a
6
+ c.
6a
+
115,
18.
3a
-
26
20. z 3
+x
+
4x 4
2
-
3a + 26.
+ 4c, 4a + 3& - 6c.
+ x + 1, 3 - z 2 -
16.
-1.
3J
4xy.
a:
2
-
a:
3
+
3z 4
-
2x 5 3x 5
,
-
a;
3
+
2o:
2
-
2x
+
1
.
-
o;
y 4-
xy
2
-
y
2
,
x*y
-
4-
/
In each problem from 25 to 60, perform the indicated operations.
26. (4s)
25. (6*)(-37/).
27. (5**y)
29.
(-
(-
2sy).
(- 2a 3 ).
3a6) (4k)
31. (3z 4- 2y)
37.
4z 2 (z 2
32. (6
(- 2).
39.
(a;
41.
4x/(-
2
-
2xy
+ xy -
2
?/
y
(x
)
36. (3a
).
-
y).
40.
2
(a;
4c).
3a) (2).
66 2 ) (3a6).
-
+ 26).
+ 3y - 4?/ ).
- y).
(x + y) (x
6) (a
2
38. 3xy(2x 2 y
2
(-
y)2a.
+
34. (4a 2
)
-
-
30. (x
+ 27/ 2 (3xy).
+ 3y) (x - y).
33. (4xy
35. (2x
(-2).
28. (3a) (2b 2 )
y
2
)
42. 3x 2 y/2xy.
2).
43. 4x*y*z/2xy*z.
44. 20z 6 2/ 4 zyiOz 6 2/2 3
45.
46. (2xy) 2/2xy 2
4a6/(-
4a6).
- 2*y)/(- s).
- So: 2 2 + 4x*y 2 )/x*y*.
- 1).
2 60; + l)/(3x
(9x
2
(x
y).
y) /(x
2
'47.
48.
49. (3xt/ 2
50. (x 2 y 3
(2xy)V2(xy).
- fay + 9?/ 2 )/(- 3y).
2
51. (x + 6x + 5)/(x + 1).
53. (x 2
3
57. (x
61.
62.
63.
64.
2/
'
y).
54.
(a:
?/
+ 3x 2 + 3x + l)/(x + I) 2
-f y )/(x -f y).
+ y) 8 /( + #)
4
3
2
2
60.
(x
y )/(x
y).
(x -f x + 3a; + 2x -h l)/(z +
Divide x* - y 2 - (x - y) 2 - (x - y) (x y)byx- y.
Divide (a + 6) +6(0+6) + 5 by a + 6 + 5.
Divide z 4 + 4z 3 + 6z 2 + 4z + 1 first by x + 1 and then by x 2 + 2z + 1.
2 2x - 3 by x + 4, and divide the result by x + 1.
Multiply x
55. (x*
59.
- 2 )/(z - y)/(x -
52.
.
.
3
y).
56.
3
(a;
.
58. (x
3
2
2).
1-20
Sec.
65. Divide x 5
3z 4
4-
25
Introductory Topics
2z 2
+ x 4 + 3z - 2z - 3 by x - 1 and add the quotient to the excess of
- 9z + 7 over 2z + 2x + 7x + 3z + 4.
3
2
4
3
2
Under what conditions will ( - x) n be positive? 6) When
Assume first that x is positive and then that x is negative.
n
For what values of n will (
an ?
a) be equal to
66. a)
67.
FACTORING
Factoring a quantity
will it
be negative?
1-20.
the process of finding quantities which,
multiplied together, yield the given quantity. When a quanA
is expressed as a product B C, B and C are called factors
tity
or divisors of A and are said to divide A. Also, A is called a multiple
of each of JS and C. These concepts are applicable to numbers or
is
when
algebraic expressions generally, but are most useful when restricted
to apply to integers or to polynomials. Such restriction will be
adhered to in this book. Thus, when an integer is to be factored,
the factors sought are to be integers. And when a polynomial
be factored, the desired factors are to be polynomials.
Let us
is
to
review the fundamentals of factoring integers (positive, negative, or zero)
First, it is clear that every integer n may
be expressed as 1 *n or ( !)( n). Such factorizations are called
trivial. If an integer n, other than 4-1 or
1, has no factorizations
other than trivial ones, then n is called a prime (number). An
first
.
integer
having a non-trivial factorization
is
called
composite.
Examples of prime integers are 3, 7, and 11; examples of com40.
posite integers are 6 and
Let n be a composite integer. Then a non-trivial factorization
m *p exists in which \m\ and \p\ are less than \n\ and greater than
1. If both m and p are primes, then n is expressed as a product of
primes. If not, at least one of m and p, say p, is composite, and so
p r-s. Hence, n = m r s. The process begun may be continued
if any one of m, r, and s is composite, and additional factors may
be found, until the process cannot be continued further, in which
case only prime factors are obtained. We may conclude that the
factoring process must terminate, since at any stage the new factors introduced are numerically less than their product. The
fundamental theorem of arithmetic guarantees that every composite integer is a product of primes, which "are unique except for
their signs or the order in which they are written. For example,
successive factoring of 156 gives
156
and the
=
39
4
=
final factors 13, 3, 2,
ization of 156 into primes
150
is
13
3
4
=
13
3
2
2,
2 are primes. Another valid factoras follows:
= 2-3- (-
13)
-(-2).
26
Sec.
Introductory Topics
1-20
Let us turn now to polynomials with real coefficients. These may
1
be polynomials in x, such as x 2
or polynomials in x and y,
such a3 x 2 + Sxy + 2y 2 or, in fact, polynomials in any number of
literal quantities. Every polynomial F has a factorization of the
;
;
form
for every non-zero number a. And, of course, the factors I/a and
aF are themselves polynomials. Such factorizations are called
trivial.
nomial.
polynomial F has no factorizations other than trivial
is called a prime polynomial, or an irreducible polypolynomial having a non-trivial factorization is called
If a
ones, then
A
F
composite or reducible.
2
Examples of prime polynomials are Sx + 2, 2x + 2y, and x + S.
Every polynomial in x of the first degree may be shown to be
prime; and certain polynomials of higher (even) degree in x also
are prime.
A
development of criteria for primeness of polynomials
beyond the scope of this book.
lies
2
2
and
4, xy + y
Examples of composite polynomials are x
factor+ yz + xu + yu, because each of these has a non-trivial
ization. Thus, we have
xz
-
x2
+
xz
+
yz
+
4
xy
y*
xu + yu
=
=
=
+
(x +
(x +
(x
2) (x
-
2),
y)y,
y) (z
+
u).
Let / be a composite polynomial in x with real coefficients, not all
Then it can be shown that there exist polynomials g and h, the degree of each of which is less than that of /,
such that
of which are zero.
/ = g.
h.
g and h are primes, then / is a product of primes. If not, we may
proceed to factor (non-trivially) one or both of g and h, and we
can continue this process until primes are obtained. The process
must terminate eventually, since each non-trivial factorization
leads to polynomials of lower degree.
The argument just presented applies only to polynomials in one
literal quantity x. However, the principle may be extended to
If
apply to polynomials in any number of
"Thus,
?,
/ = Plp2
literal quantities x, y, z,
.
where p if p2
,
,
p n are prime polynomials. The problem of carry-
Sec. 1-21
27
Introductory Topics
ing out actual factorizations for certain types of polynomials
considered in the next section.
is
A special class of polynomials deserves particular attention.
This is the class that consists of polynomials in which all the
numerical coefficients are integers. It is possible to prove a factorization theorem such as that just stated, but yielding factors which
are polynomials having only integral coefficients. When the general
theorem can yield prime factors all of which have integral coefficients, the two theorems give the same result. Otherwise, they
will give different results.
For example, both theorems applied to x 2 4
factorization (x + 2) (x
The polynomial 3# 2
2)
trivial factors with integral coefficients.
real coefficients are allowed, we have
3x 2
-
4
= (V3x +
yield the
prime
4 has no nonHowever, when other
.
-
2) (-x/33
2),
which the
coefficient \/5 is a perfectly acceptable real number.
factors with integral coefficients are desired in such a
case as 36# + 24y, we shall agree to remove and factor common
in
Again, when
numerical factors, to obtain
36*
+
24y
=
3
2
2
(Sx
+
2y).
Here the prime factors are the numerical primes 3, 2, 2 and the
prime polynomial 3x + 2y.
In what follows, whenever a polynomial has only integral coefficients, we agree to restrict ourselves to factors of the same kind.
In similar fashion,
coefficients,
we
when
shall
the given polynomial has only rational
search for factors having only rational
coefficients.
1-21.
IMPORTANT TYPE FORMS FOR FACTORING
in Section 1-18 applied "in reverse" are formulas
Success in factoring a polynomial therefore depends
The equations
for factoring.
on ability to recognize the polynomial as being a particular type of
product and as having factors of a definite form. Verify the following type forms by carrying out the indicated multiplications and
learn each form,
Type
1:
(1-42)
Common Monomial Factor. From
ab + ac = a(b + c).
2
Example 1-19. Factor 4x y
Solution:
x*y
-
6xy*
-
toy*.
= 2xy(2x -
3
(1-5)
we have
28
Sec. 1-21
Introductory Topics
for actually removing the monomial factor in
1-19
may be put this way Write the common factor, 2xy,
Example
and in parentheses following 2xy write the algebraic sum of the
2
quotients obtained by dividing successively every term of 4x y
&xy 2 by 2xy, in accordance with the distributive law.
Instructions
:
Type 2: Difference of Two Squares. From (1-36) we have
-
a*
(1-43)
-
2
Example 1-20. Factor 9z
-
Solution: 9x 2
Type
(1-38)
25y
=
2
Sum and
3:
25s/
+
(3z
=
b2
+ b)
(a
-
(a
b).
2.
-
5y) (3x
5y).
Two
Difference of
From
Cubes.
(1-37) and
we have
+
a3
(1-44)
=
63
+
(a
b) (a
-
2
+
ab
b 2)
and
-
a3
(1-45)
=
63
-
3
Example 1-21. Factor &r
Solution:
-
8x*
=
=
27 y*
+
2
+
3y) [(2x)*
+
ab
b 2 ).
(4o:
+
(2x) (Zy)
+6xy +
2
-3y)
(2x
b) (a
27y*.
-
(2x
-
(a
2
(3?/)
]
2
9t/ ).
Type 4: Perfect-Square Trinomials. From (1-39) and (1-40),
+
a2
(1-46)
2ab
+
b
+
=
2
(a
+
=
(a
2
b)
and
a2
-
2a6
2
Example 1-22. Factor 4z
+
12a?y
(1-47)
Solution:
Type
nomial
4x*
5:
may
12z?/
2
-f
9y*
=
=
(2^)
(2x
2
6)
.
4
.
9?/
2
4- 2(2z) (Sz/
+
2
32/
+
2
)
2
2
(3?/ )
2
.
)
+
Ax 2
+
Cy
(ax
+
Bxy
the two factors
is
By comparing
this
+
=
(ax
+
by) (ex
and
by)
+
(be
(ex
+
+ dy).
dy)
are multiplied
ad)xy
+
bdy
2
.
product with the trinomial
Ax 2
it is
2
found to be
acx 2
note that
tri-
:
together, the product
we
+
a
-
General Trinomial. Factorization of the general
be indicated as follows
(1-48)
If
+
62
+
Bxy
+
Cy
2
,
necessary to find four numbers
ac
=
A, be
+
ad
=
B, and bd
=
a, 6, c, d,
C.
such that
Sec. 1-21
29
Introductory Topics
The following example illustrates a trial-and-error procedure,
which often involves several steps of inspection and testing by
multiplication yet it is a method commonly employed in practical
work and is recommended here.
;
+
2
Example 1-23. Factor 6x
Here we wish to
Solution:
=
-
llxy
find a,
l(ty
6,
2
.
and d which
c,
+ 6c)zz/ +
=
satisfy the identity
-
+ llxy 10i/
a
c
have
like
and
10, obviously
signs, and b and d have
6. Possible
2, db 3, and
1,
opposite signs. Possible values for a and c are
10. By trial and error we find the
values for 6 and d are db 1, =fc 2,
5, and
+
(ax 4- by) (ex
dy)
6 and bd
Since ac
acx*
+
(ad
2
fccfr/
Sx 2
2.
= -
=
=
correct selection to be a
=
2, c
=
3, 6
5,
and d
=
6z 2
=
This selection meets the
2.
requirement because
+
(2x
(3s
5t/)
-
2y)
by Grouping. An expression which does not
one of the given type forms can sometimes be
reduced to one of these forms by a suitable grouping of terms.
Type
6: Factoring
fall directly into
The following examples
illustrate the procedure.
Example 1-24. Factor Sax - 56z
Solution:
First,
+
-
Say
106y.
-
Sax
common
group within parentheses the terms having a
Thus,
+
5bx
-
Say
IGby
=
common
Then, in each group, factor out the
-
(Sax
5bx)
+
(Say
factor.
-
quantity. In this case,
=
-
- 5b)
- IGby)
56),
2y(3a
(Sax
x(3a
5bx)
to
the
terms obtained. (This quantity
the
common
factor
out
Finally,
quantity
will often be a multinomial factor.) The result is
-
(Sax
x(3a
An
alternate
+
-
method
+
2y(3a
-
+
-
=
+
56) (x
2y).
(3a
of grouping would give us the following results
56)
56)
First,
3ax
-
56x
+
Say
-
Then,
(3ax
+ Say)
4-
(-
=
IGby
56x
-
(3ax
IGby)
+
=
Say)
3a(x
+
+
56x
(-
2y)
-
Finally,
3a(x
+ 2y) -
Example 1-25. Factor x 2
-
+ 2y) =
56(x
2
?/
4-
-
6y
(x
+ 2y)
(3o
-
56(x
-
:
IGby).
+
2y).
56).
92 2 .
Solution: By grouping the last three terms, we may rewrite the given expression
as the difference of two; squares. Thus,
2
x 2 - (y* - 6y
X 2 _ y2
ey - 9
+
=
= x - (y - 3*) 2
= [x - (y - 3z)} [x + (y = (x - y + 3*) (x + y 2
See. 1-21
Introductory Topics
30
4
Example 1-26. Factor 4x
+ y*.
3x 2 ?/ 4
+
4
rewrite the given expression
adding and subtracting x2/ we may
as the difference of two squares. Thus,
- * 2 2/ 4
2 4
4
4x 4 + 3x 2 j/ 4 + 2/ 8 = 4x + 4* ?y + y*
By
Solution:
,
(2x
=
-
+
+
+
2
[(2x
(2x
2
2
-
2
4
)
!/
4
T/
X 2 T/ 4
2
XT/
)
-
4
?y
2
XT/
)
]
[(2x
(2x
+ y*) + xy 2
+ y* + XT/ ).
2
]
2
2
EXERCISE 1-5
1
Factor each of the following:
2
2. 6x
3x + Qy
4*.
7.
-
2a
-
5.
ab.
- *.
+ 2cx
ax
+ 9ac - 16.
13.
x
2
- 25t/ 4
y
2 _ ai
92/
4a'x 4 - 81.
16 ' 16x 4
19
22*.
25. 49x
28
.
2
2/
-
4x 2
31. x
ty
37. Six"
-
2
2/
17.
49x 2
-
-
.
+ 3z)
3
.
x
2
3,5.
2
18.
.
24.
62 .
64'
67
70*.
72.
74.
.
- x.
- Sly*.
+ 26) +
36 -
.
3
-l/<
+
2x
-
8
(3c_-t_4d) .__39i _
42.
45. x 2 T/ 2
15.
48. x 2
61. x 2
54. x
2
-
-
+ 81.
+ 30.
+ 4t/
18*
11*
2
.
4xj/
3x
-
10.
+ 30x3 +_ 225x.
+ 5a - /4.
2
x4
63.
2**
+ 5* - 3.
62. x + 16* + 64.
2x - 11* - 6.
2
66, 4* + 32* + 15.
65. 6x' - 37* + 6.
18x 2 + 16* - 25.
2
69. a* + 3x + 2ay + 6.
18
2 - 1 2x
68. 40x
+ 35*yz
x
+ 0.36.
12x 2 - 10* + 15.
71. 8x'
- 21.
x' + 3x 2 - 7x
- xz + 6a + xy - 3*.
73. 2x + 3y
3cy.
ax + 26x - ex + Say + 66ty
- 3.
2
75. x 3 + x - 3x
xa6 - xyz - 2aby + 2y*z.
oa. &
61
.
33.
.
12
y
2
2/a x3.
t/
30. 3x'
16.
2166 8
14.
-
- 2ay + 3a.
x' - 5x 2 + 6*y.
4x* - 9.
25a 2 - 166 2
2
a'x 4 - 9a 4 x
2
- a'x
ax'.
27. 36*
'
58.
ax
21.
4
4-1
9.
12.
.
47. x - 8x 2 + 15.
50. 2x 2 - 3x + 1.
53 x 2 - x - 12.
^'
W" *x' 4-4x3+4
T^
2 - 40x.
iQx
5x3
+
-5$.
_ 6.
- 2x - 3.
4ax'
121.
38.
44. 8x 2
4x
6.
2
ty
-
+
"
r2
-
35. x
^a
at/
+
3.
15.
-
""
+
-
.
64.
""41. 216*
12x3
+
2
-
29. 2x'
32. a
3(2?y
3y
14.
26. a 2
.
.'
36t
+
2
5t/
23. 0.01
2 &'.
z\
+ 69
-
11.
20. 1
144a
16x 4
+
3
34. a 3
-
2
Sax
8. 2at
66c.
10. 3a6
+ 4x?y.
+ 2a.
I
&
+ 9'
- 10x + 9.
j
-r uu.
57. 1
'
.
60. a 2
2
2
2
.
2/
1-22.
GREATEST
COMMON
DIVISOR
A common divisor of several polynomials
(or integers)
is
a poly-
For example,
nomial (or integer) which divides each of them.
x + y are
x
and
2 and 3 are common divisors of 12 and 18. Also,
common
A
divisors of
greatest
common
^ (a* - y-)
common
and
x* (x*
+ 2xy + y*).
known
as a highest
more polynomials
(or integers)
divisor (G.C.D.), also
factor (H.C.F.), of two or
Sec.
1-22
31
Introductory Topics
a polynomial (or integer) with the following two properties:
a common divisor of the given polynomials (or integers)
also it is a multiple of every other common divisor of the given
is
It is
;
polynomials (or integers).
follows that, for integers, a G.C.D. is a common divisor of
greatest absolute value. It also follows that, for polynomials, a
G.C.D. is a common divisor of highest degree.
For example, a G.C.D. of 12 and 18 is 4-6 or -6, since
2,
1,
It
3,
6 are the only common divisors, and 6 and -6 are those
absolute value. This example indicates that a set of
and
maximum
of
non-zero integers will have two greatest common divisors, d and
d, one being positive and the other negative.
For polynomials with real coefficients, if d is a G.C.D., then a d
is also a G.C.D. for every real number a not equal to 0. It follows
that infinitely many greatest common divisors exist. However, for
polynomials with integral coefficients, a G.C.D, should be a polynomial of the same type. Example 1-27, which follows, illustrates
the fact that in this case a G.C.D. is uniquely determined except
for sign.
Since a G.C.D. of given polynomials divides all of them, it must
contain as a factor each of the distinct prime factors occurring as
common
the given polynomials. Since, however, a
G.C.D. must divide any common factor of the polynomials, it must
contain each of the distinct common primes raised to the highest
common power. Thus, a G.C.D. of x* (x y) 2 and # 5 (x y) (x y)
must contain the prime factor x to the third power, that is,
a
factor of
all
+
to the highest
common power.
+
contain (x + y)
no further
since
a G.C.D.,
Thus, x*(x + y) is
common prime factors occur.
When no common prime factors occur,
to the first power.
Example 1-27. Find a G.C.D.
Solution:
of 3x*y*(x*
We shall begin by writing
it rftust
Similarly,
-
4y
a G.C.D.
2
)
is 1.
-
and 6xy*(x*
-f 4y*).
each of the expressions as the product of
prime factors, as follows:
3^3(^2 _
4xy
40.)
=
(3 ) (
y*)
=
(2) (3) (x) (y) (y) (x
X ) (X )
(y) (y) (y)
(
X
+ 2 y)
(
X
-
its
2|0,
and
6xy*(x*
-
4xy
+
-
2y) (x
-
2y).
t
The
different
3, x, y,
these
and x
common
G.C.D.
2y), of which only
prime factors are 2, 3, x, y, (x + 2y), and (x
the product of
now
We
form
both
are
to
common
2y
polynomials.
factors, using for each the maximum common power. Hence, a
is
3xy*(x
-
2y).
32
Sec.
Introductory Topics
1-23.
LEAST
COMMON
1-23
MULTIPLE
A common
multiple of two or more polynomials (or integers)
one
is
containing each of the given ones as a factor. Thus, 36 is
a common multiple of 6 and 9, and x 2 - y 2 is a common multiple of
x
y and x + y.
A
common
multiple (L.C.M.) of two or more polynomials
a polynomial (or integer) with the following
properties: It is a common multiple of the given polynomials (or
integers) ; also it is a divisor of every other common multiple of
least
is
(or integers)
the given polynomials (or integers).
It follows that, for integers, an L.C.M.
common multiple of
for polynomials, an
L.C.M. is a common multiple of lowest degree.
For example, an L.C.M. of 6 and -8 is 24 or -24, since the only
and 24 and -24 are
common multiples are 24, 48, 72,
those of minimum absolute value. This example indicates that a
least
absolute value.
It
a
is
also follows that,
,
m
have two least common multiples,
and m, one being positive and the other negative.
For polynomials with real coefficients, if m is an L.C.M., then
a m is also an L.C.M. for every real number a not equal to 0.
set of non-zero integers will
many least common multiples exist. Howfor
with
ever,
integral coefficients, an L.C.M. should be
polynomials
a polynomial of the same type. Example 1-28, which follows, illustrates the fact that in this case an L.C.M. is uniquely determined
It follows that infinitely
except for sign.
Since an L.C.M. of given polynomials is a multiple of all of them,
it must contain as a factor each of the distinct prime factors occurring as a factor of any one of the given polynomials. Since, however, an L.C.M. must be a multiple of any common multiple of the
given polynomials, it must contain each of the various distinct
primes to the highest power occurring anywhere. For example, an
L.C.M. of x*(x + y) 2 and x*(x -y)(x + y) must contain the various distinct prime factors, which are x, x + y, and x - y. For x,
the highest power occurring anywhere is the fifth power for x + y,
the highest power is the second f or x - y, the highest power is the
2
5
first. An L.C.M. is therefore x (x + y) (x-y).
;
;
Find an L.C.M. of 4z 2
Example 1-28.
Solution:
We
shall first rewrite
- 4s =
6z* - 6 =
4*2
-
2
4x, 6z
-
6,
and 9z 2
18*
+
9
=
18x
+
9.
each of the expressions in factored form. Thus,
(2) (2) (x) (x
(2) (3) (x
+
-
1)
1) (x
= 2*x(x -
1),
1),
and
90 -
-
(3) (3) (x
-
1) (x
-
1)
= 3*(* -
I)
2.
Sec.
1-24
The
distinct
these are
33
Introductory Topics
prime factors are
and
2, 2, 1, 1,
22
32
x
+
x
2, 3, #,
2, respectively.
(x
1,
and x
The
1.
greatest powers for
An L.C.M. is therefore
+ 1) (x - I) 2
.
EXERCISE 1-6
In each problem from
+
4.
z
7.
9z 3 !/ 2
y,
to 12, find a
1
G.C.D.
of the given expressions.
3. 4, 7, 39.
2. 9, 21, 33.
1. 4, 14, 36.
- y*, x 6. 3a6, 12a 3 6, 6a 3 6 2
6
2 2
21x
x
x
8.
9. 4x*y*z, Sxifz*, Ux*y*z 2
3)
9, (x
15zV,
3,
10. x + 2, x 2 - 4, x 3 + 2z 2 - 4z - 8.
11. x 8 - s a - 42z, z 4 - 49z 2 z 2 - 36.
12. x 4 + 2z 3 - 3z 2 2z 6 - 5z 4 + 3z 3 z 3 + 3x* - x - 3.
-
x2
2
y
5.
.
x3
.
37.
.
?/.
,
.
,
,
,
In each problem from 13 to 22, find an L.C.M. of the given expressions.
13. 6, 8, 12.
14. 8, 45, 54.
16.
4z 2 5x 4 20z.
y02.
18.
2a
4.
20. 2x
15. xy, 6xz, 8yz.
2
17. 4x ^
19. x
-
4
5
,
2,
9^
+
x
21. (s
22. 2x 4
1-24.
6
2/
22 3
2,
,
x
6x 4
-
2
,
- 49) (x 3 - 8), (x - 7) (x + 7) (x - 22/ 4 6x 2 + 12xy + 6s/ 2 9x 3 + 92/ 3
2) (x
2
- 3, a 2 - 9.
3x - 6, x 2 4- 2x - 8.
- 4), (x - 3) (x - 2).
a
.
,
,
,
+ 4,
+ 8,
REDUCTION OF FRACTIONS
From
(1-24) under operations with fractions, it follows that a
fraction a/6, in which 6^0, is not changed if both the numerator and the denominator are multiplied or divided by the same
=
0,
quantity, provided that the quantity is not zero. That is, if k
a
or
ka
a
6
For example,
=
-
~
2
2
2
2
7
4
57*
2
>
'
~
and A =
"""
3
6
6/2
'
The fundamental principles of (1-49) and (1-50) are applied in
reducing a fraction to lowest terms and in changing two or more
fractions with different denominators into equivalent fractions with
a common denominator.
The reduction of a fraction to lowest terms, that is, to a form in
which all common non-zero factors are removed from both the
numerator and the denominator, is accomplished as follows:
First, factor both the numerator and the denominator into
prime factors.
Then, divide both the numerator and the denominator by all
their
common
factors.
34
Sec.
Introductory Topics
1-24
be noted that this reduction can also be accomplished
It should
by dividing the numerator and the denominator by their highest
common
factor.
The following examples
will illustrate the reduction of fractions
to lowest terms.
30
Example 1-29. Reduce 4<
-^
Solution:
their
to lowest terms.
Factoring the numerator and the denominator and dividing both by
common
factors,
we have
_2'3-5
30
42
The common
2
3
5
'
7
7
numerator and the denominator are 2 and
factors of the
3.
6xv 2
Example 1-30, Reduce jr--
to lowest terms.
Solution: Factoring into prime factors and dividing out
2 3 x y y
3y
*
common
we have
factors,
=
x
2
In this fraction the
common
x
x
y
factors are 2, x,
--
and
y.
_
Example 1-31. Reduce ^-r2
6z
Solution:
We
-
It is
3xy
2
- ^-r~ISy
to lowest terms.
have
2
6z 2
~
- 3^ -
IS?/
2
*
3x(x
3(s
-
-
2y) (x
2y) (2x
+ 2y)
+ 3y)
2z
4-
important to note that in a fraction of the type
3y
a
Ct
^
|
y
i
where
*C
the numerator and the denominator have a common term, any
attempt to simplify the fraction by cancelling out this common
a
y
term can lead only to an absurdity. For, quite obviously,
"^ *^
^
i _i_
y
does not equal either
or - unless a = 1 or 0. For example, if
1 j] x
x
2-1-3
we attempt one of these simplifications with the fraction ]"
+
we
reach the obvious contradiction
common
error,
factors, not
it
is
-
=
- or -
-
>
To avoid
this
important to remember that only common
terms, may be cancelled. One should make
common
common
certain that the
5453
=
6564
quantity
is
a factor of the entire numera-
tor and of the entire denominator.
1-25.
SIGNS ASSOCIATED WITH FRACTIONS
from (1-49) that we can multiply both the numerator
and the denominator of a fraction by
1 without changing the
It follows
1-25
Sec.
35
Introductory Joplcs
value of the fraction. However, if just one of these is multiplied
1, then the sign of the fraction must be changed in order to
keep its value unaltered. Thus, in effect, a minus sign before a
fraction can be moved to either the numerator or the denominator
without altering the value of the fraction. We have, as previously
by
stated in (1-22),
-
~~
n
n
1-11
~"
"~
d
,
and
x
-
d
d
-
x
y
y
(x
i
y
y)
-
x
We
see, then, that any two of the three signs associated with a
fraction, namely, the signs of the numerator, the denominator, and
the fraction, can be changed without changing the value of the
fraction. In general, the rules for changing signs in a fraction are
as follows:
an even number of factors in the numerathe denominator, or in both, does not change the sign of
Changing the signs
in
tor or in
the fraction.
Changing the signs in an odd number of factors in the numerator
or in the denominator, or in both, does change the sign of the
fraction.
2x
Example 1-32. Find the missing
7
.
.
2x
solution:
=
o
---2x
=
o
-
denominator x
Solution:
-
7
(
)
)
o
to an equivalent fraction with
fraction
y.
Since x
-
y
= -
-
(y
x),
we make
a
__
__
the following changes in signs:
a
y-x~-(y-x)~~x-y
-
y
a
a
a
y
x
-
(y
-
x)
x
-y
EXERCISE 1-7
1.
(
o
a
or
2x
=
o
2x
-
Change the
Example 1-33.
quantities in
Find the missing quantity
in each of the following equalities:
36
Sec.
introductory Topics
2.
Reduce each
**
102010
350470
1-25
of the following fractions to lowest terms.
8a 2 fe 3
.
'
'
6x*y*
C*
20*
12a<6
V
'
.
e*
'
96z 3 2/ 7 * 2
+26
a
2' 4a 2
4
-
15
18x 2
"'
a2
?/)
a;
[(2x
6x 2 - 5x
4z 2 + lOa;
,
2
'
2
-
(a ^- 6)g 4- o&
- (a: + 2a)2]
d)
- 10?/ 2
4- 3sy
2
(a:
x*y*
~
3s 8
.
K
'
-y
x
-25
- 3)2
_
- (* - 2
a:
-
7x
-
x*y*
,
'
+ 8a6
-
f
'
24*V + 40zV
-
-
6
1-26.
2
ADDITION
+y
13z
49
s
-
+
14
-
z)
81
*
x* -f 729
462
(x
'
+
y)
q'
(y*
6)
(a?
-
y) (y
-
x2)
(z
a6
-
62
-
t
y)
9a 2
S*
u*
3
-
6a 2
8y*
xy
+
'
24
'
-f
-
'
3z
'
9s 2
2
x2
+
x
+ 2y
AND SUBTRACTION OF FRACTIONS
The sum of two fractions with the same denominator is the
whose numerator is the sum of the numerators and whose
denominator is the given common denominator. That is, in (1-25),
fraction
-+=^-
(1-25)
For example,
To add
x + 2y
x.2y_
+ 3
3
3
fractions with different denominators, first change the
fractions to equivalent fractions having a common denominator,
and then write the sum of the new numerators over this common
denominator.
Ordinarily, the
common denominator which
is
chosen
is
a least
common
multiple of the given denominators, since this leaves the
fewest possible common factors in the numerator and denominator
However, the same result is obtained,
after compilation of all common factors, no matter what common
denominator is used. The method of finding an L.C.M. was develof the resultant fractions.
oped in Section 1-23.
The
difference of
two
fractions,
Section 1-2. Thus, by (1-10),
when
?
>
has been defined in
a
neither 6 nor d
o
is zero,
Sec.
1-26
37
Introductory Topics
Applying (1-22), we have
a
n
(1-52)
v
'
r
x
T-
--c =
b
d
;
The following procedure
a
c
T H--rb
d
'
suggested for adding (or subtracting)
is
fractions.
common multiple of the given denominators.
fraction
each
to an equivalent fraction having the
Change
L.C.M. as the denominator in the following way: For each
fraction, note which factors are in the L.C.M. but not in the
denominator of the given fraction. These factors may be found
by dividing the L.C.M. by the denominator of the given fraction.
Then multiply the numerator and the denominator of the given
1.
Find a
least
2.
fraction by these factors.
Write the sum (or difference) of the numerators of the
new fractions found in step 2 over the L.C.M., and reduce the
3.
resulting fraction to lowest terms.
The following examples will indicate a procedure which should
be followed until some skill in working problems has been attained.
357
Example 1-35. Express j - ^
Solution: Step 1.
the denominators.
Step
2.
-f
The methods
as a single fraction reduced to lowest terms.
Q
in Section 1-23 give us 36 as the
Divide the L.C.M. successively by the denominators
36/4
Change the given
=
9,
=
36/6
6,
=
and 36/9
L.C.M.
of
and 9 to get
4, 6,
4.
fractions to fractions having 36 as the denominator in the
following way:
3j_9_27
~
4
Step
3.
Combine the new
'
5^6_30
~~
36
9
6
9
__
6
Note that adding
9
(or subtracting) the
common denominator
1.
28
'
36
30
Since x 2
~p
-
u
4
28
25
~~
"
36
numerators and the denominators of the
it is
+ =
Zoo
not true that ~
--- 5x
2
-7-75
+
36
leads to absurdities. Thus,
3x
Example 1-36. Express X
Solution: Step
-
36
given fractions leads to absurdities. Thus,
dropping the
4
-
fractions to obtain
27
3_57_27_3028
+ 36 ~
+ ~
36
4
_
~
4
7
'
36
6
-5
A
=
"
(x
X
+ 2)
1
X
(x
2
-
-
-
A
4
=
5
it is
Also,
not true thit
as a single simplified fraction.
2)
and
-
- 2
^
L
2
=
X
X
5
>
an
38
L.C.M. of the denominators
as follows:
x2
is
3x
2
O
i
2
-
When the L.C.M.
the results are
2.
Step
(x
4),
5l_ZJ*
T+T
-
I
-
5x
-
x
x2
2
~
*2
4
be written
2
4
-x o
~
X+2)
x-2
may
expression
divided by the denominators (x
is
-A
2
*
The given
4.
1-26
Sec.
Introductory Topics
*2
""
4
x
-
4
|
+
~
2), (x
2),
and
1
lf
Multiplying the numerators and the denominators of the given fractions by these
we obtain
2 3x (a? - 2)
~ 3a 2 - 6a
z
4
+
(x
2)
(x
2)
factors,
*
'
2
_
(a
(s
+
2)
+
Or
2)
5x
a;
__
+ o;--2
z-h2
-
2
__ 2j-_4
"
2 -
x
_
~~
-
5a
x2
4
'
4
2
-
4
Therefore, the desired result is obtained in the following way:
2 - (5x - 2)
5x
2
6g)
(2x 4- 4)
(3x
3.
Step
3x
2
2)
a:
-2 =
-4
2
+
x2
2
__ 3o;
x2
9x
-
4-
6
~"
-4
-
3(x
4
-
1) (x
a;
2
-
_
2)
3(x
-
1)
*
t
z+2
4
EXERCISE 1-8
Perform each of the indicated operations and express the answer in lowest terms.
L
2
1
7
.
5'
1
~9 ~9*
2
""
7
5
9
^_
~3 + 8*
Q 3
R ,2
^7"" &H"2T'
i
xZa.^_A
^'S^G
Q ^13
3+
20""Io'
6
12
17
5
5
19
'2~T + T
t
o
10
Id. -
-1
r
1
a;
-2,3
+ ^
-- x
-
2
7^
1
(x
-
I)
x
x
/r
/*
18.
4.
x
+ 1^
+
i
x
1
A
-
2
+2
x + 3
-f 60; +
+2
7^
+
^
'
12
.
a:
t
2
x
~r
i/
z
+2
2
-
4
JK
-
8
1
.j
Oi 2
zx
>
_
^ ^2 +
X
Q T^
Q
a:
a
a;
6
XV
14.
2
-2
x
a;
1yl
~r-
-
~
Q
6
<
x*
+
x
^
Q/*
*x
.
19.
x
2
?'r
-y
2
x*
-2xy +
,
y
2
1-27
Sec.
3
20.
4
-
1
_
x2
T
x
-
2
o^x
2
x
-
T
*
_j_
x2
y
-
2
1-27.
x2
-
2x
y
2
MULTIPLICATION
-
x2
AND
+y
x
^
Til
I/
y
j_
25 '
.
y
So;
26 *
i+?+A.
x
xy
y
Til
^
__
y
.l*,-<>
a -b
23.
II
_
24 *
21
.
-
x
-^-=-+4
+
1
x
22.
39
Introductory Topics
-
x
x2
y
-
.
y
2
4xi/
+y
2xy
-
x3
2
-
x 2y
f
xy
2
-f
3
2/
DIVISION OF FRACTIONS
The product of two fractions is the fraction whose numerator is
the product of the numerators and whose denominator is the
product of the denominators of the given fractions. That is, by
(1-29), if neither 6 nor d is 0,
a c
ac
n
9cn
(JL4\J)
T 1
T~l
b d
bd
.
.
To
illustrate,
26
2-6
4
xx +
.
7~~3-7~7
3
+
x
1
x-2~
2
+
x2
3
-
-
2x
3
_________________
2
x
-
4
A special case of (1-29) is worth noting. To multiply a fraction
by a number or expression, we multiply the numerator by that
number or
expression. Thus,
b __ a
~
c
For example,
1
b __ a
~~
b
c
c
*
1
_
~"
ab
c
55
(-i).Kx = (^lI^.
x
and
The quotient of two
5
if
in
no factor in a
=
J/5
b/ d
J^
b
=
*.
be
c
illustrate,
/3 __
"
7/
Thus,
defined
is zero,
(1-30)
To
been
has
T/^J
bf a
fractions,
Section 1-2 and evaluated in Section 1-7.
denominator
i\
/
2.| = |,
2
2
5
'
7
_
~ 5*2 _
~
3
In practice,
7
we
3
10
3a?
'
2
21
/
2/
divide out
all
5
/6a
3
7
" 3^
3
__
e/
2
'
6^
2/
factors
j/^_
common
5
_
~ 3x3
2/
to the
*
-
y
7
Go;
_
""
5
j/^_ '
2x 3
numerator
and denominator before proceeding with the actual multiplication.
numerator and the denominator of each given
fraction should be factored.
If necessary, the
T2
__
OT
,
Example 1-37. Multiply &x*a
+ &r
,
O/>2
.
-p
9
o
Q-y
by
.u
_
r-^V
y
u
and simplify the result
40
Sec.
Introductory Topics
The work may be
Solution:
-
a2
2x
2x*
indicated as follows:
-
2z
+ 5x + 3
x
-
3x
'
2
-
2
_
~
9
-
x(x
9
(x
+
_
~
-
x2
i
* oo
^Example 1-38. Divide
i
-
x(x
By
Solution:
x2
x2
-
5#
-f-
4
+4
3x
-
z3
*
4
-
+ 2^-1
4x 2
4
a-
__
x2
=
a;
+
3)
-2x
3)
'
3)
4x 2
+
3) (x
a;
2
-j-
-
4s
+3
4
^-^
- 5x + 4) (2x - 1)
- 4) (x 3 - 4x 2 + x - 4)
(x
- 4) (2s - 1)
(x -l)(x
(x + 1) (x -4) (a -4) (z +l)
- 1) (2s - 1)
(s
- 4) (x + 1)
(x + 1) (x
a
(x
-
2
__
+
~~
+
-
3
3) (2x
3) (x
2)
by
(1-30),
5z
rr
,
^j
+
(2x
1)
-
2) (x
(x -f 1) (a
n
1-27
3z
2
COMPLEX FRACTIONS
1-28.
The fractions we have been discussing so far may be called
simple fractions, to distinguish them from the fractions which we
now
A
discuss.
fraction which contains other fractions in the numerator or
denominator is called a complex fraction. Since the simpli-
in the
complex fraction is essentially a problem in division,
reduce the numerator and the denominator of the complex
fraction to simple fractions and then proceed as in division.
fication of a
we
first
-
1+ ;
r
Example 1-39. Simplify-
x
-
First Solution:
T
fraction
I
x
y
The numerator
of the given
-
-
complex fraction reduces to the simple
77
I!
>
and the denominator reduces
x
+y
y
-f
to -
\-
7?
Hence, we have
xy
,
+ y) xy
(y + x)
(x
x
x
Alternate Solution: Frequently it may be more convenient to multiply both the
numerator and the denominator of the complex fraction by an L.C.M. of the
denominators of all simple fractions occurring in the given complex fraction. In this
and an L.C.M. of their denomithe
fractions are - - and -
example,
simple
nators is xy.
>
>
>
by multiplying the numerator and the denominator
by xy, we get
Therefore,
fraction
l
\
+ x)'
xy
__
xy
+y* _
(x
+
y)y
_
of the
complex
Sec.
1-28
Introductory Topics
41
EXERCISE 1-9
In each of the problems from 1 to 20, perform the indicated operations and
express the answer as a simple fraction in lowest terms.
42
Sec.
Introductory Topics
8
^4^-
33.
-f
-
x
1
~
*v
#
s
+y
__
1
35.
4
g
-
x2
+
2s
1-2*
x
_
-
-
""L1
~2.
.
4
+g
g
1_
2
34.
____
;
_
x
1-28
2#
y
x
-f 1
x
-2y
-
_
I",
2y
__
07
x
-lx +
x*
l
-y*
a?
Simplify each of the following expressions:
LINEAR EQUATIONS
1-29.
Introduction. In Section 1-13, an equation
was
defined as a state-
ment of
equality between two algebraic expressions. In this section
we shall discuss equations in one unknown of the simplest type,
called linear equations, typified by the form
(1-53)
ax
,
where a and & are
specified
=
&,
numbers and a
ing to equations in general are needed
Solution or Root of an Equation.
=
0.
Some
ideas pertain-
first.
We shall use the
term unknown
to designate a literal quantity that appears in an equation and is
not regarded as specified at the outset.
system of values of the
A
unknowns which, when substituted for them, makes the equation a
true assertion
is called
also called a root
+
3(2)
a solution of the equation. A solution is
only one unknown is involved. Thus, the
1 define a solution of the equation
x = 2 and y =
=
2y 4, since the equation
+ 2(-l) = 4.
values
3#
when
is satisfied
for these values
;
that
is,
Sec.
1-30
43
Introductory Topics
Two
Equivalent Equations.
exactly the
same
solutions.
equations are equivalent if they have
For example, 2x + 11 = 7x - 4 and
14# - 8 are equivalent, since, as will be seen, each has
exactly one root, namely, x = 3. However, 2x + 11 = Ix - 4 and
2# 2 + llz = To; 2 - 4# are not equivalent, because the latter equation
4#
+
22
=
has, in addition to the root x
= 3,
the root x
= 0.
Operations on Equations. Each of the following operations on an
equation yields an equivalent equation
Adding the same expression to (or subtracting the same expression from) both sides.
:
Multiplying
(or
both sides by the same non-zero
dividing)
number.
For any
solution of
F=G
makes
also
F+H=G+H
true,
and
therefore a solution of this latter equation. Conversely, any
= G + is one of F = G, because
solution of F +
is
H
H
F = F+
H-H = G + H-H = G.
Likewise, any solution of Fj= G is one of
is a non-zero number, and conversely.
aF =
aG, provided that
a
Transposition of Terms. Transposing a term of an equation conmoving the term from one side of the equation to the other
and changing its sign. This operation is equivalent to adding the
same quantity to both sides (or subtracting the same quantity from
both sides) For example, consider the equation
sists in
.
2x+
11
=
7x
-
4.
we transpose 2x from the left side to the
-4 from the right to the left, we obtain
11 + 4 = Ix - 2x.
If
In
effect,
1-30.
we
right,
and transpose
subtracted 2x from each side and added 4 to each side.
LINEAR EQUATIONS IN
ONE UNKNOWN
An
equation in the form shown in (1-53) is called a linear equaunknown. We shall show that such a linear equation
has one and only one root, namely,
tion in one
=
a
each side of the equation ax
=
(1-54)
If
x
x
b is divided
b
=a
>
by
a,
the result
is
44
Introductory Topics
which
Sec.
1-30
equivalent to the given equation. Therefore, (1-53)
and only one solution, namely, b/a.
It should be noted that if we allow a to be
in (1-53), there are
two possibilities. Either no solution exists, when 6^0, since for
no number x is it true that (fx = Q x = Q = b^Q; or else every
number # is a solution, when 6 = 0, since a*# = 0'# = = & for
is
clearly has one
9
all
values of
x.
An equation which is not apparently linear may frequently be
solved by the theory of linear equations, by replacing the given
equation by a linear equation to which it is equivalent. The following steps serve as a guide to the method to be used when there is
only one unknown in the equation.
1.
Clear the equation of any fractions with numerical denominators by multiplying both sides of the equation by a least common
multiple of the denominators of those fractions.
Transpose all terms containing the unknown to one side and
other terms to the other side. We may collect the terms containing the unknown on either side.
3. Combine like terms. If the equation now assumes the form
2.
all
in (1-53), it is a linear equation and can be solved
sides by the coefficient of the unknown.
by dividing both
Check the
result obtained in step 3 by substituting in the
original equation. While it is desirable to include step 4 to show
that the number x found in step 3 is actually a solution of the
4.
equation, step 4 is not a necessary part of the solution process,
since the operations performed in the preceding steps always
yield equivalent equations. The purpose of step 4 is to help
to make certain that there has been no error.
2, we generally transpose terms containing the unknown
whichever side makes solving the equation easier.
The following explanations should be noted carefully.
Students at times "transpose" coefficients of the unknown. Thus,
2x = 6 takes the erroneous form x = 6
2 by "transposing" 2 to
In step
to
The correct procedure is to remove the coefficient
2 by division, since it is a multiplier of x. Therefore, we should
divide both sides of the equation 2x = 6 by 2 and have
2x
=6
2
2'
or
the right side.
a;
=
3.
Errors of this type may be avoided if the student applies the rules
of algebra properly and checks his solutions carefully.
1-30
Sec.
45
Introductory Topics
Multiplying or dividing both sides of an equation by a polynomial involving the unknown will not necessarily yield an equivalent equation. When the operation is multiplication, the new equation thus obtained may have roots in addition to the roots of the
original equation. These extra roots are called extraneous roots.
then say that the equation is redundant with respect to the
We
When
both sides of an equation are divided by
a polynomial involving the unknown, the new equation may lack
some of the original roots. It is then said to be defective with
original equation.
respect to the original equation.
If both sides of an equation are multiplied by a polynomial
involving the unknown, the check in step 4 of the recommended
procedure is a necessary step in the solution process. An extra-
neous root can thus be identified. Dividing both sides of an equaby a polynomial involving the unknown is not a permissible
procedure, since roots that are lost cannot be regained.
It should be noted that a non-linear equation may sometimes be treated so as to obtain a linear equation which is possibly
redundant. For example, the fractions in the equations of Problems
39 to 48 of Exercise 1-10 may be eliminated by multiplying both
sides of each equation by a least common multiple of the denomina-
tion
tors of the fractions in that equation. Since this multiplier contains the unknown, the new equation may have solutions which are
not solutions of the given equation. Hence, the solutions must be
checked to see whether or not they actually are solutions of the
given equation.
ST
~Example 1-40. Solve the equation o
We
Solution:
which
is
x A 7
'=
4
o
clear the equation of fractions
by multiplying both
Collecting the terms containing x on the left side and
right,
we have
10*
Combining
like terms,
3*
=
60
I3x
=
39.
+
-
by
13,
by
15,
21.
=
3.
check, substitute 3 for x in the original equation.
4
o
Therefore, 3
is
other terms on the
we have
*
|
?l2 -
all
we obtain
Finally, dividing both sides
To
sides
an L.C.M. of the denominators of the fractions, and obtain
Wx = 60 - 3x - 21.
the root.
-
iZ
o
or
2=4-2,
The
or
result is
2=2.
46
Sec.
Introductory Topics
Example
-
1-41. Solve the equation 6x
-
3y
=
1
5y
-f-
2x
-f 11 for
1-30
y in terms
of x. In this case, regard x as specified.
Solution:
left.
The
6x
Combining
unknown y on
Collect the terms in the
other terms on the
like terms,
-
-
2x
divide both sides
by the
1
-
-
12
+
by
3y.
= %.
-
3
check, substitute
for
in the original equation,
?/
^
This equation reduces to
I2x
-
to obtain
--
'
#
unknown
_
-IT
2
g
To
=
11
coefficient of the
$x
Then
all
we have
4z
Now we
the right side, and collect
}
result is
and obtain
At
3(x
-
+
-
3)
-
2
2
=
5z
=
5 (a?
-
3)
+ 4z + 22,
which becomes
I2x
-
3z
9
-
15
+ 4^ + 22,
or
+
9x
Since this result
is
an
identity,
7
y
=
=
9x
+
7.
is
^
the solution of the original
equation, regardless of the value of x.
Example
1-42.
Find two consecutive integers such that four times the
equal to six times the second diminished
by
first is
20.
Solution: Let x be the smaller integer, and x + 1 the next larger integer. Then,
from the statement of the problem, we have
4z = 6(x 4- 1) - 20.
From
this equation,
we obtain
4z
=
6z
-f
6
-
20.
Then,
14
Hence x
=
7 and x -f
1
=
= 2x,
or
s=7.
8 are the two consecutive integers.
The student
should carefully check these values by substitution in the original statement of
the problem.
Example 1-43. The speed of an airplane in still air is 400 miles per hour. If it
requires 20 minutes longer to fly from A to B against a wind of 50 miles per hour
than it does to fly from B to A with the wind, what is the distance from A to ?
Sec.
1-30
47
Introductory Topics
Solution: Let x be the distance from A to B. From the data, the speed of the
50 = 350, and the speed of the
plane against the wind, in miles per hour, is 400
= 450.
in
with
the
miles
is
400
50
4wind,
plane
per hour,
Since distance
=
;rr
;rrrr
OOU
=
=
t.
required to fly from
A
to B,
time, in hours, required to fly from
B
to A.
time, or d
rate
= time, in hours,
rt,
we have -
Hence,,
and
jrr
=
Therefore, from the statement of the problem,
x
x
Solving this equation, we have
9z - 7x = 1050,
A
B
to
is
_
~450 "~3*
350
So the distance from
follows that
it
1
x
or
= 525.
525 miles.
EXERCISE 1-10
Solve for the
1.
4.
7.
10.
-
-
4z
unknown
=
2
3
-
2.
= 2 - 20.
4 - 3z = 6(1 + 2).
60 + 7 = 50 + 6.
30
+
problem from
in each
2x.
7
5.
3x
-
-
=
6
-
9x
7
x
=
+
6
1
to 15.
12.
-
= - 4z 4w + = 3w - i
L
x - 8 = 2x + 3.
- | = | - jx
\x
3x
3.
Cr
-
2
K
6.
J
i
= 3z 4- 4.
9.
11. lOx - 3 = 9z + 4.
12.
4
13. 5z - 7 - 8z = 4x - 17 - fcc.
= 2y + 7 + 3y.
15. 6(5 + 4*) 8.
4z
-
6
+4 +
6y
Solve for y in terms of x in each problem from 16 to 24.
16.
2x
-
y
19.
40
-
2x
22. 3(x
-
4)
Solve for
25. 5z
-
=
3
3.
+ 2 = 0.
+ 40 = -
all
3.
Q>
+ 20 =
+ 1 = 1.
2x
+ 4(0 -
17.
23z
20.
-
23.
18.
4.
21. 6z
3(x
-
+ 10 =
- 20 = 3.
-
a:
4
^
ju
14. Hty
5.
1
40
4)
= 0.
0.
*
3)
=
5.
24. 3x
+
7(0
=
ij
.
|
values of x which satisfy the equation in each problem from 25 to 48.
= 4(s - 2).
26. 8fo
-
2^
-
9(
-
4)
=
13.
27.
!^-J>
= n.
48
Sec.
Introductory Topics
1-30
-
2(x
49. Divide 98 into
2
+
3rc
two parts such that one
50. Find three consecutive integers
+ 9)
of
52. If 8 times a certain
number
is
-
whose sum
is
3
2
~~
them exceeds the other by
two consecutive integers whose squares
51. Find
81
rr
18.
84.
differ
by
13.
9 more than 5 times the same number, what
is
the number?
53.
A
is
54.
rectangular plot of ground is four times as long as
feet, what is its area?
it is
wide. If
its
perimeter
4,800
How many
pounds
of coffee at 90 cents per
98 cents per pound
will it
take to
pound and how many pounds at
make 100 pounds
of a mixture costing
96
cents per pound?
55.
At a
was 25 cents for a child and 75 cents for an adult.
from 500 admissions, how many children and how many
adults were admitted?
If
college play, admission
$210 was taken
56.
A man
57.
What
has $365 in 41 bills of $5 and $10 denominations.
each denomination does he have?
and
58.
in
are the angles of a triangle,
if
one angle
is
How many
bills of
three times the second angle
six times the third angle?
One man,
same job
together?
X
y
can do a certain job in 7 days, and another man, Y, can do the
How long would it take them to do the job working
in 15 days.
2
2-1.
The Function Concept
RECTANGULAR COORDINATE SYSTEMS
IN
A PLANE
In Section 1-5 we saw how we can associate a real number with
every point on a number scale. The real number attached to a given
point is called the coordinate of the point. This representation suggests the assumption that to any real number there corresponds
precisely one point on the scale, and to any point of the scale there
corresponds precisely one real number. This one-to-one correspondence between the set of real numbers and points on the number
scale is known as a one-dimensional coordinate system.
We shall now extend the concept of a one-dimensional coordinate
system to a system of coordinates in a plane in which two number
scales are perpendicular to each other. The two perpendicular
lines, which we shall call coordinate axes, divide the plane into four
parts, or quadrants^ numbered as shown in Fig. 2-1. The horizontal
and vertical lines are designated as the
x-axis and the y-axis, respectively, and
their point of intersection is called the
origin and
On
is
each of these axes,
number
scale
D
labeled O.
by
we
selecting
construct a
an arbitrary
unit of length and the origin as the zero
point. As in Section 1-5, a coordinate on
the x-axis will be considered positive if
it is to the right of 0, that is, to the
right of the #-axis, and will be negative
if it is to the left. A coordinate on the
IV
p IGt
2-1.
be considered positive if it is above the #-axis, and negabelow.
Just as the real-number scale of Fig. 1-1 gave us a system of
one-dimensional coordinates by which we could set up a one-to-one
correspondence between points on a line and real numbers, so the
2/-axis will
tive if
it is
49
50
Sec. 2-1
The Function Concept
system of coordinates with respect to two mutually perpendicular
axes sets up a one-to-one correspondence between points in a plane
and ordered number pairs. We use the designation "ordered" pairs
for the following reason. To designate any point, we shall agree to
give its directed distance frdm the i/-axis first, and call it the
abscissa or x-coordinate, and then the directed distance from the
and call it the ordinate or y-coordinate. The abscissa and
ordinate of a point constitute its rectangular coordinates. They are
written in parentheses as an ordered number pair, as in the nota#-axis
tion (x, y), the abscissa always being written first. By this scheme
we assign to each point of the plane a definite ordered pair (x, y)
of real numbers and, conversely, to each ordered pair (x, y) of
real
numbers there
PO.2)
T
\
\
FlG 2-2
\
assigned a definite point of the plane.
Thus, the abscissa of the point P in
Fig. 2-2 is 3, and its ordinate is 2, and
we say that the coordinates of the point
P are (3, 2). Similarly, the coordinates
of Q are (2, 0), the coordinates of R
are (0,
+-x
3), and those of the origin are
(0, 0). Marking in the plane the position of a point designated by its coordinates is called plotting the point.
The coordinate system we have constf ucted is a particular case of cartesian
is
coordinates, so called in honor of Rene
Descartes (1596-1650), who first introduced a coordinate system
in 1637. It is called a rectangular system, since the axes intersect
in a right angle. (Actually the axes may intersect at any angle, but
it is usually simpler to take them perpendicular to each other. When
the two axes are not perpendicular, the coordinate system is called
an oblique coordinate system. Oblique systems
will not
be used in
this book.)
2-2.
It
DISTANCE BETWEEN
was seen
TWO
POINTS
in Section 1-10 that
|a
6|
equals the distance
between two points on the number scale represented by the real
numbers a and 6. It follows that \x 2 - a?i| represents the distance
between the points A (xi, 0) and B(x 2 6) on the #-axis of Fig. 2-3.
Let us now consider the two points PI and P2 with coordinates
(#i> #1) and (# 2 yi). The points have the same ^/-coordinate, which
means that they lie on the same horizontal line. Hence, the distance between the points PI(XI, yi) and P2 (tf2> 2/i) is the same as
,
,
Sec.
2-2
The Function Concept
51
*K
*-JT
FIG. 2-3.
FIG. 2-4.
between A(#i, 0) and Z?(# 2 ,0). This distance is
dis#i|- Similarly, we can show that [2/2
2/i| represents the
tance between two points (#2 2/1) and (# 2 2/2) on a vertical line.
We have thus arrived at the following two important properties.
If two points PI (#1,1/1) and P 2 (# 2 2/i) have the same ^-coordinate, then the distance between them, or |PiP 2 is given by
the distance
|
#2
~
,
,
,
|,
=
|PiP2
(2-1)
-
|*2
|
xi\.
two points Qi (#1,2/1) and Q 2 (#1,2/2) have the same #-coordinate then the distance between them is given by
If
(2-2)
-
IQiQal
~
(2/2
l/i|.
The concept
of the distance between any two points in a plane is
so important that we shall now develop a formula for it. Let us
denote by d the distance between the points PI (#1,2/1) and
P 2 (#2, 2/2). That is, d is the length of the line segment PiP2
2-4. Let P 3 be the point (# 2 yi) as shown.
Since the angle at P3 is a right angle, we have, by the
in Fig.
,
Pytha-
gorean theorem,
|PlP 2
2
Therefore,
|PiP 2
=
=
2
|
-
|
|PlP3
2
|
+
- #i| 2 +
- #i) 2 +
(#2
|*2
|P 3 P2|
It/2
(2/2
~
2
.
2
yi\
2
2/i)
.
the distance d between any two points PI (#1,2/1)
P2(#2, 2/2) in the plane is given by
That
is,
= V(#2 -
d
(2-3)
This formula
is
known
2
#i)
+
(2/2
-
and
2
2/i)
-
as the distance formula.
Example 2-1. Find the distance between the points
(
3 2) and
(5, 2).
Solution: It makes no difference which point is labeled Pi. Let us label the first
one Pi and the second one P 2 Since the two points have the same ^-coordinate,
.
(2-1) applies
and |PiP 3
=
1
|5
- (-
3)|
=
|5
+ 3| =
|8|
= 8.
52
The Function Concept
Sec.
Example 2-2. Find the distance between the points
Solution: Again let us label the first point PI
points have the
same ^-coordinate,
(2-2) applies
Solution:
and
(3, 7).
P2
and the second one
and |PiP2| = |7
Example 2-3. Find the distance between the points
-
Let us designate the points as Pi(2,
the distance |PiP2|
(3, 1)
-
(2,
1)
Since the
.
=
1|
and
and /Vo,
1)
2-2
6.
(5, 3).
By
3).
(2-3),
is
-2)* +(3 -(-I)) 2
= V9 +
We
16
=
\/25
=
5.
now
consider a special application of the distance
to a point
formula (2-3). The line segment OP from the origin
P(x,y) is called the radius vector to P. By (2-3), the distance
between O(0, 0) and any point P(x,y), or the length of the radius
shall
vector to P,
is
d
or
d
(2-4)
= V(* ~
= V* 2 +
O)
y
2
+
(y
-
2
O)
2
-
Thus, for the point P(3, 2), the radius vector has the length
V3 2 + 2 2 = VIST
EXERCISE 2-1
1.
Plot the following points:
b.
a. (3,5).
d.
(-3, -6).
e.
g.
(0,2).
h.
In each of the following
between them:
a. (2, 3) and (7, 3).
c. (1, 4) and (1, 0).
e. (2, 1) and (5, 6).
g. (3/2) and (5, 7).
2.
3.
(-4,7).
(1, -3).
(-5,0).
c.
f.
i.
cases, plot the pair of points
b. (5,
-
d.
(
f.
(0,
h.
(-
3,
4,
and
(5, -2).
(-8, -6)
(3,0).
find the distance
and ( - 1, - 2).
- 3, 4).
2) and (
and
8)
3).
(5,
- 6) and (- 8, - 6).
2)
-
Find the length of the radius vector to each of the following points
:
-
a. (4, 3).
b. (12, 5).
c.
(1,
d. (5,
e.
(7,0).
f.
(-3, -2).
h.
(-1,2).
i.
(a, 6).
k.
(- \/, -
L
(m, n).
gj-
(0,
(1,
2-3.
-12).
4)._
V3).
1).
1).
FUNCTIONS
So far we have been concerned with single numbers and pairs of
numbers. Now we shall consider mathematical relations, known
Sec.
2*3
The Function Concept
53
as functions, between two sets of numbers. To distinguish between
the two sets, we shall call one of these sets the domain of definition
of the function, and the other the range set Y of the function.
We begin by defining a variable to be a symbol which may take
any value in a given set of numbers. If # is a symbol which is used
to denote any number of the domain
and y is a symbol which
denotes any number of the set Y, then x is called an independent
variable and y is called a dependent variable. If a set contains only a
single number, the symbol used to represent that number is called
a constant.
To set forth a function, the domain should be explicitly specified ;
that is, it is necessary to determine definitely just what elements
or numbers the domain contains. The same is true of the range set
Then, as soon as a definite rule of correspondence is given which
assigns to each number x of the domain one or more numbers y of
the range set, the function is specified completely. We thus have a
set of ordered pairs of numbers (x, y), where x is any number of
and y is a number of Y.
The set of ordered pairs of numbers (x,y) is called the function.
The rule of correspondence which determines the collection of pairs
(x,y) is often expressed by a formula involving algebraic or other
processes. In .such cases we usually find it convenient to refer to the
formula as though it were the function. For example, we often
2
3x + 5 when actually we mean the
speak of the function y = x
X
X
X
3x + 5. If just
determined by y = x 2
one number of Y is paired with each value of x, the function is said
to be single-valued. If more than one number of Y is paired with
some value of x, the function is said to be multiple-valued. We frequently find it possible to deal with multiple-valued functions by
set of ordered pairs (x,y)
separating them into distinct single-valued functions.
The range of values of the function consists of those numbers y
in the range set Y which actually correspond to some number x of
the domain. When the range of a function has been determined,
it is always possible to replace the original range set, which may
include numbers in addition to those of the range, by the range
itself. We shall now consider several examples of functions.
The constant function y c associates the same
number with every number x of the domain X. Hence, the range
Y of the function consists of just one number c. Since y has the
same value for all pairs (x, y), the function is evidently singleIllustration 1.
c
valued.
The identity function y = x associates with every
number x the number itself. In other words, the numbers
Illustration 2.
real
54
The Function Concept
Sec.
2-3
are y = 1, 2, 3,
respectively. The
corresponding to x = 1, 2, 3,
domain
is the set of all real numbers, and the range Y is also
this entire set. The function y = x is single-valued, because it associates just one number of Y with each value of x.
X
Illustration 3. Let us consider the linear function defined by the
is the set of all real numbers,
2. The domain
equation y = 3x
and the range Y
X
is
mines a unique y
sponding to x = 1,
also this entire set. In this case a given x deterwhich is equal to 3x 2. For example, corre2, 3,
we have y =
1, 4, 7.
The function
is single-
valued.
Illustration 4. Let the rule of correspondence be given by the
2
be the set of all real numbers, and let Y
Also, let
equation y = x
be the set of all non-negative real numbers. Then, to the number
x
2 there corresponds the number y = ( 2) 2
4; to the number
x = 3 there corresponds the number y = 9 and so on. Hence, y = x 2
associates just one number of Y with a given value of x, and defines
y as a single-valued function of x. Although this is not obvious, the
X
.
;
range of the function
is
the given range set Y.
Illustration 5. In this case let the rule of correspondence be given
by the equation y 2 = x. Here X is the set of all non-negative real
numbers, and Y is the set of all real numbers. Then to the number
x - 2 there correspond the two numbers y
\/2 and y = \/2; to
3 and
the number x = 9 there correspond the numbers y = V9
2 =
=
=
defines
as
a
two-valued
x
on.
~3
and
so
y
Thus, y
y
V9
is there a single value of y, namely,
function of x. Only for x =
y = 0.
;
The function y = \A&2 #2 with domain X cona ^ x ^ a, is a singlesisting of all real numbers x such that
valued function with range set Y consisting of the numbers
^ y ^ a. Here the range is Y, as may be proved.
x 2 is negative.
x 2 has no meaning when a 2
Note that \fa2
2
x 2 < 0,
Hence, those values of x must be excluded for which a
and we must restrict the value of x to the interval a ^ x ^ a.
(We shall consider the meaning of the square root of a negative
number in Chapter 11.) The function y = -\/a2 x 2 is single- valued
Illustration 6.
>
because \/a2
x 2 represents the non-negative square root of a 2
a2 ~~ # 2 -)
x 2 is denoted by
(The other square root of a2
x2
.
V
Illustration 7.
The function y = l/x is defined for all real num0. For x = 0, l/x is not defined, since division
bers different from
by zero
is
not permissible.
In other words, although there are
Sec.
2-4
values of y for values of x neat*
when x
55
The Function Concept
0,
there
is
no possible value for y
actually equals 0.
Usually the functions which we are about to consider are defined
all values of x 9 with the following two exceptions
Values of x must be excluded which involve even roots of
for
:
negative numbers, since these are not defined as real numbers.
Values of x must be excluded for which a denominator is zero,
since division by zero
is not a permissible operation.
use
to be made of a function will restrict the
occasion, the
values of x for which it is to be regarded as defined.
On
2-4.
FUNCTIONAL NOTATION
Since functions are mathematical entities, they may be given
such as /, #, <>. To designate the number, or numbers, y corresponding to a given number x according to the rule
letter notations,
specified
/, we us6 the notation f(x).
the function / be defined by the equation
by a given function
illustration, let
y
Then /(O) =
=*
x2
-
2x
+
As an
3.
=
6, and so on. Frequently, the symbol f(x)
used to designate the function rather than the functional values.
The context will make the meaning cleai*.
It should be remembered that the notation y = f(x) does not
mean that y is a number / multiplied by another number x. Instead
3,
/(-I)
is
an abbreviation for "/ of x."
The set X does not have to be as simple as
it is
trations. If
in the preceding illus-
X should consist of a set of ordered pairs of numbers, the
rule of correspondence would then determine a value, or values, of y
for each ordered pair of X. We would then have a function of two
independent variables. For example, the area of a triangle
is
given
by the relationship
A = /(a,6)=|a&.
Here X is the set of ordered pairs, (a, 6), of positive real numbers,
where a is the length of the altitude of the triangle and 6 is the
length of its base and Y is the set of numbers A, each of which
represents an area corresponding to a given pair (a, 6). Similarly,
;
a function of three variables, f(x,y,z),
of ordered triples (x,y,z). Thus,
is
X
2
2
/(2, 3) = 2 + 3 = 13.
= 3-2 + 2(5) =11.
Also,
if
f(x, y> z)
defined in terms of a set
if
~
x
=
x 2 + y* then
2*, then / (3, 2, 5)
f(x, y)
-y+
9
56
T/ra Function
Sec.
Concept
2-4
EXERCISE 2-2
By using the phrase "a function of" in each problem from 1 to 8, express each
given quantity, which is regarded as a dependent variable, as a function of one or
more independent variables. Where possible write the relationship both in words
and
in symbols.
7.
The
The
The
The
8.
A
9.
Given f(x)
=
2x
-
10.
Given
=
3x
+
1.
3.
5.
11.
=
area of a
2.
circle.
area of a trapezoid.
volume of a cylinder.
annual premium for a
4.
6.
life
The area of a triangle.
The volume of a sphere.
The retail price of food in a grocery
store.
insurance policy.
person's height.
g(x)
3, find /(O),
/(-
1),
5, find 0(1),
g(-
3),
/(3), /(1/2),
/(V),
3/(l), /(3)/4,
(0(2)), 0(4), 20(4), g(z
Using f(x) and 0(x) as defined in problems 9 and
1).
2)
/(
10, find
-
j^-
In problems 12 to 26, let /(a;, j/) = 2xy + 3.t - 2y, and
a 2 + 6 2 + c 2 Evaluate or simplify each given expression.
let
/(6)0(3),
>
0(a,
6,
c)
.
12. /(I, 2).
13. /(O, 0).
14. 0(0, 0, 0).
15. 0(1, 2, 3).
16. /(i, 1/x).
17. p(p, g, r).
19-
20.
18.
M
25.
g(-
if)
+
ffe
a, 6, c)
V, -).
-
g(a,
-
26. 0(a,
6, c).
6,
-
c).
A and circumference of C of a circle as functions of the radius r.
r, express A as a function of C, and also express C as a func-
27. Express the area
By
eliminating
A.
tion of
28. Express the
volume
of a sphere as a function of its surface area.
29. Express the surface arfea of a sphere as a function of its volume.
30.
Suppose that
is
U is a function of V and that V is a function of W. Show that U
a function of
W.
Determine the maximal domain of values of x for which y is defined as a function
from 31 to 61. Assume that x and y are real numbers.
of x in each problem
31. y
= x.
32. y
+
= z2
= #(2z
34. y == 3z
37. y
35. y
1.
38. y
.
40.
y
43.
y=x(x-l)(x+l).
-f-
1).
41. y
44.
= 3z.
= 2x - 3.
= x - 2.
= (3x- 1)
42. y
= |
= 4z + 5.
= x + 1.
= (2x - 3)
45. y
=
33. y
36. y
2
39. y
(a?
2/=z2+i- 2 = 9.
4- 1).
2
X\X
3/
46. x 2
49. x 2
+ y 2 = 9.
+ = 0.
47. x 2
50. 3* 2
2/
+ tf = 0.
l
48. x 2
"
-
2
2/
L)
=
*2
0.
(2x
+ 1).
2-6
Sec.
57
The Function Concept
54.,=
55 - y
=
58. y
= Vr^2.
2-5.
SOME
x
-T7
T- i
'
56. z
= 4i/2.
59.
= V4 -
y
a:
2
.
57. s
+
60.
=
y
1
=
Zy*.
SPECIAL FUNCTIONS
The following
additional illustrations of two rather unusual, but
functions
are given here to help us become better
very useful,
with
the function concept.
acquainted
The absolute-value function is defined by associating with x its
absolute value x\. The functional relation is given by j/=|a?|.
Thus, the domain comprises the set of all real numbers, while the
range comprises the set of all non-negative real numbers. For this
function we have the values 2, 3, 0, TT, \/2 corresponding, respectively, to x = 2, -3, 0, 77, -\/2.
The bracket function or greatest-integer function, represented
by the notation y = [x] is defined as the largest integer which does
not exceed x. Its domain is the set of all real numbers, and its
range is the set of all integers. Thus, if f(x) = [>], then /(-3.5)
,
=
-4, /(-I)
-
-1, /(O)
-
0,
/(2.5)
-
2,
and /(5) =
5.
EXERCISE 2-3
/W =
1.
Given
2.
Given /(x)
3.
Given /(x)
4.
Given /(x)
=
=
=
=
fa?
,
|
r
x
|
x
find/(-
3),/(2.3),
find/(3.2),/(2),
[x],
\
[x],
find/(-
3) and/(2.5).
5.
Given f(x)
Let/(x) be the function whose domain
the definition
is
3).
+ [x] - x, find/( - 3.5) and/(4).
+ [2 - x] - 1, find/(0),/(- l/2),/(l),/(3/2),
6.
[x]
an
and/(-
as follows
and/(2).
X is the set of all real numbers for which
:
if x < 0, then/(aO
\ix ;>0, then/(x)
= = x.
x\
Find/(3)and/(-2).
2-6.
VARIATION
A
particularly important example of a simple type of function
often occurring in the physical sciences is given by the formula
y
If
k
>
0,
this equation
=
kx.
shows that y increases a$ x increases, and
We usually say that y varies directly
that y decreases as x decreases.
The Function Concept
58
Sec.
is directly proportional to x. If x
also be written as follows
as x, or that y
ship
may
where k
^ 0,
2-6
the relation-
:
the constant of proportionality. This relationship
to
saying that the ratio of y to x remains constant for
equivalent
all non-zero values of x.
is called
is
The value of the constant k in any particular problem may be
determined from a known pair of values of x and y in the problem.
Thus, if the given relationship is y = kx, and if we know that y = 6
when x = 2, then k = 3. Th formula then becomes y = 3x.
We
say that y varies inversely as
tO a, if
x, or
y
is
inversely proportional
r.
V
=
(**<>>
IC
This relationship shows that ?/ decreases as x increases, and that y
increases as x decreases. But, when x = 0, the following two formulas are equivalent
^
xy = fc and y =
:
*c
Therefore, the relationship between a? and # is such that the product
of x and y is constant.
Several types of variations may be combined in a single equation
to express a certain law. For example, when y varies directly as x
and z, we say that y varies jointly with x and z 9 and we write
y
=
kx&.
Direct variation and inverse variation are often combined in
applications. Thus, according to Newton's law of gravitation, the
force F of attraction between two bodies of masses m\ and
2 varies
w
directly as the product of their masses and inversely as the square
of the distance d between them. The equation is
_
*
Example 2-4,
If
a
__
kmim,2
~~d?~'
man is paid $15 for an 8-hour day, how much would
he make
in a 35-hour week?
Solution:
Since this
is
The wages a man earns vajy
directly as the
amount
of time he works.
a problem in direct variation, we have
w =
kt.
In this formula, w represents the total wages, in dollars; t represents the time
worked, in hours; and the constant k represents the wage rate in dollars per hour.
Substituting w = 15 and t = 8 determine^ the constant k = 1.875. The general
formula then becomes
w =
1.875
t.
Sec.
2-6
The Function Concept
Therefore,
when
Hence, the
man
t
= 35,
59
we have
w =
=
(1.875) (35)
65.62.
earns $65.62 in a week.
A motorist traveling at an average rate of 50 miles per hour made
Example 2-5.
a trip in 5 hours. How long would
rate of 60 miles per hour?
it
take him to
make
the same trip at an average
Solution: Since the time required Varies inversely as the speed,
we have
'=*
In this case, k represents the distance traveled, in miles; r is the speed, in miles
t is the time
required, in hours. Substituting t = 5 and r = 50
per hour; and
determines k
=
250. Hence, the general formula
,
t
when
Therefore,
r
=
_250
r
we obtain
60,
'
_ 25
~ 250 ~
60
Hence, the time required
Example 2-6.
is
Under
is
'
6
4 hours 10 minutes.
suitable conditions the electric current / in a conductor
E and inversely as the resistance R of the
R = 10 ohms, / = 11 amperes. Find what
varies directly as the electromotive force
When E =
conductor.
voltage
is
ohms
necessary to cause 2 amperes to flow through 60
Solution:
is
110 volts and
From
the statement of the problem,
given by the formula
we
see that the
of resistance.
combined variation
fejjj
1
=
11, E
110, and
Substituting I
Therefore, the formula becomes
~~W'
R =
10 determines the constant k to be
1.
/= R
This relationship may also be expressed as follows: The required voltage E
equal to the current / flowing through the conductor multiplied by the resistance
R, or E = IR.
is
For the
specified values,
In this problem k
It is
=
1.
E=
(2) (60)
=
120. Hence,
The formula obtained
is
E=
120 volts.
commonly known
widely applied to entire and partial circuits through which
elefctric
as
Ohm's
law.
currents flow.
EXERCISE 2-4
y varies directly with x, and y = 15 wh&i x = 7, find a formula for y in
terms of x.
If x varies directly with y, and x *= 32 when y = 4, find x When y = 3.
If y is directly proportional to x 2 and y = 112 when x = 4, find y when x = 9.
1. If
2.
3.
4. If
,
y
is
inversely proportional to x,
and y
= - % when x =
1,
find y
when x
=
-3.
60
Sec.
2-6
= 6.
= 12
and
when x
=9
The Function Concept
5. If
y varies inversely with
x,
and y
=
10
= 3,
when x
# varies directly with y and inversely with
2=2, find x when ?/ = 16 and 2=4.
6. If
2,
22,
y varies directly with \/x and inversely with
and 2=2, find / when z = 25 and 2=6.
8. If
y
is
directly proportional to x
when #
9. If
=
1
and
2
=
1,
find y
2=2,
find
y when
a:
4 when y
and y
=
18
and inversely proportional to
= 2 and 2=4.
\/z,
and
2/
=
4
=
1
when x
3
y varies directly with x and inversely with
and
=
and #
7. If
y when x
find
= -
1
two spheres have radii r\ and
and 2 respectively, show that
10. If
and
r2
,
= -
2
22 ,
1
=
and y
2
when #
2.
diameters d\ and
^2,
and surface areas Si
,
ri*
11. If
d
Si.*
Sa
the two spheres in problem 10 have volumes V\ and
~"
r2 2
"~
d22
F2
,
respectively,
show
that
12. If the radii of
two spheres are 3 units and 1 unit, respectively,
and b) the ratio of their volumes.
find a) the ratio
of their surface areas
13.
What
are the ratio of the surface areas
spheres
if
the ratio of their radii
is
and the
ratio of the
volumes of two
3/2?
The diameter
14.
of the planet Jupiter is approximately 10.9 times the diameter of
Earth. Assuming that both planets are spheres, find a) the ratio of their surface
areas and 6) the ratio of their volumes.
15.
The diameter of the sun
Compare the volumes and
that the sun and
16.
approximately 109 times the diameter of Earth.
the surface areas of the sun and Jupiter,
both planets arc spheres.
By how much must
whose surface area
17.
is
is
we assume
the diameter of a sphere be multiplied to give a sphere
25 times that of a given sphere?
When the volume V of a gas remains
absolute temperature T.
so-called absolute zero, which
its
if
constant, the pressure
(Absolute temperature
-
is
P varies directly as
measured from the
-
460 F or
is approximately
273 C.) If gas
enclosed in a tank having a volume of 1,000 cubic feet and the pressure is_
54 pounds per square inch at a temperature of 27 C, what will be the temperature when the pressure is raised to 108 pounds per square inch?
is
T of a gas remains constant, the pressure P varies inversely
A gas at a pressure of 50 pounds per square inch has a volume
18. If the temperature
as the
volume V.
of 1,000 cubic feet. If the pressure is increased to 150
while the temperature remains constant,
19*
The weight
of a
of the distance
what
pounds per square inch
the volume?
body above the earth's surface varies inversely as the Square
from the center of the earth. If a certain body weighs 100
is 4,000 miles from the center of the earth, hdw much will it
pounds when it
weigh when it is 4,010 miles from the center
center?
is
.
arid
when 4,100
miles from the
Sec.
2-7
The Function Concept
61
20.
The electrical resistance of a wire varies directly as its length and inversely as
the square of its diameter. A copper wire 10 inches long and 0.04 inches in
diameter has a resistance of 0.0656 ohms, approximately. What is the resistance
of a copper wire 1 inch long and 0.01 inches in diameter?
21.
What
is
is
the diameter of a copper wire 1,000 inches long whose resistance
ohms?
10
22. According to Kepler's third law, the square of the time it takes a planet to
make one circuit about the sun varies as the cube of its mean distance from the
The mean
sun.
distance of the earth
distance of Jupiter
make one
circuit
is
92.9 million miles, and the mean
Find the time it takes Jupiter to
is
475.5 million miles.
about the sun.
j
2-7.
CLASSIFICATION OF FUNCTIONS
It is often desirable to
immediate purpose
it
group functions into
a
will suffice to consider
classes.
For our
classification into
algebraic functions and non-algebraic, or transcendental, functions.
Let us first give a more precise definition of a polynomial function
and then define algebraic functions and give some illustrations
of both.
A polynomial function of # is a function given by the relationship
n~
n
+
+ a n_iz + a
y = a G x + aix
an
where a a
an are real constants, a ^ 0, and n is a posil
n,
-
lf
,
,
-i,
The polynomial function is said to be of
degree n. The function which makes the number
correspond to
x
zero
number
is
also
called
the
but
this polyevery
polynomial,
nomial has no degree.
A rational function of x is a function which either is a polynomial function or can be expressed as a quotient of two polynomials. Thus, a polynomial is often referred to as a rational
tive integer or zero.
integral function of x.
A
polynomial, or a rational integral function, of x, y,
sum of terms of the form
z,
,
is
defined to be the algebraic
kxa y bz c
where k
is
-
,
a constant coefficient and each of the exponents
a, b, c,
either a positive integer or zero*. The degree of such a function is the degree of the highest-degree term which is present.
-
-
-
is
and 5x 2 - Ixy* + 62 define
polynomial functions of the second degree and third degree, respec-
For example, the expressions 3x 2 -
tively.
These and the expressions
x
x
7
+
5
y
and 2x
y
-
x
\/7
+ x*9 +
t
.
are
l
examples of rational functions.
*
Zero exponents will be defined in Chapter
1 for any non-zero number u.
that u
=
4.
For now, one needs only note
62
The Function Concept
A
number
an algebraic number
is
equation of the
which the
2-7
a root of a polynomial
form
a,QX
in
if it is
Sec.
n
-{-
aix
n~~*
4~ a n ix
H~
-}-
an
=
0,
an are integers, not
coefficients Oo, 0i,
all zero.
term algebraic number, we have the term
Analogous
A
function y = f(x) is called an algebraic funcalgebraic function.
tion of x if y is a solution of an equation of the form
to the
Po(x)y
n
+ Pi(x)y*~ +
l
-
-
+ Pn~i(x)y + Pn (x) =
0,
Pn (x) are polynomials in
where the coefficients PQ(X), PI(X),
and
n
is
a
x,
positive integer.
Polynomials and rational functions are special types of algebraic
functions. The functions that we have considered so far were illus-
-
,
trations
y
= \/#
ft
case,
y
= y
x
According to our definition,
algebraic functions.
an algebraic function of x because y 2 - x = 0. In this
P (a?)=l, Pi(x)=0, and P2 (#) = x. Similarly,
2,
of
is
=
""
is
an algebraic function of & because
#7/
2
-
#2
Here n = 2, P (a;) = x, PI(X) = 0, and P2 (z) = - 2 + 1.
An irrational function is an algebraic function which
=
+
1
is
not a
0.
a:
rational function.
A
number
a number which is not algebraic,
and a transcendental function is a function which is not algebraic.
Functions like the trigonometric functions, which we shall take
up in Section 3-2, belong to the class of transcendental functions.
Later we shall consider other types of transcendental functions,
namely, the logarithmic and exponential functions such as log x
and
transcendental
is
w
3-1.
Tfie
Trigonometric Functions
THE POINT FUNCTION
P(t)
The trigonometric functions that we are about
to define are functions in the sense previously described in Section 2-3 ; that is, they
are relations between two sets of numbers. The student who is
familiar with the trigonometric functions from his high-school work
cautioned to note that we are not, for the present, discussing angles
with these functions. We shall see that the concept
of a trigonometric function need not be associated with an angle;
in fact, many of the most important applications of mathematics
in modern science and engineering are concerned with trigonometric functions of pure numbers. Hence, we shall adopt the numerical point of view, leaving the study in terms of angles as a
is
in connection
secondary consideration.
Consider a circle with a radius of one unit placed at the origin
of a rectangular-coordinate system. See Fig. 3-1. Let t be any real
D
ffi)
FIG. 3-2.
FIG. 3-1.
number. Starting at the point with coordinates (1,0), we lay off
on the unit circle an arc of length \t\. If t> 0, we measure the arc
in a counterclockwise direction. If t < 0, we measure in the clockwise direction. If t = 0, the arc consists only of the point (1,0). By
this procedure, there is associated with each real number t a definite end-point P(t) of the arc
whose
63
initial
point
is (1, 0).
There-
64
The Trigonometric Functions
Sec.
31
number t, we have a definite
ordered pair (x, y) of numbers which are the coordinates of the
endpoint of the arc.
Since P(t) lies on the unit circle, it is at one unit distance from
the origin. Hence, it follows from the distance formula that
fore, corresponding to every real
x2
(3-1)
+
y
=
2
1.
By means of this equation we can find the second coordinate of the
point P(t), except for sign, if one of the coordinates is known.
To determine the number pair (x, y) for the point P(t) corresponding to a given value of t, we shall take note of the fact that the
circumference of the unit circle is 2?r = 6.2832 units (approxiin Fig. 3-2 is one-half of the
mately) For example, since arc
is equal
TT
units
in
it
is
length, and arc
complete circumference,
ABC
.
AB
to 7T/2.
now becomes apparent that P(0) is the initial point
(1, 0) ; P(TT) is the point (-1, 0)
P(?r/2) is the point (0, 1) and
P(37r/2) and P(-ir/2) both represent the point (0, -1).
It
;
;
3-2.
DEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS
The correspondence between the
of ordered pairs
(x, y)
set of real
numbers
t
and the
leads to definitions of the six
set
common
trigonometric functions, namely, the sine, cosine, tangent, cotangent, secant and cosecant.
General Relationships. We shall define the cosine of the real
t to be x, and the sine of the real number t to be y. Thus,
number
we have
(3-2)
x
=
cos
t,
y
=
sin
t.
and
(3-3)
of each of these functions (the sine function and the
is the set of all real numbers t. Since, however,
every point P(t) lies on the unit circle, neither of its coordinates
(x, y) can exceed 1 in absolute value. Therefore,
The domain
cosine function)
and
|cosJ|<;i
(3-4)
|sin|<;i.
In other words, the ranges of cos t and sin t are restricted by the
requirements 1 ^ cos t ^ 1 and 1 ^ sin t ^ 1, respectively, for all
values of t. It may be shown that the range of each of these functions is the set of all real numbers u such that
1 = u ^ 1.
The other four functions may be defined in terms of the cosine
and sine, as follows
:
(3-5)
tan
t
=
^
cos
t
(cos
*
*
0),
See.
3-2
65
The Trigonometric Functions
(3-6)
cot
t
=
(3-7)
sec
t
= -^-
55i|
sin t
cos
esc
(3-8)
=
t
(sin
t
* 0),
(cos
<
5* 0),
(sin
*
5^ 0).
t
-r
Since the cosine and sine are defined in terms of the coordinates
it is also possible to express the other functions
same coordinates. From the definitions of the
cosine and the sine given by (3-2) and (3-3) and from the definitions of the other functions given by (3-5), (3-6), (3-7), and
of the point P(t),
in terms of these
(3-8),
*
we have
^ =~ =l'
<"->
^
<""'
i
(3-9)
t
(3-10)
<"
(3-11)'
v
sec*=-J
=
=
cos
esc
(3-12)
t
= -4- =
sin
We note here that cos
=
i
t
x -
\
G*?*0),
/>
'
y
(y
^ 0).
or x, appears in the denominators of both
tan t and sec t. Hence, tan t and sec t are not defined when t is a
number for which the ^-coordinate of P(t) equals zero. For
t,
example, since the ^-coordinate of P(7r/2) or P(37r/2) is 0, it follows that tan 7r/2, sec 7r/2, tan 3?r/2, and sec 377/2 are not defined.
Similarly, it can be shown that cot 0, esc 0, cot TT, and esc TT are not
defined.
We
functions tan
conclude, therefore, that the domain of each of the
t, cot t, sec t, and esc t is the set of all real numbers
for which the denominator
is
not zero.
follows from (3-7) and (3-8) that numerical values of
or of esc t can never be less than 1. Hence, the ranges of
It also
sec
t
these two functions are restricted by the requirements
(3-13)
|sec
t\
1
and
|csc
*|
^
1.
From (3-5) and (3-6) we obtain an idea of the behavior of the
tangent and cotangent functions. It may be shown that the range
of each of these funptions is the set of all real numbers.
The Trigonometric Functions
of ?r/6, Tr/4,
and
Tr/3.
The computa-
tion of the numerical values of the trigonometric functions in general is beyond the scope of this book. However, we shall use the
methods of plane geometry
to find sin
t,
cos
t,
and tan
t
for
,= 7r/6,
66
The Trigonometric Functions
3-2
FIG. 3-5.
FIG. 3-4.
FIG. 3-3.
Sec.
and 77/8, in order to show that for certain values of t the
trigonometric functions can be found exactly without tables.
To compute the functions for the real value t 77/8, we construct the unit circle of Fig. 3-3, Arc AB is given to be equal to
7T/3, which is one-sixth of the complete circumference. Triangle
OAB is inscribed in the circle, as shown, with side BO extended
through the origin to C. Our problem now is to find the values of
x = cos (7T/3) and y = sin (77/8) as coordinates of the point B.
Since OA = OB, triangle AOB is isosceles. Hence,
77/4,
OAB = angle OBA.
angle
We note also that arc CAB ~ 77 and
arc CA = arc CAB - arc AB = 77 - 77/8 = 277/3.
Therefore,
arc
Also,
angle
Furthermore, angle
Hence, it equals the
and
OBA
9
or
angle
2 angle
AOB
sum
COA =
is
2 arc AB.
2 angle
AOB.
an exterior angle of triangle AOB.
two remote interior angles OAB
of the
COA = angle OAB + angle OBA.
Thus,
and triangle
COA
CA =
AOB = angle OAB + angle OBA,
is equilateral.
we draw BD
perpendicular to OA, it will bisect OA. Then
~ 3/4 and =
1/2 and from x 2 + y 2 = 1 it follows that y 2
y
V5/2.
Hence, cos (77/8) =1/2, sin (77/8) = V5/2, and tan (77/8) = \/3
For i = 77/6, place the equilateral triangle AOB of Fig. 3-3 in the
unit circle as shown in Fig. 3-4, where E is the mid-point of arc
AB. Then arc EB = 77/6 and OE is the perpendicular bisector of
chord AB.
If
x
=
;
.
Sec.
3-2
67
The Trigonometric Functions
= 1/2. From x* + y 2 = 1 it
Since DB is one4ialf of
chord^AB, y
2 =
=
follows that x
3/4 and x
VS/2. Hence, cos (TT/G) = \/5/2 and
sin (7T/6) = 1/2. We then have tan (77/6) = 1/V3 =
yff/3.
For = 7T/4, construct a unit circle as shown in Fig. 3-5 with
arc AB = Tr/4. Since arc AC = 7r/2, arc 5C = ir/2 - Tr/4 = ir/4.
Hence, arc AB = arc BC, and
angle
A 05 = angle BOC.
Draw
BZ) perpendicular to OA. Since the two parallels
are cut by the transversal OB,
angle
BOC = angle
angle
A0# = angle OBD,
Hence,
OC
and
and
Thus, y
2x 2
=
=
x.
2
2
Substituting this value of y in x + y = 1, we have
= 1/2. Therefore, x - cos (Tr/4) = \/2/2, and # =
V2/2. It follows-that tan (ir/4) = 1.
and x 2
1
sin (77/4)
=
Other Special Values.
In a similar fashion
we can compute
rt_
C/jf
exactly the trigonometric functions of such values of
7
t
00
as
>
Q
>
and
- -~
Functions of multiples of ?r/2
may
also be
-5-
>
computed
in this fashion, if one considers a straight line as a right triangle
in which one angle is
and, hence, one side has zero length.
Example 3-1. Calculate the values
Solution:
(0, 1).
As explained
of the six trigonometric functions of
t
=
?r/2.
in Section 3-1, the coordinates of the point P(ir/2) are
Hence, by (3-2), (3-3), (3-9), (3-10), (3-11), and (3-12),
sin (?r/2)
= 0,
= 1,
tan (7T/2)
is
cos (7T/2)
undefined,
sec (?r/2)
is
esc (7T/2)
=
=
cot (w/2)
undefined,
1,
0.
EXERCISE 3-1
1.
Determine the coordinates of each of the following points:
a. P(2ir).
d.
P(
-
y)
b.
P(-T).
e. P(4ir).
c.
P(5x/2;.
f.
P(-
7ir).
In each of the following cases, carefully draw a unit circle and estimate the
coordinates of P(t).
c. P(3).
b. P(2).
a. P(l).
2.
d.
P(-2).
e. P(4).
f.-P(-8).
68
3.
The Trigonometric Functions
Sec.
3-2
Evaluate each of the following:
K
b. cos
-r-
o
d.
g.
4.
cot-
e. sin
csc(-).
u
h.
--
llTT
/
sm (-
,
f.
e. cos
t
t
= 1/2.
= - 1/2
b. cot
t
sin
t
= -
t
1.
being positive.
Complete the following
table,
secy
7r \
).
In each of the following cases assume that
approximately the appropriate arc (or arcs).
a. sin
5.
.
f
2ir,
c.
tan
t
=
f.
cot
t
=
and draw a
figure
d. esc
1.
1,
sec
i
t
showing
= -
1.
being negative.
which shows the algebraic signs of the
trig-
onometric functions in the four quadrants.
6.
Use the equation x 2
where
^ J g 27r.
+y =
1
7.
Use the equation x 2
+
=
1
2
2
1/
trigonometric functions of
a. sin
t
d. esc
t
g. cot
t
8.
e. sec
h. tan
= - sin
= sec
(
(
9.
For each of the following
t)
given condition
b.
t
which tan
f
= 3/5.
= - 2.
= - 6/7.
Prove that each of the following equations
sec
for
t
=
cot
t,
in each of the following cases to find the other
b. cos
c.
t
t.
= 1/2.
= - 3/2.
= 2.
a. sin
a.
to find all values of
is
t.
'^
.
c.
sec
f.
tan
i.
sin
J
t
J
= 13/12.
= 4/3.
= - 3/5.
correct.
b. cos
(
i)
d. tan
(
t)
cos
f.
tan
.
cases, state the quadrant, or quadrants, in which the
is satisfied.
The sine and cosine have the same signs.
The tangent and cosine have opposite signs.
3-3.
IDENTITIES
As an immediate consequence of the definitions of the six trigonometric functions, we can establish certain relationships among
Sec.
3-3
69
The Trigonometric Functions
them which hold for every value
of
Since P(t)
t.
lies
on the unit
2
2
(3-1) holds that is, x + y = 1. But, according to (3-2) and
(3-3) x = cos t and y = sin t. Therefore, we have the equation
circle,
;
,
cos 2
(3-14)
t
+
sin 2
=
t
1.
This states that "the square of the cosine of t plus the square of
the sine of t equals unity." Since (3-14) holds for every value of t,
2
it is an identity. Note that we use the symbol cos
t instead of
2
This simplified notation is used for all positive exponents,
(cos t)
but is never used in the case of a negative exponent. Thus, cos' 1 t
does not mean the same as (cos t)~ l as we shall see in Chapter 8.
.
,
Similarly,
we can prove
that for each value of
t
for which the
functions are defined,
(3-15)
1
(3-16)
1
Proof of (3-15): By
+ tan 2 =
+ cot 2 =
t
sec 2
,
t
esc 2
t.
tan
definition,
=
t
^
cos
and sec
t
=
cos
t
t
However, these relationships have no meaning if cos t equals zero,
that is, if the ^-coordinate T)f P(t) equals zero. When cos 1 7*0,
we may divide both sides of the identity cos 2 1 + sin 2 1 = 1 by cos2 *
and obtain
sin
,
2
1
*
~~~
j
cos 2
c
cos 2
t
Iherefore,
1
Proof of (3-16): By
In this case
of sin 2
1
+
we assume
cos 2
1
1
by
+
tan 2
=
t
sec 2
definition, cot
that sin
sin 2
1,
t
sin
t.
=
t
~~
nf\o
f
sin
t
^ 0. When we
we obtain
cos 2
_
~~
t
1
t
sin
2
t
2
and esc
t
=
1
sin
t
divide both sides
t
Therefore,
1
+
cot 2
t
=
esc 2
*.
We thus have established the three identities which
here for easier reference
we
restate
:
(3-14)
cos 2
(3-15)
1
(3-16)
1
+
+
t
+
sin 2
tan 2
t
cot 2
t
=
=
1
=
sec
1,
2
esc 2
t,
t.
These fundamental! identities are very important in working with
trigonometric identities and should be remembered.
Our present work with identities will consist of reducing given
trigonometric expressions to other forms. Unfortunately, no specific
rule of procedure can be given for
making these
reductions. Profit-
70
The Trigonometric Functions
Sec.
3-3
is a matter of both practice and
Generally speaking, when we want to reduce a given
expression to some other form, it is helpful first to perform any
indicated algebraic operations and then to use some form of one of
the fundamental relationships to simplify the expression.
To prove an identity, we may proceed in any one of the following ways
1) We may work on the more complicated member of the identity
and attempt to reduce it to the simpler member.
2) We may work with both sides of the identity and show that they
making such reductions
ciency in
experience.
:
induce to the same expression.
We may form the difference of the two sides of the identity and
prove that difference to be equal to zero.
It is frequently desirable to express both sides of the given identity
in terms of sines and cosines alone, and then use (3-14) if needed.
We shall consider a few examples to illustrate the procedure in
3)
reducing expressions.
Example 3-2. Show that cos
From
Solution:
= sec
t
-f sin
t
t
tan
=
t
cos
-f sin
t
,
Adding, we have
1
+ sin
=
2 1
Since sin 2
1
+
t
sec
tan
cot
+ cos
t
2
1
=
t
tan
t
sin
tan
t
cos
+ cot
=
Example 3-4.
t
t
=
:
t
to sin
t
cost
cos
t
sin
cos
t
=
sec L
t.
t
cost
sin
and cot
sin 2
cos
t
t
-
sin
Therefore,
t
cos
+
1
=-
t
cos 2
sin
sin 2
1
t
sin
t
+
By
1
t
cos
+
cot
t
sin
t
= sin
t
cos
t.
t
Establish the identity (sec
t
-
cos
O2
=
tan 2
t(l
reducing both sides to the same expression.
Solution:
t
cos 2
we obtain
1,
t
t
2J
t,
cos
1
tan
cos
t
-
definition,'
+
we have
t
^
,
t
111
By
J
t
cos
=
cos
3-3. Reduce
Example
*
tan
t
and
1
cos
Solution:
-f sin
t
cog2
=
t
cos
Since cos 2
t.
(3-5), or the definition of the tangent,
cos
.,,.
tan
-f sin t
t
definition, sec
f I
[
Vcos
t
=
cos
T
t
-
Hence, the
left side
t
.\*
COS
/
=
/I
-
cos
I
\
cos
1
t
2
1\*
)
)
becomes
-
cos 2
t)
by
Sec.
By
3-4
The Trigonometric Functions
we have
(3-14)
cos 2 *\ 2
/I
\
The
cos
_
~~
)
t
=
t
sin
J(l
=
o i\
1)
cos 2
_
.
1
<?os
1
By
definition,
t
sin 2
.
-
o 1/1
tan 2
cos 22
be reduced as follows.
may
Hence, we have
x
sin 4
_
""
Vcos t)
right side of the given identity
tan
/sin 2 *\ 2
71
.
.
.
sin 2
1
.
2
Since both sides reduce to the same expression,
=
t
cos 2
1
~-
the identity
>
cos
.
,
sin 4
is
established.
t
EXERCISE 3-2
Prove each of the following
-
.
-
.
1. sin
cos
t
tan
identities:
~
0.
=
sin
rt
,
2.
H
esc
sin
sec
t
3.
:
cos
5. sin
9.
+
13.
I
-
(1
esc
sin
)
+
= cos 2 L= cos 2
tan
+
t
14.
2
J
2
J
1
1
^
sec 2
t
-
tan 2
=
27. 1 -f tan
esc 2 <(sec 8
2 <
=
(sec
rinf-coBJ
+
1
cos
-
1
-
J
cos
=
t
1
2 sin
1
2
cos
2
J.
2 1
J
2
1
1
2
J.
t
2
2
J.
2
1
2
1
<
t.
2
1
J
+ cos
t
i
J
2
1
4
J
t
t
4
J.
*.
t)
1
J
.
^
2.
1
t
^
cos
^
t
.
t.
+ cot O = sec esc
- sin = tan
sin t(l + tan 2
28.
^
32.
2
*
2
2
1
<
cosM
sin3
sin
:
1
3-4.
2
...
+ tan = sin
4- cos
26. (tan
.
sin
t
2
l)csc i.
cot 2
f.
=
t)
-
J
t
;
.
sec
2
cos
sin
1
cos 2
t
2 J
-I-
__ sin
~~
1).
tan|-l
tan + 1
t
cos 2
2
esc
J
1.
= cos
(csc
= 1.
cos
tan
cos
- 1) (sec + 1) = tan 2
(sec
2
= sec
sec
sin tan
-f cos
2
= tan
sin
sec
esc
esc 2
+ sec = sec esc
- cos 4 = sin - cos
sin
tan
+ cot 2 = sec csc ? -
sin
t
sec
1
-
,
Ojl
31. cot 2
22.
t.
^ sm
24.
^
20.
J.
1
1
cos
sin
18.
*
2^
29.
16.
.
J.
1
.
25. sec 2
12.
^.
cot
4
cos 2
10.
1.
2
tan
-
=
=
.
.
,
1
esc 2 1
t
t
8. sin
.
t)
cos
sec
6. (sin
t.
2
tan
cos 2
+
cos
1 H-
= 2 - cos sec
= sec esc
- sec = tan 4 + tan 2
sec
sm tan =
Sec
2
*
21.
=
=
=
t
(1
cos 2
A
4.
esc
-
sin
-~
-
15. sin
19.
+
2 tan
tan 2 J(cot 2
11. (1
17.
t
(cot
7. sec 2
~
0.
=
;
esc
t
"
1+
-f
t
cos
f*os
^
+ cos
i
.
.
=
=
x
^
i
(esc
t
-
.
3
1
cos
J.
^ ^^
f
cot
O2
TABLES OF TRIGONOMETRIC FUNCTIONS
Exact numerical values of trigonometric functions in general
cannot be found. However, by use of methods beyond the scope of
this book, the values can be computed to as many decimal places as
desired. The results of such computations have been tabulated an<l
are included in this text in the form of tables of trigonometric
functions. Table I at the end of the text contains the values of the
The Tr/gonomefr/c Funcfions
Sec.
3-4
P(t)
FIG. 3-6.
^ t ^ ir/2.
six functions corresponding to numbers t such that
=
contains
table
since
1.5708
the
ir/2
Actually,
approximately,
values of
between
and
1.60.
be the point corresponding to a given value of
t, Fig. 3-6, and let ti denote the length of the shorter arc which
joins P(t) to the o>axis. In each case in Fig. 3-6, the point P(ti)
is located by measuring the arc t\ counterclockwise from the positive half of the #-axis. We shall call ti the reference number, or
^ t\ ^ rr/2. Since t v is a real
related number, for t. Note that
number, there is associated with it a point P(t\) = (#1, 2/1). Also,
since ti lies between
and 77/2, P(ti) must lie in the first quadrant.
In each case in Fig. 3-6 the coordinates of P(t^) must be numerthat is, \x\ = Xi and \y\ = y\. Since
ically equal to those of P(t)
are defined in terms of x and y, we
functions
all the trigonometric
Let P(t)
(x, y)
;
can say that
(3-17)
|
any function
of
=
t
j
same function
of
t\.
not have the same algebraic sign,
and all functions of t\ have
in any quadrant and the
lie
whereas
P(t) may
positive values,
functions of t do not necessarily have positive values. It is important to see that the proper sign is chosen to make the equation a
true one. The algebraic sign in each case depends on the quadrant
in which P(t) lies.
The following examples and Fig. 3-7 will illustrate the method
of reducing a function of any positive or negative t to the same
These functional values
may
since P(ti) lies in the first quadrant
f unctipn of the reference
We
number
ti.
our discussion for the present to direct use of
Table I and consider ipnly values of ti which are shown there. The
process of using the table for values of ti which are not shown will
be treated in Section 3-10 when we discuss interpolation. For
shall limit
simplicity at this time,
we
shall use the
approximate value
TT
=
3.14.
Sec.
3-4
Example 3-5. Reduce the functions of t
As shown
number t\ is
Solution:
reference
of 1.14
73
The Trigonomefric Functions
may
= 2 to functions of its reference number.
in Fig. 3-7(a), P(2) is in the second quadrant,
be found from Table
by noting that only the
Thus, we have
The numerical
2, or 1.14.
TT
I.
The
and the
values of the functions
signs of the functions of 2 are determined
in the second quadrant.
and cosecant are positive
sine
:
cos 2
sin 2
tan 2
cot 2
sec 2
esc 2
= - cos (T - 2) = - cos 1.14 = - 0.4176,
= sin (w - 2) = sin 1.14 = 0.9086,
= - tan (w - 2) = - tan 1.14 = - 2.176,
= - cot (T - 2) = - cot 1.14 = - 0.4596,
= - sec (T - 2) = - sec 1.14 = - 2.395,
= esc (TT - 2) = esc 1.14 = 1.101.
Example 3-6. Find tan
Solution:
As shown
Since the tangent
is
in Fig. 3-7(6), the reference
Solution:
=
Solution:
4
=
ir
=
.86.
1.162.
5.
Here, as shown in Fig. 3-7(c),
t\
=
2?r
5
=
1.28;
and cos 5
=
To
20.
locate the point P(20),
a positive direction from
revolution
is
.86
t\
= tan
0.2867.
Example 3-8. Find cot
circle in
number
positive in the third quadrant, tan 4
Example 3-7. Find cos
cos 1.28
4.
is 2ir
=
6.28, find
we
we must proceed 20 units around the
The number of units in one complete
(1, 0).
find that
20
= 3(6.28) +
1.16.
Therefore, to locate P(20), we must proceed three times around the unit circle and
then continue for 1.16 additional units in a counterclockwise direction. This means
that
*!
t
may
= =
t
be taken as
1.16.
From
1.16.
the table,
Since P(20) or P(1.16) lies in the first quadrant,
0.4356.
we have cot 20
cot 1.16
=
=
74
The Trigonometric Funcf/ons
Example 3-9. Find
sin
= -
Sec.
(-2).
=
shown in Fig. 3-8 that
IT -f 2| = 1.14. Hence, t\ is the same as t\ in Example
negative in the third quadrant, we have
For
Solution:
t
2,
it
is
i
= -
(-
2)
= -
sin 2
= -
sin 1.14
= -
- (-
|- *
Since sin
3-5.
|
sin
3-4
2)|
t
is
0.9086.
should be noted that in Fig. 3-8 the
points P(-t) andP(t) are located on opposite sides of the #-axis, and are joined by a
It
P(2)
segment which is bisected perpendicularly by the axis. Hence the coordinates of
the two points are numerically equal, but
the ^-coordinates have opposite signs.
Therefore, by the definitions of the funcline
tions given in Section 3-2,
sin
Hence,
(
cos
(
tan
(
if t
(3-18)
>
)
t)
= sin t,
= cos t,
= -tan t,
it
follows that
sec
(t) = esc t,
(t) = sec t,
cot
(
esc
t)
=
cot
t.
0,
(any function of (-t)
=
|
|same function of
t\.
However, the algebraic sign of the function is changed for all
functions except the cosine and the secant.
Because cos t and sec t remain unchanged when t is replaced by
its negative, these functions are called even functions. The remaining functions are called odd functions, since their values change
sign
when
t is
replaced by
its
negative.
EXERCISE 3-3
1.
Construct a figure, locating each of the following points.
its related number ti. (Use TT = 3.14).
Show
the point P(t)
and
a.
P(l).
2.
With the
e. P(- 4).
d. P(- 5).
c. P(10).
aid of Table I find each of the following values, using
b. P(3).
a. sin 1.45.
b. cos 3.5.
d. tan 5.
e. esc
g. sin 28.
h. cos 60.
(-
c.
2.41).
3-5.
POSITIVE
We
P(3/2).
= 3.14.
f.
cot
(-
i.
tan
(
-
4.50).
30).
227T
k. cot
sec
TT
sec 4.75.
Sir
j.
f.
1.
4
AND NEGATIVE ANGLES AND STANDARD
sin
POSITION
have defined each of the six trigonometric functions as a
relation between two sets of numbers, employing as the independent
Sec.
3-5
The Trigonometric Functions
number
whose absolute value
variable a real
t
represents the length
of an arc of a unit
circle.
Now we
return
to
the
shall
tradi-
viewpoint and
tional
consider trigonometric
functions of angles.
Although the student is probably familiar with the idea of
FIG. 3-9.
angle from the study of geometry, we shall try to make the definition
in a plane and draw the halfmore precise. Let us select a point
line or ray a emanating from O, as shown in Fig. 3-9. We shall
call
the vertex of the ray. Finally,
we
let
A
be a point on the ray
in its initial position.
Now
rotate the ray a about^O to some terminal position 6, so that
moves along the arc indicated by the curved arrow AB.
the point
A
The ray may be rotated
in the counterclockwise sense, as in Fig.
3-9 (a) or in the clockwise sense, as in view (b). Moreover, it may
be turned through one or more complete revolutions, as in view (c)
We shall speak of the position b as the terminal ray b. We have
,
.
then an ordered pair of half -lines consisting of the initial ray a and
the terminal ray b. We can now define an angle as follows
An angle is a geometric figure consisting of two ordered rays
:
emanating from a common vertex.
With each angle is associated a number, called the measure of the
angle, which indicates the sense and amount of rotation required
to turn from the initial ray of the angle to the terminal ray. This
usually represented graphically by a curved arrow. Its
evaluation will be considered in Section 3-6.
rotation
is
We may designate the angle in Fig. 3-9 as angle AOB; or we may
use a Greek letter, such as 0, $, a, /3, or y, as the designation. The
line OA is called the initial side of angle AOB f and OB is the
terminal side. Counterclockwise rotation, as in Fig. 3-9 (a) 01
3-9 (c), gives rise tb a positive angle, while clockwise rotation,
such as the one in Fig. 3-9(6), gives rise to a negative angle.
Finally, we shall say that an angle is in standard position with
respect to a rectangular coordinate system when its vertex is at the
origin and
its initial
side coincides with the positive #-axis.
See
The Trigonometric Functions
76
Sec.
3-5
III
FIG. 8-11.
FIG. 3-10.
Fig. 3-10. When an angle is placed in standard position, the
terminal side determines the quadrant to which an angle is said
to belong. Thus, angle XOP in Fig. 3-10 is positive because it is
generated in a counterclockwise direction, and is a second-quadrant
angle because the terminal side OP lies in the second quadrant.
We note that the definition of angle does not specify that the
rotation should stop at the first arrival at the terminal side OP,
Fig. 3-10. In fact, angles of any size may be generated, since any
number of angles which end at the terminal side OP of a given
angle may be obtained simply by adding a number of complete
rotations, positive or negative, to the given angle. For example, the
same terminal side may also be reached by rotation in the opposite
direction. All angles which are in standard position and have the
same terminal sides are called coterminal angles.
In Fig. 3-11 the angle a is generated by rotation of OX counterclockwise to the position OP. The angle /3, which is coterminal with
a, is generated by adding to a one complete rotation of OX. The
angle y is a negative angle, which is coterminal with a and is
generated by rotating OX in the clockwise direction to the
position OP.
3-6.
MEASUREMENT OF ANGLES
The problem of measuring an angle is
equivalent to that of finding the measure
of the associated arc. One should, therefore,
apply the discussion of Section 3-1
and construct a unit
circle as
shown
in
Fig. 3-12.
FIG. 3-12.
measuring the angle
Let
be an angle in standard position.
Since the initial and terminal sides of the
angle intersect the circle in the points P (0)
an(j P(t), respectively, the problem of
reduces to that of measuring the appropriate
$ec.
37
77
The Trigonometric Functions
arc length t. Thus, the measure of the angle can be found in terms
of a real number in any one of several ways, depending on the unit
of measure chosen.
We shall consider first the circular system, or natural system, of
measuring angles, which is used almost exclusively in the calculus
and its applications. Its fundamental unit is the radian. This unit
may be defined as follows
A radian is the measure of an angle which, if placed at the center
of a circle, intercepts an arc on the circumference equal in length
to the radius of the circle.
In Fig. 3-13 the angle AOB is
1 radian, and the length of the
subtended arc AB is equal to the
:
/
radius r.
If the circle selected for measuring a radian is a unit circle, we
have an alternate definition of a
radian. That is, a radian is an
angle which intercepts a unit arc
on a unit circle.
\
arc =: radius
m
o
radius
F IG
.
=r
A
3.43.
Another system of measuring angles is the sexagesimal system,
or degree system, which is commonly used in ordinary calculations
involving angles. The fundamental unit of this system is the degree.
In Section 3-7 we shall study various relations between radians
and degrees, and shall develop rules which allow us to convert
from one system to the other.
In discussing angles, we frequently use the term angle, in place
of measure of an angle, and we rely on the context to make the
meaning clear. Thus, when we say "0 = 2," we mean, "0 is an angle
whose measure is 2 radians." The word radian is usually omitted
when an angle is expressed in terms of radians.
3-7.
THE RELATION BETWEEN RADIANS
AND DEGREES
Since an arc that is equal in length to the radius of a circle subtends an angle of one radian at the center, it follows that the whold
circumference, which is 2rr times the radius, subtends an angle of
2?r
radians.
Furthermore, the whole circumference subtends a
central angle of
360. Therefore,
2?r radians
= 360,
and
TT
radians
=
180.
78
The Trigonometric Functions
If the
approximate value 3.1416
1
radian
180
=
=
7T
is
used for
Sec.
3-7
TT,
180
(approximately),
.
Q 1 1A
U.141D
or
1
radian
=
57.29578
1
radian
=
5717'45" (approximately).
(approximately),
or
Also,
1
=
make
In order to
=
T^T radians
0.01745329 radians (approximately).
the conversion to radians easier
we
is expressed in degrees, minutes, and seconds,
values :
1'
=
when
the angle
give the following
0.00029089 radians,
and
1"
=
0.00000485 radians.
Therefore, one of the following rules can be used to convert from
degrees to radians or from radians to degrees
To convert from degrees to radians, multiply the number of
:
degrees by
~
,
or 0.0174533.
ioU
To convert from radians
radians by
,
to degrees, multiply the
number
of
or 57.29578.
7T
Note, However, that certain angles are commonly expressed in
terms of TT radians, in order to avoid approximate values. For
example,
180
90
3-8.
=
=
TT
45
30
radians,
7T/2 radians,
=
=
7T/4 radians,
7T/6 radians.
ARC LENGTH AND AREA OF A SECTOR
In Fig. 3-14 is shown a circle of radius r. In such a circle an
angle at the center equal to one radian subtends an arc on the
circumference equal to r. Similarly, by the definition of a radian,
the number of units in the arc s intercepted by a central angle equal
to
radians is given by the relationship
Thus,
$
(3-19)
=
re,
or
FIG. 3-14.
arc
=
(radius)
(central angle expressed in radians)
.
Sec.
3-8
Now
79
The Trigonometric Functions
A
denote the area of the sector bounded by two radii
s. If
is the number of radians in the central
the
ratio of the area A of the sector to the
of
the
then
angle
sector,
to
area of the whole circle, or Trr2 equals the ratio of the angle
the angle in the whole circle, or 27r. That is,
let
and an arc of length
,
A=A
Trr
2
'
2?r
or
A =$'* 9
(3-20)
it
'
If the central angle of
an arc or a sector
must be re-expressed
in radians before
expressed in degrees,
(3-19) or (3-20) can
is
be applied.
Example 3-10. Express 210
=
Solution: Since 1
Thus, 210
=
-
^
in terms of
radians, 210
lo
loU
TT
radians.
~ =^
= 210
radians.
O
loU
radians.
Example 3-11. Express 1215'20"
in radians.
Solution: Multiply the decimal parts of a radian given in Section 3-7 for 1,
I" by 12, 15, and 20, respectively. The results are as follows:
1',
and
0' 0"
12
15' 0"
20"
15' 20"
12
Example 3-12. Express
Solution: Since
TT
Example 3-13. Express
.20943948 radians
.00436335 radians
.00009700 radians
.21389983 radians.
radians in degrees.
-^-
radians
=
=
=
=
=
180,
^6 radians = |o (180)
3.5 radians in degrees, minutes,
=
150.
and seconds.
Solution: First, convert the radians to degrees, as follows:
3.5 radians
To
by
To
by
find the
number
find thfe
60.
The
number
result
is
(60) (0.5352)
=
(57.29578)
minutes
200.5352.
decimal part of a degree, or 0.5352,
=
32.112'.
of seconds, multiply the decimal part of
0.112'
Hence, 3.5 radians
(3.5)
of minutes, multiply the
=
60. Thus, 0.5352
=
=
(60) (0.112) seconds
= 20032'6.72".
=
6.72*.
a minute, or 0.112,
80
The Trigonomefric Functions
Sec. 3-8.
Example 3-14. The radius of a circle is 5 inches. Find the length
by a central angle of 30.
of the arc of
the circle subtended
Since 30
Solution:
by
=
>
-^
the central angle
is
~-
Also, r
=
5.
Therefore,
(3-19),
=r-0=5--=j|
o
o
Example 3-15. In a
central angle
Solution:
is
By
circle of radius
(3.1416)
=
2.618 inches,
6 inches, what
is
the area of a sector whose
60?
(3-20), the area of the sector
A =
0(36)7,-
=
6?r
O
t
is
^r
&
2
0.
Since 6
=
60
=~
,
3
square inches.
EXERCISE 3-4
In each problem from
60.
2. 45.
1
to 25, express the given angle in radians.
1.
6.
11.
16.
150.
72.
283.
13.
30.
90.
215.
14.
10.
240.
196.
18.
3010'.
19.
4621 /
3.
7.
12.
8.
12.
20.
63.
17.
4.
9.
120.
330.
321.
5.
10.
15.
.
20. 23637'.
21. 8216'.
22. 6321'17".
23. 18357'43".
24. 39244'27".
25. 9331'38*.
In each problem from 26 to 40, express the given angle in degrees.
26. 7T/6.
31. T/12.
27. 7T/4.
28. Tr/8.
29. 37T/2.
32. Sir/18.
33. 7<jr/2.
34. 5ir/3.
39. 0.763 rad.
36. 3.7 rad.
30. 47T/5.
35. 3?r/20.
37. 8.21 rad.
38. 0.34 rad.
40. 0.8136 rad.
In each problem from 41 to 56, draw the given angle in standard position and
indicate its terminal side.
41.
30.
o
45.
22^-
-
42. 7T/4.
43. ir/3.
44.
46. 27T/3.
47. 170.
48. 17T/18.
50.
630.
51.
53. 77T/3.
54.
1000.
55. 97r/4.
57. In a circle of radius
4
49.
T/2.
360.
90.
52.
56.
- 47T.
- llTT/6.
feet, find the length of the arc intercepted by an angle of
radians.
Find
the
7ir/6
angle in radians that intercepts a 5-foot arc.
A
58.
central angle in a circle of radius 15 inches intercepts an arc of 5 inches.
Find the number of radians in the central angle. Express this angle in degrees
and minutes, rounding off the result to the nearest minute.
59.
A central angle of 6214' intercepts an arc of
16 inches on the circumference of
Find the radius of the circle.
60. Find the area of a circular sector whose radius is 7 inches and whose central
angle is o) 4 radians; 6) 75; c) 3 radians.
61. The area of a circular sector is 72 square inches. Find the
angle if the radius
is a) 6 inches; 6) 9 inches; c) 5 feet.
a
62.
circle.
The area
is a)
of
128;
a
circular sector is 126 square inches.
6) 1.6 radians; c)
30.
Find the radius
if
the anglfe
Sec.
3-9
3-9.
The Trigonometric Functions
81
TRIGONOMETRIC FUNCTIONS OF ANGLES
be an angle in standard position, as shown in Fig. 3-15.
associate a real number t, which is the measure of
the angle in radians. This concept is equivalent to our previous
concept of t, when t was interpreted as the length of an arc laid off
on the unit circle by starting at the point (1, 0) and terminating at
Let
With
6
we can
P(*,y)
FIG. 3-16.
FIG. 3-15.
the point P(t). Such an association of the angle 6 with the
directed length t of an arc of a unit circle allows us to define the
similar
cosine and sine of
as cos = cos t and sin = sin t.
procedure may be followed for the other functions of 0.
A
Now consider Fig. 3-16, where we show an angle in standard
position and a unit circle. By the definition of 0, the terminal side
of
intersects the unit circle at the point (cos 0, sin 0). This is, of
We
now extend
course, the point designated previously as P(t).
with coordinates
to an arbitrary point
the terminal side of
P
(x, y).
If
The length
we drop
(x,y) to the
Therefore,
of the radius vector
OP
is
r
= V#2 +
2
2/
*
perpendiculars from the points (cos 0, sin 0) and
the right triangles thus constructed are similar.
re-axis,
x
-
=
r
cos
,
and
:
y
-
=
sin
r
1
T1
Hence, the coordinates of the point P(x,y) on the terminal side are
x
=
r
and
cos
Using these results with the
y
=
r sin 0.
definitions of the functions
from
Section 3-2, we caii express the values of the six functions in
terms of x, y, and r. Thus,
sin
(3-21)
cos
tan
=
=
=
2//r,
esc
x/r,
sec
y/Xj
cot
=
=
=
r/y,
r/x,
x/y.
The Trigonometric Functions
82
3-10.
Sec.
3-10
TABLES OF NATURAL TRIGONOMETRIC FUNCTIONS OF ANGLES
Tables of natural trigonometric functions are so labeled to distinguish them from tables of the logarithms of these functions.
Angles in Radians* In Section 3-4 Table
I
was used
to find values
of trigonometric functions of the type cos 2 or sin 27T/3.
On
the
basis of the definitions of the functions of an angle given in Section
3-9, Table I
may
also be used to find the functions of angles
meas-
ured in radians.
Example 3-16. Find the cosine
Solution:
From Table
of
cos 1.43
I,
an angle of 1.43 radians.
= 0.1403.
Angles in Degrees. Table II at the end of this text contains the
approximate values of the six functions of acute angles expressed
in degrees and minutes. It is a four-place table of the functions of
angles at intervals of 10 minutes.
To find the value of a function of an angle between
and 45,
first locate the angle in one of the columns at the left, and then
same line in the column headed by the
name of the desired function. For an angle between 45 p and 90,
locate the angle in a column at the right, and then look for the
value on the same line in the column with the name of the desired
look for the value on the
function at
Table
II
its foot.
should be referred to in working through these illustra-
tive examples.
Example 3-17. Find
sin 3240'.
and 45. Look in the left-hand column to find
Solution: This angle is between
3240', and then go to the right to the column headed sin. There find 0.5398. Hence,
sin
3240'
=
0.5398.
Example 3-18. Find cos 5620'.
Solution: This angle
is
between 45 and 90. So look in the right-hand column
5620' is above 5600', and then go to the left to the
to find 5620', noting that
column with
cos at its foot.
Thus, cos 5620'
The following examples
=
0.5544.
procedure for finding an
angle corresponding to a given value of a function.
illustrate the
83
The Trigonometric Functions
Sec. 3f-10
Example 3-19. Given tan
Solution: Since tan
is
= 4.511,
6
greater than
find
is
1,
through the columns marked tan at the foot
^ 90.
assuming that
6,
greater than 45. Therefore, search
for the given number 4.511. The
corresponding angle in the right-hand column
7730'.
or
is
7730'.
So 4.511
= tan
=
Example 3-20. Given cos
Solution:
6
=
By looking through
0.8660, find
6,
^
assuming that
7730',
^ 90.
the columns with cos at either the head or the foot,
column headed cos, use the left-hand column for
find 0,8660. Since this value is in a
the corresponding angle, which
is
30. Hence,
6
=
30.
When
either the given angle or the given value of
not printed in the table, we can find the desired value
Interpolation.
a function
is
or angle by using a method of approximation known as interpolation. We assume that the change in the value of the function is
directly proportional to the change in the angle.
Although this
assumption is not strictly valid, it gives values that are accurate
enough for many practical purposes if we limit its use to small
changes in the angle.
The process of direct interpolation is used if the angle is given
and we need to find the value of either an increasing function of
the angle, such as the sine, or a decreasing function, such as the
cosine. Inverse interpolation is used when the value of a trigonometric function is known and the angle is to be found.
Example 3-21. Find
Solution:
1820'.
sin 1812'.
This angle
From
the table
is
we
not listed in the table, but
find that
sin
1810'
=
sin
1820'
= 0.3145.
it lies
between 189 10' and
0.3118,
and
The desired value of sin 1812'
The tabular difference, that
will
is,
then
lie
between 0.3118 and 0.3145.
the difference between the two values listed in
the table, is 0.0027. Also, the difference between the angles 1810' and 1820' is
10', while the angle 1812' differs from 1810' by 2'. Since the change in the angle
from 1810' to 1812' is 2/10 of the change from 1810' to 1820', we assume that
=
the corresponding change in the value of the sine will be (0,2) (0.0027)
0.0005,
and the amount to be added to 0.3118 is 0.0005. Hence, sin 1812' =0.3123.
The accompanying diagrammatic arrangement
tabular form:
o
sin 18 10
'
presents this
= 0.3118
\
f
2
10
[
same operation
sin
1812'
sin
1820'
= 0.3118 + x
= 0.3145
1*
J
0.0027
in
The Trigonometric Functions
84
Sec.
3-10
7
Since the angle 1812' is 2/10 of the way from 1810' to 1820 the corresponding
functional value will be 2/10 of the way from 0.3118 to 0.3145. Therefore,
,
10
0.0027
r
x
This amount
and
sin
to be
is
1812'
=
(0.2) (0.0027)
=
0.0005.
added to 0.3118. Hence, the value
of the function is 0.3123,
= 0.3123.
Example 3-22. Find
cos 7348'.
Solution: The process is similar to that in Example 3-21. However, since the
cosine decreases as the angle increases, we subtract 8/10 of the tabular difference
from cos 7340'.
We find
the values of cos 7340' and cos 7350' in a column of the
The work may be
table labeled cos at the bottom.
cos
10
x
7340'
= 0.2812
cos 7348'
= 0.2812 -
cos 7350'
= 0.2784
=
,
or x
=
x
x
0.0028
(0.8) (0.0028)
10
0.0028
indicated as follows:
Hence, the amount to be subtracted from 0.2812
is
= 0.0022.
0.0022,
and cos 7348'
=
0.2790,
The inverse process of finding the angle when the given value of
a function
not printed in the table is performed in a similar
Here, since we know the value of the function, we find
the two values in the table nearest the given value, one less than
is
fashion.
and one greater. Again making the assumption that small
changes in the value of the function are proportional to small
changes in the angle, we proceed as indicated in the following
example.
it
Example 3-23. Find
Solution:
/
.
cot 6
= 0.8780.
This value of the cotangent
entries 0.8796
and 4850
if
We have,
10
= 0.8796
{
cot 48
(40
cot 4850'
*
= 48*43'.
lies
To
1U
+ xY = 0.8780
= 0.8744
= 16
W
5J
,
'
and x
=:
between the
respectively, the angles
therefore, the following tabulation:
cot 4840'
x
Hence, 9
not in the table but
is
and 0.8744. To these correspond,
\
0.0016
0.0052
4840'
Sec.
3-10
The Trigonometric Functions
85
EXERCISE 3-5
In each of the problems from 1 to 30, use Table II to find the value of the given
function. Interpolate whenever necessary.
2. cot 12840'.
1. sin 3620'.
3. sec 2340'.
4. cos
9650'.
5. sin
13210'.
7. esc
22330'.
8. sec
3930
10. sin 9840'.
.
11. cos 75CO'.
tan
cot 28350'.
/
(- 13330').
tan 62340'.
(-
14. esc
17. cot 5543'.
18. esc
19. sin
5732'.
20. cot 31G'.
21. sin
22. cot 2801'.
25. cos
28. sin
((-
23. tan
15.
).
/
271G
31237
2S10').
12. cot
(- 41620
13. sec
39210').
16. tan 29852'.
(-
6.
9.
/
24. esc
.
/
7258').
26. sin
1647').
29. esc 289OG'.
(- 4451').
(- 28033').
(- 24529').
27. tan 63602'.
.
*
30. cos 12619'.
In each of the problems from 31 to 60, use Table II to f.nd the values of between
and 360 which satisfy the given equation. Express the results to the nearest
minute, interpolating whenever necessary.
31. tan
34. cos
37. sin
40. cos
43. tan
46. sin
49. ccs
52. cot
55. cot
58. cot
= - 0.11C8.
= 0.7951.
= 0.5783.
= - 0.4147.
= 8.345.
= O.G702
= 0.9503.
= - 1.381.
= 7.COO.
= 0.1340.
32. sin
35. tan
38. cot
41.
shT0
44. cot
47. tan
50. cos
53. cot
56. tan
59. tan
= 0.3062.
= 0.0553.
= - O.G494.
= - 0.9959.
= - 0.3121.
= 0.9043.
= - 0.5090.
= 0.4230.
= - 0.1191.
= - LS.CO.
33. cot
36. sin
39. tan
42. cot
45. tan
48. cot
51. cos
54. sin
57. cos
60. tan
= 1.091.
= 0.2419.
= 1.511.
= 0.0437.
= 1.446.
= 2.398.
= 0.8519.
= 0.2491.
= - 0.1323.
= 3.235.
In each of the problems irom Gl to 72, find the value of the given function.
Interpolate whenever necessary.
Take
61. sin 0.93.
C?. cot 2.46.
63. sec
64. tan 8.71.
C5. esc 9.43.
66. cot 0.678.
67. tan 0.333.
68. cot
70. sin
- -?}
(\
6 /
TT
as 3.14.
(
1)
71. esc 0.968.
(-
69. cos
^
72. cot
(-
1.24).
0.643).
In each of the problems from 73 to 84, find the values of 0, in radians, between
2ir \\hich satisfy the given equation. Use Table I and express the results to
and
three decimal places, interpolating whenever necessary.
*
0.9759.
74. sin
0.9967.
73. cos
76. sec
79. cos
82. cot
=
= 2.563.
= 0.4010.
= 0.39.
77. tan
80. tan
83. cos
=
= 0.9413.
= 1.6.
= 0.84.
75. tan
78. sin
81. sin
84. cot
= 2.066.
= - 0.736$
= 0.91.
= 1.031.
4
4-1.
The Laws of Exponents
POSITIVE INTEGRAL EXPONENTS
When
studying the progress of algebra up to the sixteenth
one cannot help but be perplexed by either the total
absence of symbolism or, when present, the lack of uniformity in
its use. At first, unknown quantities were often represented by
words. Later, symbols made from abbreviations and initial letters
of these words were used to indicate mathematical concepts, such
as number, power, and square.
Descartes (1637) is generally credited with our present system
of exponents. He introduced the Hindu-Arabic numerals as expo4
3
nents, using the notations a, aa [sic] a a etc. The writing of a
repeated letter for the second power of the unknown continued for
century,
,
,
,
many years.
Laws for positive
integral exponents were introduced in Section
shall now establish these laws and extend
1-11, without proofs.
them to apply also to zero, negative, and fractional exponents.
recall that if n is any positive integer, a n means the product
of n factors each equal to a. In this notation, a is the base and n is
the exponent or power.
shall proceed to establish the following
laws for positive integral exponents.
We
We
We
Law
of Multiplication. If a is a real number, and
positive integers,
a m a n = a m+ n
(4-1)
if ra
and n are
.
Proof. Proof of this relationship follows from the definition of an
and the associative law for multiplication. Thus,
am = a a
a
m
(to
factors),
and
an
=
a
a
a
(to
n
factors).
Hence,
am a n
=
=
=
[a
a
m factors)] [a
m
+ n factors)
cr(to
a
a
a m+n
a(to
a
a(to
.
For example, x 3 x 5
=
a;
8
,
and y k yk+3
86
=
y
2k + 3
.
n
factors)]
Sec.
87
The Laws of Exponents
4*1
Law of Division. If a is a non-zero real number, and
> n, then
are positive integers such that
= m~ n
if
m and n
m
(4-2)f
If
a
an
\
a
T^ 0,
and
if
.
n > m, then
nm
(4-3)
a
1
n
a
n ~~ m
Proof. Proofs of these relationships follow
If
n is positive. By (4-1)
> n, then
~
a m n a n = 0(w n)+n = o, m
m
:
m
an
,
.
we have
am
mn
=a
n
Hence, dividing both sides by
,
.
a
For example,
37
~5
If
m < n, then n m
Divide both sides by a
35
an
~
=
a w+(n
=
-
(
an
.
an
a n-m
sides by a n , we have
n
n
= ( an- m \) // a n = a n n1~m
/
/
\a
a
a
Now, dividing both
an
(4-1),
~ m)
to obtain
am
am
By
is positive.
a
n~m
-
>
-
I
=
an
~
For example,
37
Law
n are
for a
Power of a Power.
a
If
_
32
'
a real number, and
is
if
m and
positive integers, then
(4-4)
(a
m
=
n
)
amn
.
Proof. This relationship can be easily proved as follows
By the associative laws for multiplication and addition, the law
of multiplication expressed by (4-1) can be extended to three or
:
more
factors.
Thus,
0/n
A
.
gn
QP
.
m
rr
(a
=.
a m+n
an)
Q?
ap
similar relationship can be written for any
That
is,
(a
m
-
)
(of)
(a
r
)
=
number
of factors.
88
The Laws of Exponents
We may now take m = p =
(a
m
(a
)
m
= r to
(a
)
m
)
(to
Sec.
get
n
factors)
=
amn
=
am+m+
+
=
a "*.
1
Hence,
(a
w
w
)
.
For example,
(z
Law
and
2 3
)
=
x 6 and (2 2 ** 1 ) 5
,
Power of a Product.
for a
=
2 10 *+ 5
.
a and 6 are real numbers,
If
m and n are positive integers, then
if
n
(4-5)
(ab)
=
anb n
.
Proof. In proving this relationship, we make use of the associative and commutative laws of multiplication. Thus,
n
(ab)
=
=
=
(ab)
[a
(ab)
.....
(ab) (to
n
a ..... a(to n factors)]
factors)
b ..... b(to
[6
n
factors)]
n n
a b
.
For example,
Law
6 T^
0,
for a
and
Proof.
if
Power of a Quotient. If a and
n is a positive integer, then
By applying
(a\
(b)
n
a
6 are real numbers, if
the law for multiplying fractions,
an
a ,,
a
f
N
=b'b
.....
we have
,
^ (ton
factors)
=^-
For example,
2
/3xy^\
\ 22 /
_
_3^y^
"
4
9o;
2
""
2
2
2fe
i/
4
An
exponent affects only that quantity to which it is attached.
2
2
2
Q
Thus, -5x(y*) = -5xy , whereas (~5xy*) = 25x y*.
So far we have defined a n only when n is a positive integer. We
shall
now introduce zero, negative integer* and rational powers in
way that they will obey the same laws which were proved
such a
for positive integral exponents.
4-2.
MEANING OF o
We shall define the zero exponent by the equation
o
(4-7)
A few illustrations are
=
1
(a 7* 0).
:
=
5,
(o
-
to)'
a
1,
Sec.
4-2
If
The Laws of Exponents
a in (4-2)
is
89
m = n, we get
not zero and
an
In this case, the quotient on the
left
equals
1,
while the value of the
term on the right is a. Since a
1, by definition, the law of
division holds for n = m, as well as for m > n and n > m.
The student should note that (4-7) gives the only possible definition of a if the law expressed by (4-2) and (4-3) is to hold for the
zero exponent, as can be seen from the foregoing discussion.
We shall show that the definition a = 1 is consistent with the
five laws of exponents in Section 4-1 that is, we shall show that
;
when any exponent
these laws also hold
is zero. In the following
explanations, where a quantity occurs in a denominator, we assume
that it is not zero. Also, the exponents are assumed to be non-
negative integers.
Let us, for sake of discussion, suppose that n
a m~
'
Then we have
.
OP
a
.
a m+n
=
an
am
=
=
m+0
= am
=a
-
.
ao
~*~
=
in (4-1), that
n
.
a.
.
-_
x
am
=
.
^
Hence, the law of multiplication holds when n =
0.
procedure will verify the law if
Now let us suppose that n = in (4-2), that is, in
A
0.
similar
m
1
=
a
=
fr
=
am
~n
am
and
.
=
~
am
=
aw
.
1
Hence, (4-2) holds when n = in (4-3) we have
If
m
0.
,
=
""
an
=:
~~
an
an
and
,
~~
an m
=
~~
an
~
=
"~
an
m
It is clear that in (4-2)
cannot be zero, and in (4-3) n cannot
be zero. Therefore, the law of division holds.
Suppose that n = in (4-4) that is, in
,
(a
m
n
)
=
a mn
.
Then
(a
If
m=
m
n
)
=
(a
n
)
=
m
)
=
1,
we have
n
n
(a) = l =
in (4-4)
w
(a
and a mn
=
a mt
=
a
=
1.
,
1,
and amn
= a' n =
Hence, the law for a power of a power holds.
a
=
1.
The Laws of Exponents
90
Now consider
(4-5), which
n
n=
4-2
is
(ab)
If
Sec.
=
a nb n
.
0,
(a6)
n
we
Finally,
=
(a&)
let
n=
=
anb n
and
1,
in (4-6), that
/a\ n __ a^
-
W
Then
in
"
a"
,
and
1>
is,
1.
6
"
t
= a6 = 1-1 =
a
=
1
=
= L
t
The demonstrations just given prove that the five laws of exponents, originally stated for positive integral powers, are true for
all non-negative integral powers, and that the law of division is
true even
4-3.
when
the exponents are equal.
NEGATIVE EXPONENTS
In order to extend the meaning of exponents to negative integers,
define crn by the following relationship
we
:
or*
(4-8)
=
1
(a*
0),
where n
is a positive integer.
Several illustrations are:
(
v
a
K-2
5
~
fa.)-2
'
=
52
~
-
6z)
(a
10- 3
25
-
in
1U 3
-
innn
1UUU
>
2
V
in Section 4-2, we shall show that our definition is consistent
with the five laws of exponents.
Let us first note that (4-8) is true even if n = or if n is a negative integer. If n = 0, then
As
a
_ a - - I-!-!.
-_()-!a0
1
j
If
n
=
p,
where p
is
fln
a positive integer, then cr n
=
ap
=
1
/
a
an
cr p
We shall use this result in the proofs that follow.
In order to extend (4-1),
n be a negative integer,
that a
0. Then
say,
let
m be a non-negative integer, and let
n=
p.
In this case
=
#m # n
=
dm
cpp
=
am
1
a*
=
am
a*
it is
also
assumed
Sec.
4-3
If
The Laws of Exponents
m S p, we have, by
am
a
(4-2) for non-negative exponents,
W
= - = am
/y
n
91
~~
=
p
am+(-p)
_.
am+n^
a"
Ifm<p, we have, by
(4-3) for non-negative exponents,
~~~
~~ /tW
A
similar demonstration establishes
m < 0, or in case m < 0, and
n<
in case
(4-1)
n^
and
0.
Proof of the extension of (4-2) rests on the validity of the law
of multiplication just established. If a =Q, and if
and n are
integers (positive, negative, or zero), then
am
= a m 1n = am or n = am~ n
a
an
m
.
Although (4-3) is now an immediate consequence, it is not really
needed, in view of the general validity of (4-8). The demonstration just given allows the law of division to be stated as a single
relationship as follows
:
^=
fjTn
(4-9)
aw
~n
(a 7* 0).
Thus, a single law applies, regardless of whether
m > n, n > m,
or
m = n, where m and n are arbitrary integers.
Now
consider the law for a power of a power.
^ 0. Then
0, while
p, where p
In (4-4)
let
m
n
Also,
a m-p
a mn
Hence, (4-4) holds in this case.
If
p, where p i^ 0, while
m
(a
m
n
(ar
)
p
Y
(
-.
a-mp
nS
=
=
a mp
we have
= a ~ p)n =
and a mn
0,
(
J
a~ pn
=
holds. If both m and n are negative, a similar proceand the extension of (4-4) holds.
To extend the law for a power of a product, let n = p in (4-5),
where p ^ 0. Then
Again (4-4)
dure
is
used,
n-
-
So (4-5)
_
_
*
is verified
nn--?-/)-
*
11
-
1
for negative integral values of n,
of a quotient, assume
To verify the exltended law for a power
that n = -p in (4-6), where p ^ 0. Then
n
h)
(a\
=
/a\~~ p
(h)
=
1
P
=
/a p
*/hP
~
bp
~P'
,
anci
an
fr
^
arp
IT*
=
bp
"
"5
The Laws of Exponents
92
Sec.
4-3
Hence, (4-6) holds for negative integral values of n.
Thus, the laws of exponents hold for positive integral exponents,
zero exponents, and negative integral exponents. In Section 4-5
we shall
From
consider the case of fractional exponents.
the general validity of (4-8), it follows immediately that
a factor of the numerator or the denominator of a fraction can be
moved from the numerator to the denominator, or vice versa, provided only that we change the sign of its exponent. For example,
a2
x3
x^z 2
a 2 b~ 3 = TQ
and
o =
SCIENTIFIC
NOTATION
We are now in a position to introduce certain simplifications
when operating with very large or very small numbers, as are
customarily used in scientific writing. Any positive number that is
greater than 10 or less than 1 may be written compactly by
expressing it in standard form, that is, by writing it as a number
that lies between 1 and 10 multiplied by a suitable positive or nega4
tive integral power of 10. Thus, 27,000 would be written 2.7 10
4
Similarly, 0.00031 would be 3.1 10~
.
.
Example 4-1. The speed
number in scientific notation.
Solution:
of light is 186,000 miles per second.
The given number 186,000 may be
10 5
written as 1.86
Express this
.
mass of an electron is 9.11 10~ 28 grams. How many
zeros would he required between the decimal point and the first non-zero digit, 9, if
Example 4-2. The
the
rest
number were written
Solution:
in decimal notation?
28 means that we would have to move the decimal
its present position. We would thus have to place
The exponent
point 28 places to the left from
27 zeros to the left of the 9.
Example 4-3.
If
the sun
light to reach the earth
Solution:
As given
10 7 miles from the earth,
9.3
is
in
Example
4-1, the speed of light
9 3 . iQ7
4-5.
long does
it
take
is
1.86* 10 5 miles per
500 seconds
=
meaning of exponents from integers
to
the required time
8 minutes 20 seconds.
second.
how
from the sun?
Therefore,
is
i
=
"b
fi
.
i?^
5
10 2
=
RATIONAL EXPONENTS
We
shall
now extend
rational numbers.
the
Here again we
shall
make
the extension in such
Sec.
45
93
The Laws of Exponents
a way that the laws for positive integral exponents will be
preserved.
Suppose that a is a real number and that n is a positive integer.
Let us assume that a 1/n has meaning and that (4-4) applies. Then
it would be true that
1
(4-10)
(a /")"
=
a'
1
/^ =
=
a1
a.
This says that the nth power of a l/n would have to be a, or in other
words that a l/n would be what is called an nth root of a. For
example, (4-10) would yield
(a
2
1 /2
)
=
a,
and
(a
1
/
3 3
)
=
a.
l/n
Real nth Roots of a. Before defining a
let us examine the situation with respect to the existence of nth roots of a given number
a. The following results may be proved with the help of the theory
of equations.
Case I. If n is an even integer and a is a positive real number,
there are two real numbers that satisfy the equation r n = a. One
,
is the positive nth root of a, which is denoted by tya. The
the negative nth root of a, which is denoted by
^/a. We
may also denote these two numbers together by ^/a.
Case II. If n is an even integer and a is a negative real number,
of these
other
is
no real nth roots exist, since no even power of a real number can
be negative.
Case III. If n is an odd integer and a is a positive real number,
there is one real (positive) value of r such that r n = a. In other
words, if n is odd, there is a real positive nth root, which is denoted
by^a.
Case IV. If n is an odd integer and a is a negative real number,
a. That is,
there exists one real (negative) value of r such that r n
if
n
is
odd, there
is
a real negative nth root, which
is
denoted by
y*.
Case V. If n is any positive integer and a is zero, there is only
one real nth root, and this root is zero.
Thus, the definition of an nth root of a is valid under all conditions except when a is negative and n is even. In this situation, no
real nth roots exist. (However, the introduction of complex numbers in Chapter 11 will allow us to eliminate this exception.) We are
now ready
for the following definition.
If a is a non-negative real number and n is a positive
lfn
a
designates the non-negative nth root of a, or ^/a. If a
integer,
is negative and n is an odd positive integer, then a 1/n designates the
real nth root of a, or ^/cT.
When a is negative and n is even, a 1/w is undefined:
Definition.
The Laws of Exponents
94
Sec.
4-5
Meaning of am/n. Let a be a given real number, n a positive
m/n has
meaning, and if (4-4) holds
integer, and m an integer. If a
m/n
= a (1/n)TO = (a 1/n ) m Under these
for fractional powers, then a
1/n
m/n would be the mth
It is
power of a
assumptions, then, a
.
.
natural to state the following definition.
is any integer such that
Definition. If n is a positive integer, if
a
in
and
if
is a real number which
the fraction m/n is
lowest terms,
m/n
n
is even, then a
is assumed to be non-negative when
designates
1/n
the mth power of a , that is, the mth power of ^/a. Hence,
1/n m
amln =
m
(a
(4-11)
.
)
If the fraction m/n is not in its lowest terms, it is first reduced to
lowest terms, and (4-11) is then applied.
When a is given, the value of am/n depends only on the value of
and n.
the fractional exponent, not on the particular values of
m
Thus,
2 4/2
=
22
=
26
4,
'
8
=
23 '4
and
,
(-
2)
2/6
= (-
2)
1 /3
=
^^2.
In the last example it would be incorrect to apply (4-11) directly,
1/6
has no meaning.
(
2)
It may be shown that, if a is positive,
since
amfn
(4-12)
The proof
We
is
=
<\/a.
omitted.
omit the details of the procedure for showing that
the five laws of exponents hold for rational exponents and nonnegative bases. The reader is cautioned against using the laws for
negative bases, since some fail under certain conditions.
To summarize the results now established, we restate the laws of
exponents here for easy reference. It is assumed that a and b are
and n are rational numbers.
non-negative real numbers, and that
Furthermore, if either a or 6 appears in a denominator or raised
to a negative or zero power, it is assumed to be different from zero.
shall also
m
Law
Law
m
of multiplication: a
of division:
Law for a power
Law for a power
am
an
=
an
~
am n
=
a m+n
.
.
= awn
n n
n
(ab) = a b
m n
of a power: (a )
of a product:
.
n
-J
(a\
Law for
reciprocal:
Zero power: a
=
ar n
1.
=
an
5=
an
7-
.
Sec.
4-5
95
The tows of Exponents
Note that the radical notation can be replaced by the simpler
and much more convenient exponential form. Everything that can
be done with the radical notation in the simplification of roots of
numbers and in operations involving roots can be done much more
naturally by means of the exponential notation. A few illustrations
of the meanings and uses of exponential forms follow
:
x
1'2
32 2 / 5
=
(\X32)
= Vx,
(32)
2
=
1 /5
22
if
^
*
= =
(-
0;
X/32
27)
= -
1/3
2;
= \/^~W = - -tf/5 5 = - a 2
3;
;
-
4;
4
4x 2
The following examples illustrate the applications of the laws of
exponents to the solution of problems involving radicals.
(16s
Example 4-4. Compute the value
of
\/2
\/2,
1 '2
yi6x*
)
and write the
result in expo-
nential form.
Solution:
Using exponential notation and the laws of exponents, we have
2i/3
.
Example 4-5. Remove
Solution:
=
21/4
all
2 3/12
2 4 /*
=
2 4/12+3/12
=
2 7/12
possible factors from the radical
We may proceed as follows:
= (2 3 4 z 4 2
2/
=
Example 4-6. Use the laws
1/3
)
=
2 1/3
2 1/3 3 1/3 x 1/3 2/ 2/3
3x
=
.
$
3 4/3
3
of exponents to express
^/y by using only
^/x
one radical.
Solution:
Changing to
we have
fractional exponents,
Example 4-7. Rationalize the denominator
in the fraction
-J7=f
Solution: To write an equivalent fraction in which no radical appears in the
denominator, we proceed as follows:
__
Example 4-8. Rationalize the denominator
Solution:
radical
We
use the relationship (a
+ b)
-=
of the fraction
5
(a
b)
= a2
v3
b 2 to
from the denominator. Thus,
1
5
- V3
=
"
5
1
'
5,-
V3
5
+ \/3 =
" 5 + V3 = 5 + V3
22
25 - 3
+ V3
remove the
96
The Laws of Exponenfs
.
4-5
^2
-
/a?
Sec.
x2
+
2
__
Example 4-9. Change
-
to a simple fraction.
^
Solution: Write the expression in exponential form, as follows:
a'2
-
x2
Then, multiplying the main numerator and the main denominator
have
_
^
(a
Example 4-10. Express
(x
2
+
<r2
-
2
/
2
(a
2
-
.r
2
-
x 2 ) 112
,
3 '2
2
)
+ Zx 2 (x + a
3 /2
2
(a
^2
z2) 3
+a
by
2
)
1 '2
2
)
in a factored form,
Solution: Rewrite the expression as
(x
Removing
+a
2
common
the
(x
2
+a
112
2
(x
)
factor (x 2
112
2
)
[x
2
+
+ a2 +
2
)
+
a2
a2) 1
-j-
/
2
,
3^ 2 ]
Zx 2 (x 2
+
a 2 ) 112
.
we obtain
=
(x
2
+a
1 /2
2
)
(4z
2
+
a 2 ).
EXERCISE 4-1
In each of the problems from
eliminate
all
zero
1
to 20, perform the indicated operations
and
and negative exponents.
3
1.
3x*y.
2.
ari".
3.
(jV
5-
Trb'
6-
UJ*
7.
(9)".
-
4.
10- 2
.
8.
9.
13.
(a;V)*(a:-V
16. x-i 4- y-1.
Lz^l!!!.
-
10
4 2
)
.
14. (x 1 '^- 8 ) 4
+ I/)-
17. (x
^-
1
-
1 '8
)
.
15. (x*' 2 -hi/ 1/2 ) 2 .
18.
(a;
+ 2/)- 1/3 (x + j/)i/.
;
20
Write each of the following expressions in exponential form. Remove all possible
from the radical and, wherever necessary, rationalize the denominator.
factors
21.
V80.
s- V* 2 -
Sec.
4-7
1)1/2.
4-^4.
-
49. (2
4-6.
97
The Laws of Exponents
x2) 5
/
2
2)2
+ s 2 (2 -
z2) 3
/
-
50. Or 2
2
.
3)
I/2
-
x*(x
2
-
3)-
1 '2
.
THE FACTORIAL SYMBOL
The product of
positive integers from 1 to n inclusive is called
"n factorial" or "factorial n" and is represented by either of the
symbols n or /.n. Thus, if n is a positive integer,
all
!
n\
=
1
2
-
l)-n.
1
2
3
r!
=
[(r
-2-3 .....
1
(n
For example,
3!
=
6!
4-7.
=
-
=
3
5!. 6;
j
=
5!
6;
=
71;
4
-
5
=
120;
l)!]r.
THE BINOMIAL THEOREM
The statement known as the binomial theorem enables us to
express any power of a binomial as a sum of terms without performing the multiplications.
By actually performing the indicated multiplications, we
+ 6) =
3
(a + b) =
4 =
(a + 6)
2
(a
These formulas
may
+ 2ab + 6 2
a + 3a
+ 3ab + 63
3
4
a + 4a 6 + 6a 2 6 2 + 4a& 3 + 6 4
a2
+
=
a2
(a
+ 6) =
a3
(a
+ *)4
2
&)
3
,
3
2
2
fe
,
:
+ ?a6 + ~fc
2
,
3
2
2
+ja 6 + j^|a6 + |ff^&
= a4 +
+ if
Applying this suggested rule to (a
(a
+
6)5
=
a*
+
.
be rewritten in the following manner, so as
to suggest a general rule 1
(a
find that
a4 6
+
a3 6 2
^
+
6)
+
S-4-8-2
1
2
3-4
we
5
,
,
obtain
f
a
4
5>4>3>2>1
^1.2-3-4-5
1 The
justification for writing the expressions on the right in this form will
be found in Chapter 17, where the binomial coefficients are "given in terms of
the combination formulas.
The Laws of Exponents
98
3
10a b
we
simplification of coefficients,
Upon
+
2
2
10a 6
3
-I-
5ab*
+
ft
which
3
,
obtained by multiplying (a
+
&)
4
Sec.
get (a
by (a +
6)
b)
same
the
is
+
5
= a5 +
result
47
+
5a4 b
that
as
.
Each of the expressions on the left is of the form (a + b) n in
which the exponents 2, 3, and 4 of (a + 6) are special values of n.
If we let n denote the exponent of (a + 6) in each of the expres,
sions on the
n
+
we
note that the expansion of (a
terms with the following properties
1
In any term the
1.
the
left,
first
term
sum
an and the
is
4-
b)
term
is
bn
contains
:
of the exponents of a and b
last
n
is n.
Also,
.
The exponent of a decreases by 1, and the exponent of 6
increases by 1, from term to term.
3. The denominator of the coefficient in each term is the fac2.
exponent of b in that term.
4. The numerator of the coefficient in each term has the same
number of factors as the denominator. Specifically, wherever 1
torial of the
appears in the denominator, write n directly above it in the
1
numerator wherever 2 appears in the denominator, write n
;
directly above
the
it
in the
number above
1 is
numerator; and so on. Thus, in (a
5, and the number above 2 is 4.
Assuming that these properties hold for
values of n,
all
+
6)
5
,
positive integral
we have
(4-13)
(a
+ b) n =
a*
n(n
+ ^i a^b +
"^ a - b
2 2
z
i
1
*
^
*
O
This result is the binomial formula. So far we have verified this
formula only for n = 2, 3, 4, and 5. In Chapter 16, we shall prove
the binomial theorem, which states that the formula is true for all
positive integral values of n.
The following example shows the procedure for the expansion of
Example 4-11. Expand
Solution:
and n
(x
-
6
2?/)
by the binomial theorem.
-- -- -
required expansion will be obtained by letting a = x, b
begin by setting up the following pattern of n + 1 terms
=
The
= 6. We
x6 +
(
+
-
2y)
x*(
+
x*(
4
2t/)
+
-
2y)
x(
2
+
2y)
x*(
+ ( 2).
2y)
:
2y,
Sec.
4-8
The exponents
numerators and denominators of the
simplifying,
(x
-
and
of b in the second, third, fourth,
fifth
Hence, remembering that the last term
respectively.
By
99
The Laws of Exponents
22/)
we obtain the
terms are
is
6n ,
1, 2, 3, 4,
we may
fill
and
5,
in the
coefficients as follows:
following result:
= z6
6
GENERAL TERM IN THE BINOMIAL EXPANSION
4-8.
we wish
any particular term of the expansion of
without considering any of the other terms, a study of the
binomial formula in (4-13) Section 4-7 will reveal the' following
If
(a
+
b)
facts
to write
n
:
In every term the exponent of 6 is one less than the number of
the term. Thus, in the (r-f l)th term, the exponent of 6 is r.
(The expression for a particular term is simplified slightly if the
number
of that term
is called
r
+
rather than
1,
The sum of the exponents of a and
the (r
+
l)th term the exponent of a
The denominator of the
since
it is
n
coefficient
+
l).
obtain, then, for the
(r
1) (n
-
n(n
r
(n
2)
-
1) (n
n
in each term.
+
(n
2)
For
r.
coefficient in the (r
tors as the denominator. In the (r
We
is
is
the factorial of the exponent of
The numerator of the
n(n
6
r.)
+
l)th term
is rl,
b.
has the same number of fac-
+
l)th term,
it is
the product
l)th term of the expansion
-r+
1)
alMftr
r!
Example 4-12. Find he
Solution:
6 is r
= 5.
=
Here a
sixth term of (3x
= -
3z, 6
y
2
,
and n
Hence, the exponent of a
8<7
z
'
6
Q
'
5
.
'
4
is
oxr(v
. \ox)
=
n
-
)
;
8
2
)
.
Since r
8.
5
if
y
=
=
3.
+
1
=
6,
the exponent of
Therefore, the sixth term
is
1
00
The laws of Exponents
Sec.
4-8
EXERCISE 4-2
In each of the problems from
1 to 16,
reduce the given fraction to lowest terms.
9.
In each of the problems from 18 to 32, expand the given expression by the
(Hint: In problems 29 thru 32, first consider the first two terms
binomial theorem.
in parentheses as a single quantity).
+ y)*.
18, (x
19. (x
-
I)
7.
20. (a
-
21. (2a*
26).
/ll/3\6
^r+V
(7/2
iC
30.
+ 2y +
(
2
)
.
31. (x*
+x +
I)*.
32. (o
-
362)3.
2. (*+
)
'
2r
-
a
-
I)
3
.
In each of the problems from 33 to 42, find the indicated term.
33. (1
-
35. (a
+ 26) 12
(x5
t
s)
8
,
2\
X/)
8th term.
,
5th term.
34.
36. (a
8
>
4th term.
+ n)",
(m
38. (x
- 6),
- 2/)
l
,
10th term.
3rd term.
term involving y 4
v*\ n
-- ~
]
(xy
41.
(V^ - vV) 12
,
middle term.
42. (2x
-
term involving
y
7
?/)
1
,
*
.
,
middle terms.
5
Logarithms
A LOGARITHM
assume here that the laws of exponents stated
DEFINITION OF
5-1.
We
shall
in
Chapter 4 for rational exponents ark valid also for irrational
exponents. The definition of a base raised to an irrational power is
beyond the scope of this book. However, let us make the assumption that, if b and x are real numbers, with b positive, a corresponding number designated by" b exists. Without giving an
x
let us assume that all laws of expoexplicit rule for computing b
nents established in Chapter ~4 are valid generally for real powers.
Finally, let us assume that, corresponding to any two positive real
numbers & and n, where b
1, there exists a unique real number x,
such that n = b x We can then give the following definition.
33
,
=
.
Definition.
from
x
=
1,
then x
log n.
&
If
n=
b x,
where 6 is a positive real number different
called the logarithm of n to the base b.
write
table
shows
both
of
forms
several
following
equiva-
We
is
The
lent statements.
We shall restrict n to positive numbers, since negative numbers
do not have real logarithms.
For any positive base &, we have 6 = 1 and 6 1 = 6. Hence, it follows from the definition of a logarithm that
'
log b 1
=
and
log& b
=
l.
f
Another valuable, relationship results from combining the two
equations n = b* and x = log & n. Replacing x in the first equation by
its value from the second equation, we have
For example, 2 10** 8 =
8,
and 10 10*i<> *
101
= x.
1
02
5-1. Find n,
Example
,
Sec. 5-1
Logarithms
if
= 2.
n
logs
Solution: Write the given equation in exponential form, as follows:
Hence, n
Find the base
5-2.
Example
= 32
n
= 9.
= 2/3.
4
6, if log*
.
Solution: In exponential form, the given equation
to the 3/2 power and recall that 6 > 0. Then
=
(52/3)3/2
Therefore,
is 6 2/3
=
4.
Raise both sides
= 43/2,
&
6=8.
Example 5-3. Find
= x.
32
x, if logi/ 8
we have
Solution: Writing the equation in exponential form,
Express 1/8 and 32 as powers of
(2-3)*
=
we have - 3x
From
this,
5-2.
LAWS OF LOGARITHMS
=
2
5
2~
or
,
= -
and x
5,
=
and get 1/8
2,
1/2
3J!
=
=
3
2
2~ 3 and 32
,
=
2 5 Hence,
.
5
.
5/3. Therefore, logus 32
=
5/3.
Since a logarithm is an exponent with respect to a given base,
the rules for operating with logarithms are the same as the laws
of exponents. These laws, expressed in terms of logarithms, have
the following form.
Law
I.
rithms of
The logarithm of a product equals the sum
The logarithmic form is
of the loga-
its factors.
(5-1)
log&
(m
n)
=
log&
Proof: To prove this equation,
x
logb m
=
m+
Iog6 n.
let
and
y
=
log& n.
n
=
Then
m=
Multiplying,
Hence,
bx
we have
and
bv
.
mn =b*+v.
.
(mn)
=
x
+y=
log&
m+
log& n.
Law II. The logarithm of a quotient equals the logarithm of the
dividend minus the logarithm of the divisor. The logarithmic
form
(6-2)
is
Iog6
=
Iog6
m-
Iog6 n.
Sec.
5-2
103
Logarithms
Proof: The proof follows Let
x = logb m
and
:
Then
m=
Dividing,
we have
,,
o*
_
Hence,
,
iog& (
\
Law
III.
m^ =
)
71
x
-
/
The logarithm
=
,
and
?i
=
b<-.
y
=
n
=
y
Iog6 w.
,
o^.
m
logb
logb n.
power of a number equals the expo-
of a
nent times the logarithm of the number that
is,
;
(5-3)
=
logo (n*)
Proof: The
first step in
* logb
n.
the proof is to
x = logb n.
Then
n
=
let
.
b*.
Raise both sides to the kth power and obtain
n k = (b x) k = b kx
.
This relationship, when written in logarithmic form, becomes
Iog6 (n
Replacing x by
its value,
k
)
we have
k
log& (n ) =
The student should note
and (log&n)
=
kx.
k Iog6 n.
k
carefully the difference between log& (n )
fc
.
Law IV. The logarithm of a root of a number equals the logarithm of the number divided by the index of the root that is,
;
\fn = ^
Iog 6
(5-4)
log b n.
Proof: This equation follows as a corollary of law
definition of a fractional exponent,
law
we have
= n l/k
<tyn
III.
.
tyn =
\
Example
A/51
5-4. Express loga
Solution: Iog 2
=
Iog 2
= Iog
Iog6 (n
2
-~^- as a
VSl S1^
+
1/fc
)
Iog6 n.
linear combination of logarithms.
= Iog
4
Iog 2 3
lo g2
=
17 1
/
2
-
2
(3
4
Iog2 3 .
17V'
-
the
Hence, by
III,
Iog6
By
Iog 2 3*
104
Sec.
Logarithms
Example 5-5. Express 2
Solution: 2 logio 3
-
=
J
logio 3
logio
x
-
~ logio $
+ logio y as a single logarithm.
+ logio y = logio 3 2 - logio 2 + logio
= logio (32 y) - logio x"*
a:
Example 5-6. Transform the equation
logo
5-2
+ y = Iog
x
1
'
sin x into
2/
an equation
free of logarithms.
By
Solution:
transposing,
y
=
we get
,
loga sin
*
-
=
,
logo
Change to the following exponential form
x
sin x
,
logo
---
:
sin
x
EXERCISE 5-1
In each of the problems from
1.
5.
9.
2 = 8.
10 3 = 1000.
100- 6 = 10.
3
=
2.
2
6.
10- 3
10.
y
1
to 12, write the equation in logarithmic form.
64.
3. 3 4
=
7.
0.001.
= e*.
=
81.
256 1/8
11. 10"
= 2.
= x.
=
4.
10
8.
216 l/3
12. 10
1.
= 6.
= x.
10 * *
In each of the problems from 13 to 21, write the equation in exponential form.
13. logs
16. log*
= 2.
64
= ggg
19. logio 10,000
4.
= 4.
14. logs
125
=
3.
15. Iog 2
^=
17. Iog 7
343
=
3.
18. Iog 9
729
20. logio 0.0001
In each of the problems from 22 to
22. Iog 9 3
25. log*
= Z.
= x.
x = -
29. log, 100
= x.
=
=
6.
3.
3/2.
33, find the indicated value of x.
26. logo.s
= 4.
31. logo 243
21. Iog 4 8
4.
24. Iog 4 x
23. Iog 2 64
4=2.
28. log* 81
= -
-
32. Iog 64
x
= -
= -
27. Iog 3 x
1.
30. log,
2.
~=
7
33. log* x
^
= 0.
= 1.
=
5.
2.
In each of the problems from 34 to 39, use the laws of logarithms to write the
expression as a single logarithm.
34.
191
ODD
logfr
2-3 log& 5 + log& 7.
36. \ log, 7
+ 1 Iog
38. 3 logs 2
+ logb 13-2
4
4
+ 1 log* 3.
log& 5.
35. log& 4 -f log&
37.
-
5 log 23
TT
+
-
log&
3+3 log&
2^
12 log* =f
&
r.
Sec.
5-4
39.
z
105
Logarithms
- Vu 2 -
logb (u
-
a2 )
&
logb (w
+ Vu 2
a2 )
+
log& a.
40. Find the logarithm to the base b of the area of a circle in terms of the logarithms
of
and the
TT
The time T
41.
where g
log?,
T
radius.
for a
pendulum
The area
Find
Iog6
a) (s
is
T =
IT
\/ -
>
where
s is
and
c is
given by the formula
the semi-perimeter
^ (a
mean G
of
n
positive
numbers
x\,x*,
b,
and
.
+6 + c).
r.
xn
is
defined
relationship
log*
5-3.
c),
6,
K in terms of the logarithms of combinations of a,
,
Show
oscillation
g.
b) (s
positive geometric
by the
make one
of a triangle with sides of length a,
Vs( s
The
to
a constant representing the acceleration due to gravity, a) Find
terms of the logarithms of IT, I, and g. b) Find log& I in terms of the
K=
43.
/
is
in
logarithms of w, T, and
42.
of length
that
G =
n =
G
\/XiX2
-
-
xn
.
SYSTEMS OF LOGARITHMS
As we mentioned
in Section 5-1, any positive number 6 different
be used as a base in a system of logarithms. However,
only two bases are widely used in practice.
The common, or Briggs, system of logarithms, named for Henry
Briggs (1556-1631), employs the base 10 and is used for ordinary
from
1
may
computations.
The natural, or Napierian, system of logarithms, named for
John Napier (1550-1617), is generally used in calculus and theoretical work, and employs the more convenient irrational base
=
2.71828
In this book, when the base is not indicated, it is understood to
be 10. Thus, log n means logic n, and the word logarithm will mean
common logarithm unless otherwise stated.
e
5-4.
In
COMMON LOGARITHMS
Table 5-1, we begin with
a list of powers of 10, give equivalent
from
these determine the form of the logalogarithmic forms, and
rithm of a number that is not an exact power of 10. It should be
mentioned that the logarithm is an increasing function; that is,
as n increases, log n increases. Another way of stating the conditions is to say that if a > b then log a > log 6.
106
See.
Logarithms
5-4
TABLE 5-1
From
true
Table 5-1
it
can be seen that the following statements are
:
The logarithm of an integral power of 10 is an integer.
The logarithm of a number which is not an integral power of 10
consists of two terms or parts an integral part, called the charac:
teristic;
and a positive or zero decimal
which
determined from a table of mantissas.
is
part, called the mantissa,
Thus, since log 10 = 1 and log 100 = 2, we may expect the logarithm of any number between 10 and 100, that is, a number between
IO 1 and IO 2 to be 1 plus a positive decimal part. For example, we
shall find that the logarithm of 35.4, which number lies between 10
,
and
100, is equal to 1.5490, to four decimal places. In this case, the
characteristic is 1
5-5.
A
and the mantissa
RULES FOR CHARACTERISTIC
is
.5490.
AND MANTISSA
study of Table 5-1 reveals that the characteristic changes as
the position of the decimal point changes in the sequence of digits
0035400. The first entry in the column headed "Logarithm of the
number"
is
log 354
=
2.
+
decimal.
Sec.
5-5
Logarithms
1
07
In the number 354, or 354.0, the decimal point is two places to the
right of the first non-zero digit, 3 (reading from left to right) ;
the corresponding characteristic is 2.
The second entry
is
log 35.4
=
1.
+
decimal.
In this number, 35.4, the decimal point
first non-zero digit (reading from
the
is
one place to the right of
left to right)
;
the corre-
sponding characteristic is 1.
Similarly, we note that the zero characteristic corresponds to the
position of the decimal point immediately following the first nonzero digit. This position of the decimal point is called the standard
We may now formulate the following rule for
position.
characteristics
:
Rule for Characteristics.
tion,
the characteristic
If
the decimal point is in standard posiFor every other position of the
zero.
is
decimal point, the characteristic is equal to the number of places
the decimal point has been shifted from the standard position. The
characteristic is positive if the shift is to the right, and is negative if the shift is to the left.
We
shall
now
same for
see that the mantissa remains the
all
numbers having the same sequence of digits. Let us again consider
the sequence of digits 0035400. Any number containing this
n
sequence can be written 3.54 10 where n is a positive or negative
integer or zero and depends on the position of the decimal point.
Suppose that we consider the form log 3.54 = 0.5490. Then the
,
logarithm of any number containing this sequence
log (3.54
10
n
)
= log 3.54 + log
= n + log 3.54
= n + 0.5490.
10
is
n
Thus, a shift of the decimal place in the number affects only the
characteristic n, and the mantissa remains the same for the same
sequence of digits.
EXERCISE 5-2
In each of the problems from
1
to 16, find the characteristic of the logarithm
of**
the given number.
1.
34.63.
5. 0.1340.
9. sin
6341'.
f
34630.
2. 3.463.
3.
6. 2637.
7. 0.00346.
10. 0.000001.
11. 378364.
14. sin 8453'.
15. cos 6143'.
4. 268.1.
8.
12.
tan 428'.
7 821
~~'
lUjUUU
'
13, cot 8113'.
16. sec
b
24
8'.
1
08
Sec.
Logarithms
5-5
In each of the problems from 17 to 24, place the decimal point in the sequence of
7314 corresponding to the given characteristic.
digits
17. 3.
18.
21. 6.
22.
5-6.
HOW
-
2.
19. 0.
5.
23.
-
20. 1.
3.
24-1.
TO WRITE LOGARITHMS
As
stated in Section 5-4, the mantissa of a logarithm is always
positive or zero, whereas the characteristic may be a positive or
negative integer or zero. A positive characteristic or a zero characteristic can readily be combined with a given mantissa. For
example, the logarithm of 354 is written 2.5490. But when the
characteristic is negative, say
fc, where 1 ^ k ^ 10, it is more convenient to write it in the form (10
k)
logarithm of 0.00354. The characteristic
Let us consider the
but the mantissa is
34- 0.5490.
regarded as positive. We could write log 0.00354 =
For convenience in computation, however, we write log 0.00354 in
the form
and so
(10
- 3) +
0.5490
-
10
=
10.
is
7.5490
3,
-
10,
or 17.5490
-
20,
on.
Note. It would be incorrect to write log 0.00354 =
3.5490, for
this notation means
3
0.5490 and would imply that the mantissa
is negative. To perform certain computations, it is convenient to
write the logarithm 7.5490 - 10 in the form -2.4510, which equals
2
0.4510. It is important to note that the decimal part of the
number
since
5-7.
it is
HOW
2.4510 is not the mantissa of the logarithm of 0.00354,
not positive.
TO USE A TABLE OF MANTISSAS
The following examples will illustrate the procedure in finding
the logarithm of a number with the aid of a table of mantissas. The
student should work through each example, determining the characteristic from the position of the decimal point in the number and
determining the mantissa by referring to Table III at the end of
this book.
Example 5-7. Find
Solution:
The
log 46.7.
characteristic
is -f 1.
To
find the mantissa, locate 46 in the
column in the table headed N, and then go to the right to the column headed 7.
Here we find the mantissa .6693. So the complete result is log 46.7 = 1.6693.
Interpolation. If the number consists of more than three digits,
the mantissa is found from Table III by means of interpolation.
Since the method of interpolation is the same as that described in
Section 3-10 for the table of trigonometric functions, there will be
no further discussion of
it
here.
Sec.
5-7
Find log 0.03426.
5-8.
Example
The
Solution:
characteristic is
The mantissa
2.
number 3426 has more than
since the
109
Logarithms
three digits. It
found by interpolation,
is
of the
lies
way between
the mantissas of 3420 and 3430, as shown in the accompanying tabulation:
Number
Mantissa
10
13
Since the difference between the mantissas of the two numbers in the table
we have
x
This
is
rounded
and the amount
off to 8,
Hence, the mantissa
is
=
is 13,
A
to be
added
=
.5348 and log 0.03426
to 0.5340
8.5348
-
is
given
by x =8.
10.
Finding Antilogarithms. The number which corresponds to a
given logarithm is called the antilogarithm. That is, if log n = x,
then n is the antilogarithm of x and is written antilog x.
5-9. Find n,
Example
if
log
n
=
1.8710.
Search through the body of Table III to locate the mantissa .8710.
the columns headed N and 3, is 743. Since the
Solution:
The corresponding number, from
characteristic
is 1,
n
=
Example 5-10. Find
To
We may
antilog 7.5349-10.
The mantissa
Solution:
.5353.
74.3.
.5349
is
not in Table III but
these correspond, respectively,
indicate the
work
in tabular
numbers whose
lies
between .5340 and
3420 and 3430.
digits are
form as follows:
Number
Mantissa
3420
.5340
9
3420
10
+x
.5349
3430
From
this,
we
see that
=
13
.5353
*
.
-JL
~
10
13
'
Therefore, x
6.9, or 7 after rounding off. Hence, the sequence of digits in the
desired number is 3427. Since the characteristic is - 3, the untilogarithm of
7.5349-10
is
0.003427.
110
Sec.
Logarithms
5-7
EXERCISE 5-3
In each of the problems from
number.
26. sin 1018'.
1 to 30, find
27. tan 4133'.
the
28. sec 6416'.
In each of the problems from 31 to
common
logarithm of the given
29. cos 8214'.
30. cob 3116'.
50, find the antilogarithm of the given
number.
31. 1.6665.
32. 4.4857.
33. 9.4183-10.
34. 0.0645.
35. 2.7024.
36. 7.7388-10.
37. 9.4409-20.
38. 6.3404-10.
39. 1.8401.
40. 3.9552-10.
41. 2.4658.
42. 1.9501.
43. 9.7367.
44. 4.9960-10.
45. 8.7863-10.
46. 9.8821-20.
47. 0.6584.
48. 3.0150.
49. 5.0300-10.
50. 0.1504.
Solve for x in each of the following equations:
= 4.
51. 10*
54. 10
57.
60.
-
1
= x.
55.
^/10
-
a;.
61. 10 1 -*
53. 10 2 *
2.019.
= x.
58. 10 1 314
^10*=
10-*' 2 = 0.0123.
5-8.
=
52. 10"
56.
=x.
10^
=
7.132.
= x.
= 0.003146.
10 *- 3 = 0.6735.
59. 10-*
= 0.2346.
62.
2
LOGARITHMIC COMPUTATION
The fundamental laws of logarithms given
in Section 5-2 are
applied in the following examples to illustrate the application of
logarithms to computation.
Example 5-11. Find the product
Solution: Let
x
=
(0.0246)
(1360). Then
log x
log 0.0246
(1360).
(0.0246)
=
log
+ log
0.0246= 8.3909-10
log 1360
logo;
=
=
=
Hence, by interpolation, we have x
3.1335
11.5244-10
1.5244.
= 33.45.
1360.
Sec.
5-8
Logarithms
Example 5-12. Evaluate
Solution: Let x
=
(0.506)-
=
-'
(0.506)
/
1
'
3
.
05(
L
logs =logl
1
log
1/3 log 0.506
log x
Therefore, x
=
1.255
by
1 1 1
.
1/3
-
Then
log (0.506)
1
'
3
= log 1 - (1/3) log 0.506
= log 1 - (1/3) (29.7042-30)
= log 1 - (9.9014-10).
= 10.0000-1Q
= 9.9Q14-1Q
= 0.0986.
interpolation.
Alternate Solution: Let Z
=
(0.506)
-1 /3
Then
.
- (1/3) log 0.506
= - (1/3) (29.7042-30)
= - (9.9014-10)
= - - 0.0986)
log x
s=
(
=
Therefore, x
=
1.255
by
0.0986.
interpolation.
n
i
r 10
^
*
Example 5-13. Evaluate
i
=P
(0-352) (1.74)2
-^0.00526
Solution: Let x denote the desired value.
Then
= log 0.352 + 2 log 1.74 - (1/3) log 0.00526.
We find that log 0.352 = 9.5465-10, log 1.74 = 0.2405, and log 0.00526 = 7.7210-10.
log x
log 0.352
2 log 1.74
log numerator
=
=
=
9.5465-10
(1/3) log 0.00526
0.4810
10.0275-10
= 10.0275-10
log denominator = 9.2403-10
log x = 0.7872.
Interpolating, we have x = 6.126.
log numerator
Example 5-14. Evaluate
Solution: Let x
=
( 7=7
Then
.
)
\174/
log
x
=
1.14 [log 253
log 253
log 174
-
Then
log x
Therefore, x
=
1.532.
=
-
log 174].
= 2.4031
= 2.2405
1626
1.14 (0.1626)
'
= 0.1854.
= (1/3) (27.7210-30)
= 9.2403-10.
112
Sec.
Logarithms
5-8
EXERCISE 5-4
In each of the problems from
1
to 30, perform the indicated computation using
logarithms.
1.
d*
2. (13.25) (26.80).
(3.142)(2.718).
29.34
4. (0.8134) 1/3 .
68^5*
6.
7.
\/(0.003468)
(16.83)
836
ft
fi
.
-
3
'*.
1
"
42,860
o
9.
J/ (8,321,000)
4
r
10. (4.313)(3,068)(0.000642).
(36,250)
11. (63.84)2(0.0134).
12. (8.364)(321.5)
13.
14.
15.
V(168.3)
(14.21).
-K-
42.63).
2
V(213.6) (43.98) 2
.
V(23,310) - (20,180)2.
(Hint: Factor the radicand.)
2
is
*O
567.87
3
1
*
5
7
19
8' 16
4
no>
4
(1,083) (0.0813)
31
1
19.
2
v/ (21.36) 3
3,642^
"/i
4/3188"
K 0.8103
20
32* 64'
3
38.63
f
4
21
A/ <8L68
V (8.013) (0.034)
(0.0
>
22. (1.08) 10 .
24. 3,648 (1.03) 3 6.
23. (1
25.
-
v
27.
(0.8123)-
3 '4
26. 0.083
.
0.08614.
29, [(3.864)-3.i3
31. If e
=
30.
2.718, find log
Problem
412 .
28. log 16.84
+
32. Find the geometric
'
e,
log <\A, log
mean
c
i
e',
and
-
(83.14)-*
^483.6.
-
(0.8134)
2 '*
IT*.
of 564.3, 8634, 0.1349, 8.316,
and
(Hint: See
42.61.
43, Exercise 5-1.)
33. Find the area of a circle of radius 6,381 feet.
34.
The volume
3621
of a sphere is
V =
35. Find the radius of a sphere
36.
5
irr*.
Find the volume of a sphere of radius
feet.
whose volume
is
8423 cubic
feet.
Find the length of a pendulum which makes one oscillation in
= 980 centimeters/sec 2 (Hint: See Problem 41, Exercise 5-1.)
g
.
1
second,
if
Sec.
5-9
113
Logarithms
37.
Find the area of a triangle with sides 6,384
(Hint: See Problem 42, Exercise 5-1.)
38.
The
stretch s of a wire of length
=
relationship s
>
|r
where g
I
is
5,680 feet, and 2,164 feet long.
feet,
and radius
r
by a weight
m
the gravitational constant
is
given by the
and k (Young's
modulus) is a constant for a given material. Find how much a copper wire of
length 120 centimeters and of radius 0.040 centimeters will be stretched by a
12
weight of 6,346 grams, if g = 980 and k is 1.2 10 for copper wire.
39.
The current
henrys
t
i
flowing in a series circuit with a resistance of
seconds after the source of electromotive force
RilL
given by the relationship i = Ie~
circuit before the short circuit. If i
seconds, and
40. If
n
is
L =
,
where /
=
is
is
R ohms
the current flowing in the
R = 0.1 ohms, t = 0.25
=
2.718.)
1*2
n.
However,
Stirling's
a positive integer, n\ has been defined as the product
it is difficult
to
L
10 amperes,
0.05 henrys, find /. (Take e
When n is very large,
and
short-circuited is
compute
this product.
formula gives (approximately) n\ = n n e~n \S2irn. Use this formula to estimate
91, and compare the result with the true value which you should calculate
exactly.
Do
-
-
log 2
+
the same with 30!.
log
TT
+
-
(Hint:
log (n!)
=
n
log
n - n
log e
+
log n).
41. If the rate of depreciation r per year is constant, the scrap value S after n years
n
Find the
of a machine with first cost C is given by the formula S
r)
C(l
of
value
10
a
machine
after
which
cost
scrap
$10,000, if 20
years
originally
=
per cent per year
5-9.
is
.
written off as depreciation.
CHANGE OF BASE
sometimes desirable to change from one logarithmic base to
another. Suppose there is available a table of logarithms to some
known base b (say 10, for example), and we wish to find the logarithm of a number n to some other base a. We then let x = log& n
w
whence, by definition, n = b
Similarly, if we let y = Iog n, then
we have n = ay
It follows that a v = b, and our problem reduces to solving this
equation for y. Taking the logarithm of both sides to base 6, we
have
y log& a = x log* b.
=
6
1.
But Iog6
Therefore,
It is
;
.
.
t
y
J
=
x (:
Vlogfe
/e
e\
(5-5)
a/)
=
log&
n
(|
)
\logb a/
1 1
4
Example 5-15. Find
,
Sec.
Logarithms
Solution:
By
log* 125,
where
e
i
It is usually easier to multiply
10K
125
result can
0.4343.
Thus,
by using a
table to the base 10.
lSio
= -p
125
logic e
e is
a fairly
=
n xuo
2.3026,
than to divide. Since division by logio
frequent operation in practical work,
and the
2.7183,
-
(5-5),
loge
6
=
5-9
it
should be noted that
be obtained by multiplying by 2.3026 instead of dividing by
log. 125
=
(2.0969) (2.3026)
= 4.828.
EXERCISE 5-5
Find each of the following logarithms by using a table of common logarithms:
1. log, 10.
2. log fl 100.
3. Iog2 e.
4. log e
5. log, e.
6. log* 10.
7. Iog 2 64.
8. Iog 20 1000.
9. Iog2o 100.
13. Iogi 25 1000.
TT.
10. logioo 64.
11. log, 8.
12. logo.i 50.
14. log* 20.
15. logo. 02 0.04.
16. logiooo 100.
O
6-1.
and Vectors
Right Triangles
ROUNDING OFF NUMBERS
Numbers that arise in the applications of trigonometry are
usually not exact, but are sufficiently accurate for a given purpose.
Numbers of this kind are called approximate numbers, and the
degree of accuracy of such a number
significant figures
it
contains.
indicated by
is
Reading from
left to
how many
right, the
number are the digits starting with the first
non-zero digit and ending with the last non-zero digit, unless it is
definitely specified that the zeros on the right are significant. Thus,
in the numbers 2.405, 0.002405, and 240500, the digits 2, 4, 0, and 5
are significant figures. The zeros after the 5 in 240500 may or may
significant figures in a
not be significant figures.
When it is desired to indicate whether final zeros are significant
or not, scientific notation is often used. Thus, in 2.405 10 5 the
5
the final two zeros are
last significant figure is 5 in 2.40500 10
,
,
;
regarded as significant.
To round
off a number in which the last desired significant figure
the units place or in any decimal place, drop all digits that lie
to the right of the last significant figure. It is sometimes necessary
also to increase the last digit in the retained part by 1.
is in
dropped part is less than 5, the last digit
unchanged. If the first digit in the
5
is
than
or if that digit is 5 and it is followed
greater
dropped part
by digits other than 0, the last digit in the retained part is
increased by 1. Whten the dropped part consists of the digit 5
alone or the digit 5 followed only by one or more zeros, we shall
use the following procedure as an arbitrary rule in this book: If
If the first digit in the
in the retained part is left
the last digit retained
even,
it is left
is
odd, this digit
unchanged. This
rule,
115
is
increased by 1;
although popular,
is
if it is
inferior
1 1
6
Right Triangles
and Vectors
Sec.
61
common-sense rules in many cases. For example, if .245, .165,
and .725 are to be rounded off to two decimal places and then
added, it would be more sensible to round off two of the numbers
in one direction and two in the other direction.
To round off a number in which the last significant figure will lie
to the left of the units place, first drop all digits to the right of the
place occupied by the last significant figure, and replace each
dropped digit to the left of the decimal point by a zero. Also,
either leave the last digit of the retained part unchanged or
increase that digit by one, in accordance with the directions just
given for dropping only a decimal part. For example, if 2533.62 is
to be rounded off to three significant figures, the result is 2530 and
if 487,569 is to be rounded off to three significant figures, the result
to
.485,
;
is
488,000.
There are
working
two
rules that are generally adopted by computers in
with approximate numbers in order to guard against
retaining figures that may indicate a false degree of accuracy
1.
In adding or subtracting approximate numbers, round off the
answer in the first place at the right in which any one of the
:
2.
given numbers ends.
In multiplying or dividing approximate numbers, round off
the answer to the fewest significant figures found in any of
the given numbers. The numbers entering a problem involving multiplication or division may be rounded off before the
computation is begun. If these numbers are rounded off,
they should have one more significant figure than the answer
is to
have.
While these rules point in the right direction, it should be mentioned that rounding off computed quantities to as many significant
figures as there are 'in the given
numbers does not necessarily
produce the degree of accuracy implied by the results. The subject
of accuracy of computation with approximate numbers is somewhat
complicated and beyond the scope of a book at this level.
6-2.
TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES
One
of the simplest, yet important, applications of trigonometry
is in the solution of right triangles.
right triangle has, in addition to the 90 angle, five other parts. These are two acute angles
A
and three sides. If we know the length of any two sides, or either
acute angle and any one side, the triangle can be solved; that is,
the unknown parts can be found.
63
Sec.
Right Triangles
and Vectors
1 1
To
solve problems involving parts of triwe shall find it helpful to be able to
express the trigonometric functions of an
7
~
A
angles,
acute angle A of a right triangle ABC in
terms of the sides of that right triangle. To
derive suitable relationships, let us place the
acute angle in standard position, as shown in
Fig. 6-1.
In the right triangle ABC the side AC, which
FIG
is
-
^
the abscissa of
becomes the side adjacent to the angle A the side CB,
which is the ordinate of B, becomes the side opposite to angle A
and the side AB, which is the radius vector to B, becomes the
the point
J5,
;
;
hypotenuse.
If
we
the lengths of the side adjacent, the side
let
opposite, and the hypotenuse be represented by the symbols 6, a,
and c, respectively, we may express the six functions of the acute
angle A in terms of a, b, and c as follows
:
,
n
A
sm A =
+ .
(6-1)'
^.
,
cos
(6-2)
v
'
/c
ox
,
tan
(6-3)
/a
..
esc
(6-4)J
sec
(6-5;
ff
>
A
AA =
=
c
Side adjacent
b
~
=
r
hypotenuse
-r
side opposite
-7-5
?
=
A
hypotenuse
.^
,.
=
^
,
cot
A =
A
side adjacent
-r-;
.
side opposite
'
c
a
T
>
c
a
>
c
T
6
side adjacent
(6-6)
>
o
side adjacent
hypotenuse =
r
AA = -~-
A =
a
-
hypotenuse
side opposite
r
,
A =
side opposite
-r
r^
=
b
a
using (6-1) to (6-6), we can express the trigonometric funcan angle of a right triangle without reference to any
coordinate system, since the ratios of the sides remain the same
regardless of the position of the triangle.
By
tions of
6-3.
PROCEDURES FOR SOLVING RIGHT TRIANGLES
When
is
solving a right triangle in which two parts are known, it
advisable to arrange the work systematically and to follow A
definite
1.
procedure consisting of the following steps
a figure reasonably close to scale, and indicate the
known
2.
:
Draw
parts.
Write an expression containing a trigonometric function
which involves the two known parts and one unknown part.
1 1
8
Right Triangles
and Vectors
3.
Find the selected unknown part from
4.
Find
all
unknown parts
other
Sec.
6-3
this equation.
of the triangle
by a similar
procedure.
Check
5.
all results.
Whenever possible, select a trigonometric function that gives a
by means of a multiplication rather than a division.
solution
In the following illustrative examples,
the acute angles are represented by the
letters A and 5, and the right angle is
denoted by C, while the small letters a,
6, and c, respectively, represent the sides
a -658
opposite them.
FlG> 6 _ 2<
Example 6-1. Solve the
The triangle
Solution:
is
triangle
ABC,
if
A =
drawn approximately
2820' and a
=
658.
to scale in Fig. 6-2.
The unknown
parts are the angle B and the sides b and c.
6140'.
90 - 2S20'
Since A
90, we have B
To find the side b, we may apply either (6-3) or (6-6), since both equations
+B =
involve the
because
it
=
unknown
b
=
A
and the known parts
and
We
a.
shall use cot
A=
-
means of a multiplication rather
we have
enables us to proceed to the solution by
than a division. Since
A =
2S20' and a
=
658,
Then
b
In this example,
To
we take
find the side
c,
we
=
=
658 cot
2820
(658) (1.855)
c
we may use the
=
=
1220.59.
A =
shall use (6-4), or esc
Hence,
check,
=
This result
b equal to 1221.
CSC 28020'
To
/
is
rounded
off to
four digits.
**
-
We
have, therefore,
rrJjg.
658 esc 2820'
(658) (2.107)
relation cos
=
1386.
1221
A = r^^ =
loot)
0.8802. Hence,
A =
2820'.
2
2 a2
Checking by means of the Pythagorean theorem yields the result b = c
2 =
2 =
=
=
6
whereas
1490841.
(c
a) (c + a)
(1221)
(728) (2044)
1488232,
These values of b 2 agree when they are rounded off to three significant figures,
=
Note. In most situations where
we must
solve triangles,
we
are
dealing with measured quantities, which are necessarily approximate. Therefore, our answers can be no more accurate than the
See.
6-3
and Vectors
Right Triangles
1
we begin with. If the original data are
approximate, our answers must be rounded off to
the degree of accuracy indicated by the data.
In example 6-1, for instance, the answers may
be given as b = 1220 and c = 1390, both rounded
19
data
off to
,
three significant figures.
FIG. 6-3.
Example 6-2. Solve the
Solution:
The
triangle
ABC,
shown
conditions are
if
b
= 250 and
c
in Fig. 6-3. Since -
= 371.
=
C
OKf|
=
j
o/
=
0.6738. Therefore,
A = 4738
/
and
B =
90
cos A,
- 4738' =
we have
cos
A
4222'.
I
To
hence,
find a,
we have a
we may use
either
unknown a with
choice of combining the
-r
=
tan
A
or -
o
=
A
sin
c
.
We
either 6 or c;
shall illustrate, in order,
the computation with each of these equations, thus providing a check on our work.
Using
Hence, a
we have
(6-3),
= 250
tan 4738'
=
^=
tan 47=38'.
(250) (1.096)
=
274.0, or 274
when rounded
off
to
when rounded
off
to
three figures.
Using
(6-1),
we have
=
sin
47 38 ''
371
Hence, a
=
371 sin 4738'
=
(371) (0.7388)
=
274.09 or 274
three figures.
EXERCISE 6-1
In each of the problems from
1.
a
3. b
5.
o
7.
a
9.
a
11. b
13. a
15. 6
17.
= 12, A - 33.
= 62.4, B = 7110'.
= 3.187, 6 = 6.249.
= 4.318, B = 6716'.
= 9.863, A = 3621'.
= 78.21, A = 4317'.
= 123.6, b = 783.1.
= 2.312, B = 4057'.
1
to 16, solve the right triangle.
2. b
4.
a
6. 6
8. b
10. 6
12. a
14.
a
16.
a
= 168, A = 3816'.
= 42, c = 76.
= 63.21, B = 8336'.
= 827.6, c = 963.4.
= 16.32, B = 8710
= 43.21, c = 63.75.
= 36.83, A = 5744'.
= 389.3, 6 = 62.34.
/
.
?
A
wire stretches frorn. a point on level ground to the top of a vertical pole. It
touches the ground at a point 15 feet from the foot of the pole and makes an
angle of 63 with the horizontal. Find the height of the pole and the length
of the wire.
18.
A ladder 40 feet long rests against a vertical wall. If its footjs 5 feet from the
base of the wall, what angle does it make with the ground?
120
19.
Right Triangles
A ladder 65
and
Vectors
Sec.
6-3
a window 35 feet above the
held in the same
to
the
other
side
of
the
is
moved
the
and
street, it will reach a
top
position
window 28 feet above the ground. How wide is the street from building to
feet long is placed so that it will reach
ground on one side of a
If the foot of the ladder is
street.
building?
20.
The grade of a hill
the tangent of the angle the
is
hill
makes with the
horizontal.
Find the grade of a hill which is 275 feet long and which rises 120 feet.
21. To find the width of a river, a surveyor sights on a line across the river between
two points A and B on opposite banks of the river. He then runs a line AC
perpendicular to AB. He finds that AC is 250 feet and angle ACB is 4217'.
How wide is the river?
22. Find the length of a side of a regular hexagon and the radius of the inscribed
circle, if
An
the radius of the circumscribed circle
is
10 feet,
560 feet while flying upward for 2,387 feet along an inclined
straight-line path. What is the angle of climb?
24. A pendulum 4.5 inches in length swings through an arc of 28. How high does
the bob rise above its lowest position?
25. A man 6 feet tall is walking along a straight horizontal path directly away from
a lamp post 10.5 feet high. How far is he from the post at a certain instant
23.
airplane rises
when
his
shadow
is
5 feet long?
Horizontal
B
"""*,,
Horizontal
FIG. 6-4.
6-4.
ANGLES OF ELEVATION AND DEPRESSION
Let the line
an observer
AB
in Fig. 6-4 be a level or horizontal line,
at the point
see an object at the point C.
A
and
let
If the
C is above the horizontal line AB, then the angle BAG measured up from the horizontal to the line of sight AC is called the
angle of elevation to C from A. If the object C is below the horizontal line AB, then the angle BAG measured down from the'
object
horizontal to the line of sight
to C from A.
AC
is
called the angle of depression
Example 6-3. From a point on the ground 300
the angle of elevation to
Solution:
In Fig.
= tan 2210'. Hence,
ing
is
122 feet high.
its
top
is
22
10'.
How
feet
high
is
from the base of a building,
the building?
a
6-5, we have b = 300 and A = 2210'. By (6-3),
300
a = 300 tan 2210' = (300) (0.4074) = 122.22. So the build-
Sec.
6-5
Right Triangles
and Vectors
121
30
W-
B
\22IQ'
S
FIG. 6-7.
FIG. 6-5.
6-5.
BEARING IN NAVIGATION
AND SURVEYING
In marine and air navigation and in surveying, the direction in
which an object is seen is expressed by the bearing or azimuth of
the line of sight from the observer. The bearing of a line is the
acute angle which its direction makes with a meridian or northsouth line. Such angles are sometimes called quadrant angles, or
quadrant bearings.
we
first
To describe the bearing
write the letter
the letter
E
or
W. The
of a given direction,
or S, then the acute angle, and finally
letters depend on the quadrant in which
N
the given direction falls. Thus, the bearings of the lines OA, OB,
45 W, respectively.
and OC in Fig. 6-6 are N 60 E, S 37 W, and
N
from its bearing only in that the
at north in a clockwise
azimuth is the angle measured from
and 360.
direction. An azimuth may have any value between
Thus, in Fig. 6-6 the azimuths of the lines OA, OB, and OC are
60, 217, and 315, respectively, measured clockwise from the
north. This method of measuring directions is coming into more
The azimuth
of a line differs
frequent use than that of quadrant bearings. We note also that the
is often used instead of azimuth. Thus, we may speak
term bearing
of the bearing of an object regardless of whether
or bearing as here defined.
we mean azimuth
Example 6-4. A ship heads due east from a dock at a speed of 18 miles per hour.
After traveling 30 miles it turns due south and continues at the same speed. Find
*
its distance and bearing from the dock after 4 hours.
Solution: In Fig. 6-7, let A be the point at which the dock is located, let B be the
point where the ship turns south, and let C be the position of the ship after 4 hours.
orj
Since the
number
of hours required to travel
of hours spent in travel
from
B
to
C
is
4
from
- 5^ =
7
^
A
to
B
is
^5
lo
=
Hence, a =' 18
f*
~
>
the
number
o
^
o
= 42.
1
From
the figure,
tan0
= 42
B
=
and
b
=
30 sec 5428'
Hence, the distance from the dock
14428' or S 3532' E.
is
Sec.
=
~~
(30) (1.721)
=
and
30
Therefore,
6-6.
and Vectors
Riaht
Triangles
Right Trianales
22
sec 6
6-5
5428',
=
51.6.
52 miles and the bearing of the line
AC
is
PROJECTIONS
desirable to consider direction along a line segment.
underThus, if Pi and P2 are the end points of a segment, we shall
P
directhe
stand PiP 2 to mean the directed segment from PI to 2
Often
it is
,
by the order in which the end points are named.
The non-negative length of the segment PiP2 is denoted by |PiP2 |.
as a
Frequently a directed segment PiP2 may lie on a line, such
tion being specified
coordinate axis, on which a positive direction has been specified.
Then the positive direction on the line may agree with the direction
from PI to P2 or the two directions may be opposite to each other.
The directed length of the segment PiP 2 is equal to |PiP 2 when the
directions agree or when PI and P 2 coincide and is equal to -|PiP 2
when they disagree. Since the context will make the meaning clear,
,
|
|
we
shall designate the directed length of the
by PiP2
segment PiP2 also
.
M
B
FIG. 6-9.
FIG. 6-8.
We recall that the projection of a point on a given line is the foot
in
of the perpendicular dropped from the point to the line. If
B
the
if
is
and
line
the
on
of
the
projec6-8
is
I,
PI
Fig.
projection
tion of P 2 on I, then the directed segment from A to B is the
A
projection on I of the directed segment PjP 2 We draw PiM parallel
between I and
or perpendicular to P2 , to show the angle
I,
.
to
Sec.
6-6
We
on the
Right Triangles
shall
now assume
line L
and Vectors
123
that a positive direction has been specified
P\M = AB, it follows immediately from
Then, since
trigonometry that
AB = PiP2
|
where
|
cos
0,
the acute angle between the positive end of I and the
positive half -line determined by the directed segment P\P* In Fig.
6-8 it is considered that I is positively directed toward the right.
The
is
result just given can be applied, as seen in Fig. 6-9, in find-
ing the projections of PiP2 upon the coordinate axes. The directed
lengths of the projections upon the cc-axis and the 2/-axis are,
respectively,
(6-7)
AB = PiP2
(6-8)
CD = PiP2
|
|
where
|
cos 0,
sin 6,
OX
and PiP2 as shown in Fig. 6-9.
the angle between
the coordinates of the end points of the segment PiP2 are
6
is
When
known, the projections
of these coordinates.
AB
From
AB =
(6-9)
x2
,
CD
are readily expressed in terms
the definitions of horizontal and vertical
and
distances given in Section 2-2,
C
|
it
follows that
and
xi
CD =
y*
-
yi.
P,(3,2)
(0,2)
B(7,Q)
A (3,0)
\67
P2 (7,-5)
0(0, -5)
FIG. 6-11.
FIG. 6-10.
Example 6-5. What jare the projections of the segment PiPa on the
Pi = (3, 2) and P 2 = (7, - 5)?
Solution:
=
-5-2=-7.
Also, since
AB = x 2 - %i = 7 - 3 =
Since AB is + 4, we know that AB
In Fig. 6-10,
CD = -
7,
we know that
CD is
directed
4,
is
and
CD =
t/
axes,
2
-
if
2/i
directed to the right.
downward.
1
24
and Vectors
Right Triangles
Sec.
6-6
Example 6-6. A ladder 12 feet long leans against the side of a house and makes
an angle of 67 with ground. Find its projections on the ground and on the side
of the house.
The conditions are represented in Fig. 6-11. Let
of the ladder. The required projections are found as follows.
The projection on the ground is given by
Solution:
x
The
=
I
=
cos 6
12 cos 67'
projection on the side of the house
I sin
12 sin 67
y
=
=
=
12 (0.3907)
given by
= 12 (0.9205)
I
=
12 be the length
= 4.7.
is
=
11.0.
EXERCISE 6-2
1.
Two points A and B are 5,000 feet apart and at the same elevation. An airplane
10,000 feet directly above point A. Find the angle of depression from a
and the airplane's distance
horizontal line through the airplane to point
is
B
from point B.
2.
The Washington monument is approximately 555 feet high. Find the angle of
elevation to the top of the monument from a point that is 621 feet from the
base of the monument and at the same elevation as the base.
a kite is 130 feet above the ground and 150 feet of string is out, find the angle
of elevation to the kite, assuming the string to He on a straight line.
3. If
4.
5.
Find the angle of elevation to the sun
63 feet long on horizontal ground.
if
a flagpole 95 feet high casts a shadow
A boat leaves its dock and heads N 52 W for 4 hours at 14 knots (1 knot = 1
nautical mile per hour = 6,080.4 feet per hour). It then turns and heads
N 38 E for 3 hours at 16 knots. Find the boat's
from the dock.
6.
The grade
rise
7.
8.
of a certain railroad
bod
is
0.1095.
final
bearing and distance
How many feet does a locomotive
while traveling 175 feet along the track?
An approach must be built up to the end of a bridge which is 40 feet above
ground. If the approach is to have a 10% grade and the original ground is
assumed to be level, how far from the end of the bridge must the approach start?
A surveyor wishes to find
the distance between two points
A
and
B
separated
by a lake. He finds a point C on the shore of the lake such that angle ACB is
90. He measures AC and BC and finds that AC is 640 feet and BC is 285 feet.
How far apart are A and B?
9.
A smokestack is
175 feet from a building.
angle of elevation to the top of the stack
base from the same window
From a window of the building the
2810 The angle of depression to
/
is
.
2430'. Assuming that the ground is level,
find a) the height of the window above the ground and b) the height of the
smokestack.
its
10.
is
N
Two ships leave the same port at the same time. One travels 42 E at 25
knots. The other travels S 48 E at 33 knots. How far apart are the two ships
after
4 hours?
Sec.
6-7.
6-7
Right Triangles
SCALAR
We
AND VECTOR
and Vectors
125
QUANTITIES
shall at this point find
it
necessary to distinguish carefully
between two kinds of quantities, namely scalar quantities and vector
quantities.
A
scalar quantity is a quantity whose measure can be fully
described by a number. It is a quantity which can be measured on
a real number scale. For example, temperature is a scalar quantity, measured on the scale of a thermometer. Also we shall define
the scalar components of the segment PiP2 to be the projections on
the coordinate axes, or the directed lengths x 2
Xi and y 2
3/1
given by (6-9) in Section 6-6. The student should note, however,
that the scalar components of P2 Pi are not equal to those of P\P2
The components of PJP\ are x\ x 2 and yi y 2 and are the negatives of the respective components of PiP 2
.
.
A
vector quantity, or simply a vector, is a quantity possessing
both magnitude and direction. A vector may be represented by any
one of a set of equal and parallel line segments. Algebraically,
a vector
described by the scalar components of any segment
all such segments have the same components. We
shall, in fact, call these the scalar components of the vector and shall
is fully
representing
it
;
./
o
FIG. 6-13.
FIG. 6-12.
enclose
vector
them
in brackets. In Fig. 6-12, three representations of the
The arrowhead indicates the order in
[4, 3] are shown.
which the end points of the line segment are named.
We may denote a vector by a single letter in boldface type, say v,
or may represent it geometrically by any one of the segments, such as
AB, OP, or CD. Here A, O, and C represent the initial points of the
three segments, while the terminal points at the arrowheads are
B, P, and D. Actually, any other segment with the same magnitude
and direction could have been selected to represent the vector v.
Instead of using a single letter in boldface type to denote a vector
represented by a segment PiP2 we may also use the notation PiP2.
,
1
26
Right Triangles
and Vectors
Sec.
6-7
We
have seen that a vector is unambiguone
of a set of equal and parallel line
ously represented by any
segments. Hence, any point may be taken as the initial point
of a segment which represents a vector. If the origin is so chosen,
the coordinates of the end^point P(x,y) are actually the scalar
components of the vector OP. We can therefore give the following
simple definition of the magnitude of a vector and its expression
Properties of Vectors.
in
terms of scalar components
:
The magnitude, or length, of a vector v
of
one
the
segments representing v.
any
Definition.
Let v
where
Since
is
the length of
=
v\
[yi, v2 ] be the vector represented by OP in Fig. 6-13,
and v 2 are scalar components. Then we have |v| = \OP\.
OP is the hypotenuse of a right triangle,
=
v
(6-10)
|
|
VV +
v2
2
.
In case the scalar components are given by the coordinates (x, y)
of the end point P of the segment, the magnitude of the vector is
From
it follows that two vectors are
their
and
if
equal
only
respective scalar components are equal.
For example, u = [u lt u2 ] equals v = |>i, v2 ] if and only if HI = Vi
and u2 = v 2
the definition of a vector,
if
.
= [0, 0] to be the zero vector.
also define a special vector
It corresponds to the exceptional case in which
2 coincides with
We
P
PI and
may
be considered as represented geometrically by a segment of length zero, that is, by a point. The zero vector may be
regarded as having any direction whatsoever.
v = [ Vi, v 2 ] is
If a vector v. = [v lf v 2 ] is given, the vector
defined to be the negative of v. Thus, ifj^ift = v denotes a vector
represented by the segment PiP2 then P2 Pi = v denotes a vector
having the same length as PiP2 but oppositely directed, namely,
from P2 to PI. We note that
,
- (-
A
If
is
unit vector
u=
[fa,
u^
is
is
Y)
=
v.
defined as a vector
any non-zero
whose magnitude
vector,
then
-,
is
unity.
= \T^I
r^]\
>
a unit vector.
Multiplication of a vector by a scalar is performed by multiplying
the magnitude of the vector by the absolute value of the scalar,
maintaining the direction of the vector if the scalar is non-negative
and reversing it otherwise. Thus, by k[ui, u^] we mean the vector
Sec.
6-7
Ri ghf Triangles
[kui, ku2],
Using the equation
[MI,
and Vectors
u2
=
]
any vector v = Oi,^2 ] as proportional
= V^i 2 + ^2 2 Changing v to
choose
A:
-
normalizing
Sums and
k
1
27
[~ yl we can express
?
i
to a unit vector v/fc if
a unit vector v/fc
we
is called
v.
Of the many questions which
Differences of Vectors.
arise in the study of vectors, the one of greatest importance for us
at present concerns the addition of vectors. The sum, or resultant,
of two vectors is defined to be the vector which has for its scalar
components the sums of the scalar components of the two vectors.
Thus, the sum of the vectors u and v is given by the relationship
u
(6-11)
It is
+
=
v
[MI
+ vi, u2 + v
2 ].
important to note that some of the laws of the algebra of
also hold for vectors. Thus, the commutative law is
numbers
u
+
v
=
v
Also, the associative law of addition
(u
And, for every vector
+
v)
+w=
u
+ -
v)
+ u.
is
+
(v
+ w).
v,
v
(
=
0.
That these laws are satisfied for vectors can be shown geometrically
or can be seen from (6-11) by virtue of the known laws of addition
of real numbers.
In terms of components, the rule for multiplication by a scalar is
given by
In consequence of this definition, the following algebraic laws are
satisfied
:
+ v) =
=
(k + m)u
fc(u
fc(mu)
=
fcu
fcu
+ &v;
+ mu;
(fcw)u.
To find the sum of two vectors geometrically, we proceed as
indicated in Fig. 6-14. This graphical representation of the sum
of two vectors by means of the triangle construction was probably
suggested by the behavior of physical quantities represented by
vectors, such as forces, displacements, velocities, and accelerations.
Their addition is effected by the triangle law or the parallelogram
law.
128
Right Triangles
and Vectors
Sec.
6-7
1C
o
FIG. 6-15.
FIG. 6-14.
The following
2.
steps indicate the actual procedure.
segment representing one of the vectors, say u, with
its initial point at the origin of the coordinate system.
Place the initial point of a segment representing the second
3.
vector, say v, at the terminal point of the first segment.
Draw the segment from the origin to the terminal point of the
1.
Select a
segment for
u 4- v.
The
v.
This segment represents the resultant vector,
v of two vectors
difference u
u
If
u
=
[ui, u%~\
and v
[y\, v%]
U
=u+
v
V
=
,
manner
defined in a
is
analogous to that used in defining the difference of
numbers. Thus,
v).
(
then
[U\
Vi,
U2
V2\.
To find the difference u v of two vectors geometrically, we
proceed as indicated in Fig. 6-15. The steps are as follows
1. Place segments representing u and v with their initial points
at the origin.
v at the
2. Place the initial point of a segment representing
terminal point of the segment representing u. Note that the
segment representing v is parallel and equal in length to
that for v, but has the opposite direction.
3. Draw the segment from the origin to the terminal point of
This segment represents the vector
v.
the segment for
:
u
v.
Example 6-7. Find the sum and
Solution:
The
difference of the vectors
required vectors are
u+ v = [5
-
2,
-
3
+
1]
=
[3,
-
2],
and
u-? = |5-(-2), -3-l] =
[7,
-4].
u
=
[5,
-
3]
and
Sec.
6-7
Right Triangles
Example 6-8. Express u
Solution:
The
=
[4, 3]
and Vectors
129
as proportional to a unit vector.
expression representing the vector
is
Q~1
-
>
[A
note that the magnitude of
-r
>
From
Section 6-7,
.
To
check,
is
-r
To add vectors analytically,
of the vectors to be added.
that
-
we
first find
the scalar components
we know
to*)
segments representing u and v make angles a
and /?, respectively, with
the #-axis, the scalar components are given by the
projections upon the coordinate axes. Thus, as shown
if
FIG. 6-16.
in Fig. 6-16,
ux =
|
u
|
cos
a
cos
]8 ;
=
t
I
u
I
sin a,
and
vx
=
|
v
|
vy
=
|
v
Then the components of the resultant
algebraic sums
X=
Ux
+V
Y =
x,
Thus, the magnitude of the resultant
|R|
Uv
+
R is
|
sin
will
]8.
be given by the
Vy.
given by
=
Also, the direction angle 6 satisfies the relationship
tan 6
= Y
To determine the quadrant of
in mind and use the correct signs
-
=1
1
j=
important to keep
and Y. For example, tan
correctly, it is
of
X
might lead one to an incorrect value 45 for 6 instead of
the correct third-quadrant angle 225.
The student should note that this (analytic) procedure and the
geometric procedure (parallelogram law) give the same results. This
can be shown by appropriately combining the scalar components of
the vectors u, v, u + v, and u - v in Figs. 6-14 and 6-15, where
illustrated the geometric procedure.
we
130
Right Triangles
and Vectors
Sec.
6-7
o
W-
-*-E
o
^*
C
\25
D
\
s
FIG. 6-17.
FIG. 6-18.
A block weighing 500 pounds rests on a smooth plane making an
with the horizontal, as indicated in Fig. 6-17. What force, parallel to
the plane, is necessary to hold the block in position?
Example 6-9.
angle of 25
The force due to the weight mayjbe represented
downward through A, such as vector AC. The component of AC which is parallel to the inclined plane and which must be overcome is
represented by the projection AB of AC on the inclined plane. Since angle
Solution: Let the block be at A.
by a vector acting
vertically
BAG = 90 - 25 = 65,
AB = 500 cos 65 =
(500) (0.4226)
= 211.3.
Hence, a force of 211 pounds must be applied parallel to the inclined plane to keep
the block from sliding. (It
is
assumed that three
Example 6-10. Find the magnitude and
direction of the resultant of a force of
E
110 pounds acting in the direction S 4627'
the direction
2914' W.
significant figures are appropriate.)
and a
force of 100
pounds acting
in
N
Solution: The given forces are represented by vectors in Fig. 6-18. Let F denote
the magnitude of the resultant force and 6 the angle it makes with OE.
Since 90 - 4627'
4333' and 90 - 2914'
6046', the components Fx and
=
Fy
=
of the resultant are found as follows:
Fx =
110 cos 4333'
= (110)
= 79.73
Fv = = = -
(0.7248)
-
100 cos 6046'
(100) (0.4884)
-48.84=30.89;
+ 100 sin 6046'
+ (100) (0.8726)
+ 87.26 = 11.47.
110 sin 4333'
(110) (0.6890)
75.79
Now
tan*=^
Therefore B
= 2022',
the magnitude of
=
and
F is 33
11.47
=
~~
30.89
F=
v '"' *"'
0.3713,
11.47 esc 2022'
pounds, and
its
^
and
"""
=
direction
= F
6 ~~
esc v
11.47
(11.47) (2.873)
is
N 6938' E.
= 32.95. Hence,
Sec.
6-8
Right Triangles
and Vectors
131
EXERCISE 6-3
1.
Draw a diagram showing three different segments representing each of the
given vectors. Find the magnitude of each vector.
b. [12, 5].
a. [3, 4].
-
c.
[-2,
d.
4].
h.
[-
3,
-
-
5].
e.
[1,
2.
Express each of the vectors in Problem 1 in terms of a unit vector.
In each of the following cases, add the given vectors. Find the magnitude and
3.
f.
1].
[5, 3].
g. [0, 4].
[0,
1].
direction of the resultant.
and [2, 3].
b. [4, - 2] and [1, - 10].
and
d.
3].
[- 6,
[2, 0]
[1, 2], [4, 3], and [0, 7].
In Problem 3, parts (a), (6), and (c), subtract the first vector from the second,
and find the magnitude and direction of the resultant.
In Problem 3(d), subtract twice the third vector from the sum of four times the
first vector and twice the second vector, and find the magnitude and direction
a. [1, 1]
c.
4.
5.
of the resultant.
6.
A force
pounds acts at an angle of 63 with the horizontal. What are the
and horizontal components of the force?
a ship sails N 48
at 30 knots, what are its westward and northward
of 40
vertical
7. If
W
components?
8.
One
force of 28
pounds acts vertically upward on a
43 pounds acts horizontally on the particle.
resultant force,
9.
10.
and what
is its
What
particle.
is
Another force of
the magnitude of the
direction?
A
barrel weighing 160 pounds rests on a smooth plane which makes an angle
of 22 with the horizontal. Find the force parallel to the plane necessary to
keep the barrel from rolling down the plane.
Three forces act on a particle. One of 50 pounds makes an angle of 25 with the
horizontal; a second of 60 pounds makes an angle of 50 with the horizontal;
of 75 pounds makes an angle of 230 with the horizontal. Find
the magnitude and direction of the resultant force.
11. Four forces act on a body. The forces are 30, 45, 50, and 65 pounds, and they
and the third
make
angles with the horizontal of 25, 160, 240, and 330, respectively.
Assuming that all the forces lie in the same vertical plane, find the magnitude
and direction
required force
of the force necessary to hold the body in equilibrium. The
equal in magnitude and opposite to the resultant of the given
is
forces.
12.
A
an angle of elevation of 37. Its initial velocity is 2,500 feet
second.
the
Find
horizontal and vertical components of its initial velocity.
per
6-8.
shell is fired at
LOGARITHMS OP TRIGONOMETRIC FUNCTIONS
in this chapter, we have considered the use of a table Qf
natural trigonometric functions and have solved various problems
involving right triangles. In many problems, however, the computation is greatly facilitated by the use of logarithms to perform the
numerical operations. For this purpose the values of the logarithms
of the trigonometric functions are required. Table III might be
used to obtain the logarithms of the functions found in Table II,
So far
132
Right Triangles
and Vectors
Sec.
6-8
but the work is considerably lessened by the use of Table IV at the
end of this book, which gives the logarithms of the trigonometric
functions at once.
Table IV
a four-place table giving the logarithms of functions
to 90, For the sine and cosine
from
is
at intervals of 10 minutes
of
and 90, the tangent of any angle between
any angle between
and 45, and the cotangent of any angle between 45 and 90,
than 1; hence, the logarithms of
10 after the
these functions are negative, and we must write
tabulated entry. For the sake of uniformity, 10 has been added
to each of the other entries in the table. In using the table,
therefore, 10 must be subtracted from every entry.
The method of using Table IV is similar to that described for
Table II and will be illustrated by the following examples.
the value of the function
Example 6-11. Find
log sin 2310'.
Since this angle
Solution:
is less
we
given in Table IV,
is
find that log sin 23 10'
= 9.5948-10.
Example 6-12. Find
From Table IV we
Solution:
10'
The
x
log cot 5127'.
tabular difference
= 0.7(26) =
18.2,
obtain the values for the following tabulation:
log cot
5120'
log cot
5127
log cot
5130'
is
26. Since
;
-
9.9032
-
10
=
9.9032
-
10
=
9.9006
-
10
5127'
6
if
Hence,
to 5130',
log tan 6
.0018
=
=
=
9.7816
-
+ x) = 9.7827 log tan 3120' = 9.7845
log tan
10'
x
way from 5120'
9.7827-10.
The positive part 9.7827 lies between the entries 9.7816 and 9.7845
The procedure for finding 6 may be indicated as follows:
log tan 3110'
TT
of the
= (9.9032-10) = 9.9014-10.
Example 6-13. Find the acute angle
Solution:
-^
26
x
and we have
log cot 5127'
in Table IV.
is
-
II
31(10
f
.
>
and
8
=
3110'
+
55(10')
10
11
10
10
29
Sec.
6-9
Right triangles
and Vectors
1
33
EXERCISE 6-4
In each of the problems from 1 to 15, find the value of the given logarithm.
2, log sin 2120'.
3. log cos 8620'.
log sin 4820'.
4. log tan 8830'.
5. log cot 1020'.
6. log sec 4350'.
7. log sin 1326'.
8. log sec 4857'.
9. log tan 4114'.
10. log esc 7832'.
11. log cot 6843'.
12. log cos 1818'.
/
13. log esc 8316'.
14. log cos 18'.
15. log tan 5134
In each of the problems from 16 to 35, find the angle (or angles) 6 between
1.
.
and 360.
= 8.8059-10.
= 9.9959-10.
= 0.4625.
log tan 9 = 9.8483-10.
log esc 6 = 0.4081.
log sin 9 = 9.9567-10.
log cos 9 = 9.9755-10.
log sec 9 = 0.1967.
log cos 9 = 9.1860-10.
log sin 9 = 9.9974-10.
16. log sin 6
18. log cos 6
20. log sec
21.
22.
23.
24.
26.
28.
30.
32.
34.
6-9.
= 8.3661MO,
= 9.1697-10.
= 0.4882.
log cot 6
log tan 9 = 0.1430.
log sec 6 = 0.3586.
log tan 9 = 9.7648-10.
log cot 9 = 9.8666-10.
log esc 6 = 0.3370.
log cot 9 = 1.5976.
log sec 9 = 0.3870.
17. log tan 6
19. log sin 6
25.
27.
29.
31.
33.
35.
LOGARITHMIC SOLUTION OF RIGHT TRIANGLES
The solution of a right triangle by means of logarithms
the same as by natural functions, except that
is
exactly
for the actual numerical computation a table
of logarithms of the natural functions is used
in conjunction with a table of logarithms of
The following example
numbers.
will illus-
trate the procedure.
Example 6-14. Solve the
The values
Solution:
B =
90
-
2140'
=
ABC,
right triangle
which
in
A =
known parts are indicated in
To find the side a, we have
of the
6820'.
tan 2140'
=
^
2140' and b
Fig. 6-19.
= 8.43.
We see that
>
8.4o
Hence,
a
= 8.43 tan 2140',
or
log a = log 8.43
Arrange the work as follows:
log 8.43
.
Therefore, a
To
=
find side
log tan 2140'
log a
-f log
=
=
=
tan 2140'.
0.9258
(Table III)
9.5991-10
(Table IV)
10.5249-10
3.35.
c,
(Table III)
we have
=
c
cos 2140'.
1
34
Right Triangles
Hence,
and Vectors
Sec.
6-9
8 43
.
C
and
log c
'
cos 2140'
= log 8.43 -
Jog cos 2140'.
Arrange the work as follows:
log 8.43
cos
2140'
log
log c
Therefore, c
=
=
=
10.9258-10
(Table III)
9.9682-10
(Table IV)
0.9576
= 9.07.
(Table III)
EXERCISE 6-5
In each of the problems from
I.
4.
7.
1
= 100, A = 31.
A = 8817', c = 108.1.
J3 = 279', a = 36.13.
8. 6
a railroad track
30
6
9. If
rises
2. c
5. c
to 8, solve the given right triangle.
= 3.45, a = 1.76.
= 6.275, B = 1845'.
= 98.34, B = 1848'.
feet in
3.
A =
6.
a
=
= 63.4.
= 396.3.
2520', a
645.3, b
a horizontal distance of one mile, find the
angle of inclination of the track.
10.
A
II.
A force of 628 pounds acts at
force of 341 pounds and another force of 427 pounds act at right angles to
each other. Find the magnitude of the resultant force and the angle it makes
with each of the forces.
180, and a force of 237 pounds acts at 270. Find
the direction and magnitude of the resultant.
12. An airplane is flying due east at a speed of 485 miles per hour, and the wind
is blowing due south at 33.6 miles per hour. Find the direction and speed
of the phane.
13.
The westward and northward components
of the velocity of an airplane are
363 and 487 miles, respectively. Find the direction and speed of the airplane.
14. The eastward and southward components of the velocity of a ship are 10.4 and
16.8 knots, respectively. Find the speed of the ship and the direction in which
it is
15.
A
moving.
Find
16.
17.
pounds is acting at an angle of 4713' with the horizontal.
horizontal and vertical components.
force of 2673
its
A
force of 162.4 pounds is just sufficient to keep a block at rest on a smooth
inclined plane. If the block weighs* 783.1 pounds, find the angle at which the
plane is inclined to the horizontal.
Two tangents are drawn from a point P to a circle whose radius is
14,32 inches.
the angle between the tangents is 3228', how long is each tangent segment?
18. Two buildings of the same height are 11,640 feet apart. When an airplane is
If
19.
8,000 feet above one of them, what is the angle of depression to the other one?
cable which can withstand a pull of 10,000 pounds is used to pull loaded
trucks up a ramp. If the angle of inclination of the ramp is 3616', find the
A
weight of the heaviest truck which can be safely pulled up the
the cable.
20.
The angle
is
4528'.
How high
of elevation from one point
From a
is
on
level
ground to the top of a flagpole
point on the ground 25 feet farther
the pole?
ramp with
away the angle
is
3956'.
Trigonometric Functions of
Sums and Differences
7-1.
DERIVATION OF THE ADDITION FORMULAS
Heretofore, we were concerned with relationships between trigonometric functions of a single angle. We shall now establish
certain fundamental identities involving two angles, in terms of
the functions of the single angles. The following identities express
functions of the sum and difference of two angles in terms* of the
functions of the separate angles.
(7-1)
sin (a
j8)
=
sin
a
(7-2)
cos
/3)
=
cos
a cos
(7-3)
sin (a
/3)
=
sin
a
(7-4)
cos (a
/3)
=
cos
a cos
(7-5)
tan (a
p)
=
-
^
p)
K/
=
+
(a +
,
+
,
tan (a
v
(7-6)
We
.
^
cos
cos
tan
tan
1
+
+
oj
tan
cos
a
sin
a sin /3,
cos
a sin
sin
a sin j8,
/3
/3
ft
+
tan
tan
1
|8
a
a
/3,
|8,
]
tan
tan
sin
p
j8
tan
/3
now prove
the formulas for the sine and cosine by using
the derivation developed by E. J. McShane. 1
shall
o
FIG. 7-2.
FIG. 7-1.
1 E.
"The Addition Formulas for tho Sine and Cosine,"
J. McShane.
American Mathematical- Monthly, Vol. 48 (1941), pp. 688-89.
135
Trigonometric Functions of
136
Sums & Differences
and a be any two angles with the same
Let
shown
On
in Fig. 7-1.
their terminal sides
we
Sec.
initial side
7-1
OW,
choose points
P
as
and
Q, respectively, each at unit distance from 0.
d represent
Let
distance
the
from P to Q. We shall now make
two computations for d 2 using
first OW, and then OP, as the
0[cos(-0).
sin
(a
-0)]
,
OW
is used as the x-axis
When
of a coordinate system, as shown
in Fig. 7-2, we find that the coordinates of P and Q are (cos 0, sin /?)
and (cos a, sin a), respectively.
Hence, by the distance formula,
d2
=
=
Since
a
cos a
cos 2 a +
+
cos 0) 2
(sin
2 cos a cos
(cos
a
sin 0) 2
+
2
sin
2
a =
cos
=
d2
cos
+
2
+
2
sin
2 (cos
2
FIG. 7-3.
a
a
sin 2
=
2
1
a sin
+
sin 2 0.
,
+
cos
2 sin
sin
a
sin 0).
Let us now use OP as the #-axis, as shown in Fig. 7-3. Then the
coordinates of P are (1,0) and those of Q are [cos (a
/3),
sin (a
0)
d
2
]
Hence,
.
=
=
=
-
[cos (a
cos
2
2
-
-
0)
(a
j8)
2 cos (a
-
+ sin 2
2
I]
(a
'2 cos (a
-
j8)
0)
+1+
sin 2 (a
0)
0).
2
Equating the two expressions for d yields
2
2(cos
+
a cos
a sin j8) =2
sin
2 cos (a
|8).
Therefore,
cos (a
(7-4)
0)
=
This establishes (7-4).
Setting a = 90 in (7-4),
cos
we
cos (90
(7-7)
If in (7-7)
we
let /3
=
90
-
sin (90
(7-8)
From
(7-7), (7-8),
sin (a
+ 0) =
=
=
a
+
cos
sin
a
sin
]8.
find that
-
0)
=
sin 0.
we have
- 7) = cos 7.
y,
and (7-4), we obtain
cos [90
cos [(90
cos (90
- (a + 0)]
- a) - 0]
- a) cos +
sin (90
-
a) sin 0,
Sec.
&
Trigonometric Functions of Sums
7-1
137
Differences
or
sin (a
(7-1)
+
/3)
=
+
a cos /3
sin
a sin
cos
/3.
This establishes (7-1).
=
Since cos (-)8)
cos
/3
and sin
(-/?)
=
-sin
,
we
have, as a con-
sequence of (7-4),
cos (a
+
/3)
=
=
cos [a
cos a cos
=
cos
(
|8)]
(
j8)
+
a sin
sin
(
j8)
(
]8)
or
cos (a
(7-2)
sin
To prove
a cos
TT
]8)
each
/3,
a
a.
cos
sin
a
cos
j8)
(
+
cos
|8
We may obtain
cos
a
]8.
a
sin
a cos B
5
cos a cos p
sin
numerator
the
of
sin
/3.
+
= sm
COS
fl
a sin ]8
r
~#
sin a sm p
cos
and
denominator by
a sin )8
a cosjg _ tan a + tan ff
1
tan a tan /3
sin a sin ff
__
cos a cos 8
a -+ tan pQ
/o\
= tan
+ ^|8)
(a
v
y
^5
1
tan a tan p
cos
^
'
/
,
.
manner from (7-3) and
(7-5).
Example 7-1. Find the exact value
Solution: Substituting 45
sin 75
for
= sin
= sin
of sin
75.
a and 30 for
(45 + 30)
45 cos 30
+ cos 45
V2
V3
in (7-1),
_\/2
"2*2^2*2
.-=
Therefore,
75
'
'-
cos
(7-6) in a similar
sin
and (7-1)
C7
_,
,
tan
sin
use the relationship tan 6
term
we have
sin a cos j8
cos a cos j8
cos a cos j8
cos a cos B
(7-5)y
v
from
sin
j8
fi)
^
or
sin
,
Hence,
t*
cos
follows that
it
=
=
we
(7-5),
,
tan
Dividing
(a:
a
We then obtain
m sin (a +
(a + p) =
7
r^\ ~
cos (a + p)
and (7-2).
cos
]8)
from (7-1)
Similarly,
(7-3)
+
=
V (V
4
3
+
D-
1
we obtain
sin 30
(7-4),
Trigonometric Functions of Sums
138
Example 7-2, Find the exact value
angle such that sin
=
a
and
,
O
of tan
+
(a.
Differences
/3)
a
if
a third-quadrant angle tan
is
=
It follows that tan
4.
= -
a
~
x
m =
+ 6)
/
4
+
'
Using
12
(7-5),
= -
7-^-7
,
tan (a
3
j
7-1
Sec.
a second-quadrant
is
ft
5
=
\JL
Let a be a second-quadrant angle for which y
Solution:
Hence, x
3
=
&
=
3
and
=
r
5.
we have
16
^
EXERCISE 7-1
In each of the problems from
to 8, find the exact value of the given function.
1
135.
15.
105.
1.
cos 75.
2.
tan 105.
3. sin
5.
tan 195.
6.
cos 195.
7.
tan 15.
is
a second-quadrant angle, and
a
9. If
sin
a
a third-quadrant angle and
is
= -
3
and cos
-r
O
=
ft
/3
5
rz find
>
lo
sin (a -f
= r and a - B = 45, find tan 0.
= 180, find tan 0.
If tan a = 3 and a +
3
= - find sin (a +
If tan a = j and tan
4
10. If
j(S),
4. sin
8. cos
cos (a -f 0), and tan (a -f
|8).
a
tan
,4
11.
1
12.
>
j8
jft
o
j8)
and cos (a
+
)8),
where a and
are first-quadrant angles.
13. If cos
a.
= - and
=
cos
J
>
o
find sin (a
j9)
and cos (a
j8),
where a and
are acute angles.
14. If sin
a
=4
-=
and cos (a
15. If sin
a
17.
^r
LZ
>
cos
j8
= 24
^rr
~
>
+
ft)
>
cos
sin
j8
|3
= -1
and a and
is
obtuse,
7^
A/2
= -jp
>
a
Id
and tan (a
45)
and a and
-
(sin
is
obtuse,
22. cos
are both obtuse, find sin (a
+
j3
is
acute, find sin (a
/3)
0)
and
]S
are obtuse, find sin (a -f
j3)
and
and
is
in the third quadrant, find
0).
identities:
a
19. cos (a
cos a).
Sin(a
sin
/3
and
>
5
=
Prove each of the following
20.
a
>
^O
2
O
18. sin (a
>
j9).
If cos a =
cos (a + |8).
3
If sin a = -
tan (a
5
=
tan #
-f 0).
= ^3
cos (a
16.
>
O
7?=cot|8-cota.
a sin
p
2 a = cos 2 a - sin 2 a = 2
21. tan
cos 2
a -
1
=
1
-
(a
\
2
TT)
= -
cos a.
=i-*^.
+-)
4/
1 ~ tan a
sin 2 a.
7-2
Sec.
o
no
23. sin 2
sin
25.
(
-
Trigonometric Functions of Sums & Differences
a
a
cos (a
s
97
fo (
a
sin (a
9Q
29
cos
(
a
=2n sin a cos a.
--
sin (a
""
0)
rin (a
f
+
a
a
tan
+ 0)
---~ = tan a + tan p.
-
cos
a;
_
cos^a
jS
tan
0) ~~
P)
4-
+
+ tan a tan
I
'
tan
a
a
cot
ft
$
tan
"
1
'
_
2
p
cos
(A
C
cos
2
+
35. sin 2
7-2.
1
cot
(3
/?)
ft
cos (a
^
^
sec a.
ct
sin (a
cos
l^TTtSJ
cos 2
cos
+ B + C) = sin A cos cos (7 -f cos A sin 5
A cos B sin (7 sin A sin 5 sin C.
sin 3 a = sin 5 a cos 2 a
cos 5 a sin 2 a.
= cos 2 ft cos (a - 0) cos (a 4- 0) = cos a - sin
32. sin
34.
+
1
g) sin (a
cos 2
33.
"
a - tan
39
cos 2
:
26.
ft
tan
ft)
'
tan a,
.
-ft)
-
a
----=
a
n 2a
sin
- 0)
+-_
/3) + sin (a
~_ =
+ p) + cos (a - p)
* an 0= + tan
+
_
si
<n
24.
1
~~
a
cos2
sin 2 a.
2
~
+
a
sin 2
'
THE DOUBLE-ANGLE FORMULAS
we
=a
in (7-1), (7-2), and (7-5), we obtain functions
of twice a given angle in terms of the functions of the angle itself.
If
Thus,
let ft
we have
the following identities
sin
(7-9)
(7-10)
2a =
a
2 sin
cos 2
a =
n
tan 2
a =
cos
2
:
cos a,
sin 2 a,
a
and
/-
* *
\
(7-11)
We may
2 tan
Q!
tan 2
a
obtain two other useful forms for cos 2
2
using in turn cos
by
forms are given by the
a
(7-12)
sin 2
1
a and
sin 2
a=
1
a from (7-10)
cos 2 a. These
identities
-
cos 2
a =
1
cos 2
a =
2 cos 2
2 sin 2 a,
and
(7-13)
The following
applications of
1.
an indication of the possible
illustrations give
(7-9),
a -
(7-10), and
The student should
(7-11).
study them carefully.
sin 4
a =
sm a =
oj
cos
-
=
sin 2 (2 a)
.
sin 2
rt
cos 2
/a\
(
^
)
/a;\
^
j
=
2 sin 2
=
2
=
cos 2
.
sm
9
a cos
a
c <> s
a
-
2 a,
a;
o
.
'
9
sin 2
a
g
1
Trigonometric Functions of Sums & Differences
40
Sec.
7-2
Example 7-3. Find the exact value of sin 120 by means of a double-angle formula.
We
Solution:
use (7-9) to obtain
sin 2(60)
sin 120
=
=
/1\
(2)
\
V
Example
7-4.
2 sin 60 cos 60
_
-
Derive a formula for cos 3 a in terms of cos a.
Solution: Applying the identity for cos (a -f
and the double-angle formulas,
/3),
we have
cos 3
a
= cos (a + 2 a) = cos a cos 2 a sin a sin 2 a
= cos a (2 cos a 1) sin a (2 sin a cos a)
= 2 cos a cos a 2 sin a cos a
= 2 cos a (cos
sin a)
cos a
= 2 cos a (2 cos a 1) cos a
= 4 cos a 3 cos a.
2
3
2
2
2
a;
2
3
_.
-
Example
r
7-5.
,1
-r*
i
,
sin 3
,
cos
rz
Prove the identity
sm
30
_
-r-
2.
cos
Solution: First combine the fractions on the left side
to the right side. Thus,
sin
30
sin
__
__ sin 3
cos_3_0 ~"
cos
-
sin
20
_
~~
sin
_
~~
-
sin (3
__
"~~
result
0)
cos
sin
2_smJ!0
2
sin 2
cos
sin
cos 3
cos
sin
~~
7-3.
-
cos
and then reduce the
THE HALF-ANGLE FORMULAS
Functions of an angle in terms of the functions of twice that
angle can be obtained directly from (7-12) and (7-13) If cos 2 a =
1
2 sin 2 a is solved for sin a, we obtain
.
sin
a =
/I
A
d=
4/
Also,
=
by solving cos 2 a
cos
2 cos 2
a =
a
a
1 for cos a,
we have
+cos2.
.,
,
'
cos 2
^
r
2
Since these formulas may be equally well regarded as expressing
functions of half an angle in terms of the functions of the given
angle itself, the same relationship is retained if the identities are
written
,~
* A \
(7-14)
sm
OL
^
=
d
/I
cos
a
A / *
T- cos
a
A
and
1K x
(7-15)
v*
/
cos
- =
.
db
Sec.
Trigonometric Functions of Sums & Differences
7-3
141
These are the so-called half-angle formulas for the sine and cosine.
From (7-14) and (7-15) we obtain, by division,
,
^x
tan
(7-16)
^
'
a
sin
__
~~
2
cos
The algebraic signs
mined by the quadrant
If
we
in
A
The student
/
y
2
(7-15), and (7-16) are deter-
1
-
(1
+
,
numerator and the denominator of
we
cos 2
obtain
a __
-
coga)2
(7-17)
dropped the
same
A
sin 2
/
y
(1
+
a
coga)2
will note that, in deriv-
(7-18), we have
sign in each formula.
The validity of this step should be
verified by consideration of the signs
cos a. Thus,
of tan a/2, sin a, and 1
tan a/2 and sin a necessarily have the
ing
a
I
of a/2.
the right hand side of (7-16)
a _
-
cos
V + cos a
(7-14),
rationalize, in turn, the
tan
*/l
a/2 __
~
a/2
and
sign, while 1
.
cos
a
is
r-24
non-
negative.
Example
Find the exact value
7-6.
of
tan 22.5.
Solution: Since the exact values of the functions of 45
(7-16), (7-17) or (7-18). Selecting (7-18),
x
rtrteo
tan 22.50
_
we obtain
_
= xtan 45 = 1~ cos 45
V2
are
known, we
may
use
1
Trigonometric Functions of Sums & Differences
42
Example 7-7. Given tan 2
Find sin a and cos a.
Solution: Since 2
a
is
a.
r
576
=
-=-
where 2 a
>
=
-=-
is
=
and y
7
a second-quadrant angle.
is
24
an angle whose tangent
the terminal side of the angle with x
= V49 +
24
= -
7-3
Sec.
*
24, as
we may
shown
find a point
in Fig. 7-4.
on
Thus,
25. Therefore,
B ina
4
= ,4/1+7 '25
>|/
2
=5-
~7/25
= -3
and
j
cos
a.
-
/I
l\/
'
Example 7-8. Given that tan -
=
O
w, find sin
__ 1
~~
2
1
2
Substituting u for
/
cos a, we find that
i
,
i
i
tan -
we get u
>
^
2
=
If
^
J-
cos
;
1
If
we
substitute this value of cos
COS
i
1
-
=
a
a
cos
a
we
solve this equation for
in
terms of
w.
we have
cos a
-f cos a
Solution: Squaring both sides of (7-16),
ex
a and
-f-
QJ
u 2-
u2
in the relationship sin 2
a
+ cos
2
=
a.
1,
we obtain
Note that we can drop the i sign, just as we did
a always have the same sign.
in (7-17)
and
(7-18), since
tan a/2 and sin
EXERCISE 7-2
In each of the problems from
1
to 8, find the exact functional value
by using an
appropriate double-angle or half-angle formula.
I.
sin 22.5.
5. sin
9. It is
known
cos 15.
2.
67.5.
6. cos
that cos Q
=
3. sin
67.5.
12
?
and
7.
6
is
lo
a. sin 26.
b. cos 0/2.
e. sin 6/2.
f.
10. It is
known
that sin 6
40
and
a. sin 26.
b. cos 6/2.
e. sin 6/2.
f.
II. It is
known
that tan 6
c.
6
is
3
and
a. sin 26.
b. cos 0/2.
e. sin 0/2.
f.
cos 20.
tan 26.
g. sin 36.
6
cos
tan 60.
is
Find
:
d. cot 26.
h. tan 46.
positive in the third quadrant.
c.
cos 26.
= -
tan 26.
g. sin 30.
90.
4.
8.
positive in the second quadrant.
cos 26.
= -
120.
tan 67.5.
Find
:
d. cot 26.
h.
tan 40.
positive in the fourth quadrant. Find:
c. tan 20.
d. cot 20.
g. sin 30.
h.
tan 40.
Sec.
7-4
Trigonometric Functions of
Sums & Differences
1
43
In each of the problems from 12 to 28, write the given expression in terms of a
a multiple of 6. Make use of appropriate formulas to reduce the
answer to as few terms as possible.
single function of
Prove each of the following
29. sin
40-4
31. sin 20
^
1-
=
tan
2
(1
cos 0(sin
+
0/2
cos
7-4.
identities:
0-2 sin
cos 20) tan
3
=
30. cos 40
0),
32. 1
0.
= 2(1- 2cos 0).
4
+ sin
8 cos 4
0-8 cos 9+1.
=
+ cos 0) 2
20
(sin
= sin
+ sin 30.
AS
SUMS,
AND SUMS
sin
PRODUCTS
OF
TWO
FUNCTIONS
EXPRESSED
.
2
cos
sin
2
EXPRESSED AS PRODUCTS
By adding and
subtracting corresponding
members
of
(7-1),
(7-2), (7-3), and (7-4), we obtain
(7-19)
sin (a
+
]8)
(7-20)
sin (a
+
/8)
(7-21)
cos (a
+
]8)
(7-22)
cos (a
+
j8)
+ sin
(a
sin (a
+
-
cos
-
-
(a.
cos (a
-
j8)
=
2 sin
a cos j8,
|8)
=
2 cos
a sin |8,
j8)
=
2 cos
a cos ]8,
]8)
= -
2 sin
a sin /3.
we
reverse these identities, they become the following product
formulas, which express given products of sines and cosines as
sums or differences
If
:
(7-23)
sin
a cos ft =
(7-24)
cos
a sin /3 =
(7-25)
cos
a cos
(7-26)
sin
=
^
z
[sin
(a
+ 0) + sin
(a
I
[sin
(a
+ 0) - sin
(a
\ [cos (a
4
a sin /3 = -
[cos
-
-
+ 0) + cos (a (a
ft],
0)],
ft)],
+ 0) - cos (a -
j8)].
1
44
Sums & Differences
Trigonometric Functions of
Sec.
7-4
To obtain the sum formulas, which express given sums or differences of sines and cosines as products, we first let
a+
Then, solving for a and
=
=
]8
a
and
x
j8
=
y.
we have
/?,
*-*
and
/3=^'
Substituting these values of a and /3 in (7-19), (7-20), (7-21),
and (7-22), we obtain the sum formulas. These are
sin y
=
2 sin
sin y
=
2 cos
+
cosy
=2 cos
-
cos
= -
(7-27)
sin
x
+
(7-28)
sin
a
-
(7-29)
cos x
(7-30)
cos x
a
cos
-y
sin
(^-y^)
-^
cos
2 sin
cos 5
a
sin
as a
sum
of sines.
Example 7-9. Express
sin 3
Solution: Using (7-24)
and replacing a by 5 a and
cos
5a
3a
sin
=
+ 3a) -
~[sin (5a
2
Solution: Using (7-29)
+
n
cos 20
11
T^
=
2 cos
xi
7-11. Prove the
Example
r
n
r
Z
A .
Solution:
sin
7x
cos 7x
-I-
8a
-
sin 2a],
cos
by
20,
we have
=
2 cos 30 cos
=
tan
0.
2i
sin
sin
=
cos 7x
/7a?
(
+
;
5z\
2
5x
=cos 5z
.
Bin
/)
,
x.
/7
(
\
ft
COS
2 cos _
6x sin x
2 cos 6# cos x
__
;
?/
OA
r
cos
\
sin
_
^fl
-^ +--
x-x
i
2 COS
,
o/j
I
identity
rt
rt
= ^[sin
we obtain
cos 26 as a product of cosines.
and replacing x by 40 and
A
4.6 -f
3a)]
a,
-^
Example 7-10. Express cos 40
cos
-
sin (5a
by 3
_
tan x
EXERCISE 7-3
In each of the problems from 1 to
two sines or two cosines.
10, write the given expression as a
sum
or
difference of
2 sin 60 cos 40.
4. sin
28 sin 20.
8. cos
1. sin
30 cos 40.
2.
2 sin 40 cos 20.
3.
5. cos
40 cos 20.
6.
2 sin 65 cos 15.
7. sin
9.
sin 50 sin
0.
10. sin 110 sin 30.
cos 40.
21
cos
31.
7-4
Sec.
Trigonometric Functions of Sums
&
145
Differences
In each of the problems from 11 to 20, write the given expression as a product of
and cosines. Hint: In problems 18, 19, and 20, note that cos 6 = sin (90 - 6).
sines
11. sin 30
13. sin 60
17. sin
80
40
19. sin
40
15. cos
+ sin 20.
+ sin 30.
-
+
+
+
cos
16. sin
-
sin
18.
+ cos 38.
20.
25.
cos 44.
cos
sin
cos
.
A* -r
esc 40
sec 40
cos 20 -f
sin
27.
cos
28. sin
29. sin
30. cos
sin
31.
a
a
+
cos40~
sin
-f cos
33.
=
__
tan
tan
v<*"
a
9A
&*
a
s*n
L
,
-f
_L
e\/\
cot
a
ft
= COS
+ -| ) -
sin
(0
+ 0) -
cos
(-|
cos
+
=
1
rTTJ
sin
+
7;
-
= V2
=
-|)
0)
(-|
sin 0.
\/3 cos
= - V3
0.
sin 0.
-^
2
- ft) =
- /3) =
(a
sin (a
cos
-|)
sin 2
cos
2
a a -
sin 2
sin
= cos
= cos
2
j8
2
j8
ft
2
ft
-
(
a
tan
0.
cos 2 a.
sin 2 a.
;
=
cos
-f
tan
80.
cos 33.
cos
cos
i
sin 29
26.
-
(0
ft)
j.
2
ft
(0
ft)
oo
22. tan
'
T)
cos 30
+ -j) + sin
+ cos
sin (a +
cos (a +
(^
p
^'
(0
sin
32.
ft
~
cos
-
sin
identities:
= COB 30.
sin 20 4- sin 40
25.
30
sin 64
20. sin 65
= V2
,
23.
cos 40.
+ sin 20.
40
14. sin
Prove each of the following
21. sin
-
12. cos
n
~~
ft)
gin
p
*
Graphs of Trigonometric
O
8-1.
Functions; Inverse
Functions and Their Graphs
VARIATION OF THE TRIGONOMETRIC FUNCTIONS
In Section 3-2, the trigonometric functions were defined in terms
of the coordinates (x, y) of the
point P(t), w here the number t
represents the directed length of
the arc of a unit circle measured
r
from the point (1,0). Later, in
Section 3-9, an equivalent definition was given in terms of an
(-1,0)
angle 9 in standard position. We
shall now consider the variation
of the trigonometric functions in
the four quadrants as f, and with
it 9, increases from
to 2n. From
Fig. 8-1, we can read off the variations shown in Table 8-1.
FIG. 8-1.
For example, by noticing the changes
tinuously from
to
27T,
we
find that sin
t
in y as
varies
t
increases con-
from
to 1 in the
in the second, from
1 in the third,
to
quadrant, from 1 to
and from -1 to in the fourth. Similar considerations lead to the
results for the cosine and tangent.
first
Recalling that esc
t
-
sin
>
we know
that
if
either of these f unc-
t
tions increases the other decreases.
Hence, the variation in esc t
can be determined from the variation in sin t. Similarly, we may
learn about the variation of sec t from that of cos t, and about the
from that of tan t.
Example 3-2 that tan
variation of cot
We found
that tan
t
in
t
has no value when
t
=
?r/2 is undefined,
7r/2.
146
which means
For the sake of
easier tabu-
See.
8-2
147
Trigonometric Functions; Inverse Functions
TABLE
8-1
Variation of Trigonometric Functions
lation of this result in Table 8-1
we have employed
the
much
used symbols oo (infinity) and -co. These symbols merely signify
that in the neighborhood of Tr/2 or one of its odd multiples the
value of tan t is very large numerically. They are not to be used as
numbers.
8-2.
THE GRAPH OF THE SINE FUNCTION
To construct a graph representing the variation of the sine, we
x denote a real number or the value of an angle measured either
let
in radians or in degrees, and we let y denote the corresponding
value of the function. Corresponding values of x and y are plotted
FIG. 8-2.
148
Trigonometric Functions; Inverse Functions
Sec.
8-2
as points on a rectangular coordinate system. We can infer the
general appearance of the curve from the results summarized in
the preceding section, but an exact representation is more readily
obtained by using a table of sines.
Let us now construct the graph ofy = sin x from x = 7r/2 to
x = 2?r. The following values, found by the methods of Section 3-2,
are used to obtain the curve in Fig. 8-2.
While the choice of a scale is arbitrary, a better proportioned graph results if the same unit of length is used on both
axes. The unit so selected will represent the number 1 on the y-axis
and one radian on the #-axis. In terms of this unit a suitable length
can then be marked off on the x-axis to represent 2-rr or 360.
8-3.
THE GRAPHS OF THE COSINE
AND TANGENT FUNCTIONS
Using the table of trigonometric functions, the student should
table of corresponding values for y = cos x and one for
make a
FIG. S-3.
Sec.
84
149
Trigonometric Functions; Inverse Functions
7T/2
27T
37T/2
**tzn
x
FIG. 8-4.
y = tan #, similar to that used for y = sin # in Section 8-2. Study
the graphs in Fig. 8-3 and Fig. 8-4 on the basis of the tables you
have made.
If we compare Fig. 8-3 with Fig. 8-2, we see that the graph of
=
cos x may be obtained from the graph of y = sin x by moving
y
the graph ofy = sin x to the left a distance of rr/2 units. This fact
can be checked by using the relationship cos x = sin (# + ir/2),
from which
it
follows that cos
=
sin 77/2, cos vr/G
=
sin 27T/3,
and
so on.
8-4.
PERIODICITY, AMPLITUDE,
AND PHASE
The trigonometric functions are among the simplest of a large
which are periodic. As a preliminary to defining
periodic functions, we shall call attention to some examples of
phenomena which recur periodically, such as the rotation of the
class of functions
its axis, sound and water waves, the vibration of a
and
many other vibratory and wavelike phenomena. The
spring,
behavior of the object involved in a phenomenon of periodic nature
earth about
determines the type of function that is required to represent it
properly. We note particularly that, because of the recurrence
characteristic of such a phenomenon, the values which the function
assumes in any interval of given length are also taken on in any
other interval of the same length. This statement apparently indicates that a function of x is periodic with period p if, for every
value of x, the function returns to the same value when x is
increased by p.
More
specifically,
we
state the following.
1
50
Trigonometric Functions; Inverse Functions
A function f(x)
Definition.
zero
number p
is
Sec.
said to be periodic if there
is
84
a non-
for which
f(x
+ p)
=/(x)
for all numbers x in the domain of f(x). Any such number p is
called a period; the smallest positive number p satisfying the
requirement
the period.
is called
Evidently,
if
p
the period of f(x), then
is
np
is
a period, for
every integer n.
Periodicity of Trigonometric Functions. No matter which of the
is considered in the definition of the trigonometric
two viewpoints
functions,
we
shall see that, if
any trigonometric function
g t ^ 2ir, then
of (t + 2?r) = same
function of
t.
According to the definitions given in Sections 3-1 and 3-2, P(t)
and P(t + 27r) represent the same point on the circumference of
the unit circle. To locate P(t) we start at (1,0) and proceed
around the circle in the proper direction a distance of \t\ units. To
locate P(t + 27r) we continue another 2?r units from the point P(t)
This merely adds another complete revolution, and we arrive at the
same point P(t).
If we consider the definitions of the functions in terms of angles,
+ 2n is coterminal
as given in Section 3-9, we note that the angle
with
and that any trigonometric function has the same value for
.
coterminal angles.
Period of Sine, Cosine, Cosecant, and Secant. From a study of the
1
it is apparent that sin x assumes all values between
In
2ir
units.
and +1 as x, starting from any value, varies through
other words, the graph of the function repeats itself during each
interval of length 27r, for positive and negative values of x. Or
sine curve,
stated
more
concisely,
sin (x
This
is
+
2?r)
=
sin x.
equivalent to saying that sin x
is
a periodic function of
x,
remains now to show that 2ir or 360 is
the smallest positive number p for which sin (x + p) = sin x and
for which cos (x + p) = cos x.
and that
2?r is
a period.
It
Since, by definition of a period p, sin (x + p) = sin x for any value
x 9 we shall select for the purpose of our proof the particular value
x
= ir/2. We
then have
.
sin
But
since sin f -5-
+
pj
/7T
(-
=
*
\
+ P) =
cos p,
it
.
sm
7T
= L
1
2"
follows that cos
p
=
1.
Sec.
8-4
Trigonometric Functions; Inverse Functions
151
Hence, p must be an even multiple of TT. The smallest even multiple
TT is 27r, which must also be the smallest positive period of the
of
sine function.
In a similar manner,
we
find that 2rr is also the period of the
Because of the reciprocal relationships existing
between the sine and cosecant and between the cosine and secant,
27r is also the period of the cosecant and the secant.
cosine function.
Period of Tangent and Cotangent.
tangent, we write
+ p) =
tan (x
and we
let
x
=
0.
A
of the tangent.
of the cotangent.
To
tan x
find the period of the
}
Then tan p =
similar
0, and we find that TT is the period
argument shows that TT is also the period
We
have just seen that the period of sin x is
27T.
We shall now determine the period of sin bx, where 6 is a
positive constant. That is, we want to know the smallest positive
change in x which will produce a change of 27r in bx. If p represents this change in x, we can find p from the relationship
Period of sin bx.
b(x
Solving for p,
+
p)
we immediately
P
=
bx
+ 27T.
find that
=
27T
T
-
Thus, the period of sin bx is equal to the period of sin x divided by
b, that is, 2;r/&.
Similarly, it can be shown that 27T/6 is also the period of cos bx,
esc bx, and sec bx. It is also true that the period of tan bx and
cot bx is 7r/b.
FIG. 8-5.
1
52
Trigonometric Functions; Inverse Functions
Sec.
8-4
Let us consider the graph ofy = sin 2x shown in Fig, 8-5. Since
the period of sin 2x is 2ir/b = 2rr/2
TT, the function will assume
to TT that sin x takes
the same range of values in the interval from
in the interval from
to 27r.
Note that the graph of y = 2 sin x is similar in form to that of
=
sin x, which is shown in Fig. 8-2. The period is 2ir for both
y
curves. However, for any given value of x, the corresponding value
of y in y = 2 sin x is twice as large as is the corresponding value of
y in y
=
sin x for the
same value
of x. In the
graph ofy
=
a sin
x,
where a > 0, the greatest value of y is a, and the smallest value of y is
a. The constant a is called the amplitude of the function or of the
graph. Thus, the amplitude of the graph of y = sin x is 1, while that
of the graph of y = 2 sin x is 2.
a sin x, where a is a real number,
In general, for the function y
the amplitude is equal to |a| and the period is equal to 27r. Also, in
general, for the graph of y = a sin bx, where a and b are real, the
amplitude is |a| and the period is
Phase Angle. Since the graph of y = cos x may be obtained from
the graph ofy = sin x by shifting it to the left a distance equal to
7T/2 units, we say that the graph ofy = cos x differs in phase by
sin x. The amount of horizontal dis77/2 from the graph ofy
placement of two congruent graphs, amounting to Tr/2 radians in
this case, is called the phase difference, or the phase angle. The
amplitude and the period are the same for y = cos x as for y = sin x.
y
cos
2x
- sin (2x +
7T/2)
= sin
2 (x
+
7T/4)
FIG. 8-6.
Now
8-6.
consider the graph of y
This curve
in Fig. 8-5
may
- sin
2(x
+ ~)
in Fig.
ofy =
sin 2x
evidently be obtained from that
7r/4 units to the left in the x direction.
=
sin (2x
+ ~)
4U
sin
2x
by a shift of
Hence, the graph of y
ofy =
= cos
2x by
7r/4 radians.
differs in
phase from that
Sec.
84
A
153
Tr/gonomefric Functions; Inverse Functions
simple method for finding the phase displacement
is
the point near the origin for which the function sin
equals zero
that
;
is,
to locate
(2x+^)
to find the smallest numerical value of x that
makes the quantity x + j zero. We have then sin2(#-f ~) =0
when x + - = or when x = 7r/4. Hence, the phase difference is
and the
7T/4 radians,
sign of x
is
shift of the
graph
toward the
is
left since
the
negative.
can be shown that, in general, the phase displacement of any
trigonometric function of (bx + c), where b > 0, is to the right or
It
left
by
The
radians (or degrees).
\c/b\
ment depends on whether c/b
direction of the displace-
negative or positive.
conclusion
that for the graph of y =
we
arrive
at
the
Finally,
a sin (bx + c) the amplitude is \a\ and the period is 27T/6. Also, its
phase differs from that ofy = sin x by c/b.
Because of
usefulness in
its
is
many
we
applications,
shall illustrate
by means of an example a procedure for reducing an expression
+ B cos & to the form a sin (9 + a)
of the form A sin
.
Example
Reduce
8-1.
A
+B
sin
+ a).
to the form a sin (0
cos
Solution: Write
A
+B
sin
= \/A 2 + B 2
cos
(~ _ 2
\\/A
The absolute values of the coefficients
An
'
-f
B*
B2
=
+
sin
+
VA + B
2
=
and
\/A 2
-f
B2
2
cos 0^
/
cannot be greater
1, and the sum of their squares is 1. Hence, they may be taken as the cosine
and sine, respectively, of some angle a. Therefore, the expression A sin -f B cos B
becomes
than
+ B2
where
A =
a cos
(cos
a and
a
sin 6
=
J5
Example 8-2. Express 4y
and sketch the graph.
Solution; Since
A =
sin 26
If
1
-f-
and
+
sin
a
cos 0)
= \M 2
cos
angle and
a
may
=
^
and
=
sin 26
B =
a
+ a),
sin (6
- V3 cos 26 in the form y =
\/3,
\/3 cos 20
sin
B2
a sin a.
=
we have \/^ 2
2^ sin 20
we identify this result with the expression \/A*
we have
4-
=
be taken equal to
~~
ir/3.
-f
-
^
B2
-h
cos
(cos
It follows that
#2 =
2.
+ a),
Therefore,
$
20Y
a sin 20
a
a sin (60
is
+ sin a cos 20),
a fourth quadrant
154
Trigonometric Functions; Inverse Functions
We have,
finally,
y
The amplitude
-
-
_
= i(sin 26 - V3
of this function
is
r
4
=
cos 29)
its
>
|
period
Sec.
8-4
.
-
sin
is TT,
^20
and
yJ
its
phase angle
is ir/6.
FIG. 8-7.
Since the interval from (TT/G, 0) to (77T/6, 0) is one period in length, the part of the
curve obtained for this interval may be repeated indefinitely in both directions to
give the complete curve.
we have employed
The curve
is
shown
different units of length
in Fig. 8-7.
(Here for convenience
on the two axes.)
EXERCISE 8-1
In each of the problems from
and phase angle
to 24, find the period, amplitude,
1
of the trigonometric function.
1.
3 sin
0.
4. cos =
7.
4 tan \ 8
o
10. sin
3.
2.sin|.
13. cot 60.
6.
2 cot
8.
si
9.
cos0-
19. esc (ir6
22. 6
-
+
7).
cos 40.
|
4
11. 3 sin 50.
12. 5 tan 7T0.
14. sec 90.
15.
+ 4).
17. tan (TT
i
0.
= sin 7 6
4
3
37r
16. 3 esc
cos
5.
o
7T0.
^
20. cot (2?r
23. 5
+3
sin
\
40.
18. cos (30
TT).
(26
5 cot
-
-
o
,
-
21. 2
+ sin 0.
24. 4
+2
2).
cos (20
\
+ -|
o
,
In each of the problems from 25 to 30, sketch the graph of the given function
by constructing a
25.
28.
y
= cos
y
= 5 cos \ x
x
6
table of values.
26. y
= tan
29. y
=
^
2 tan
z
|
y
=
2 sin
30. y
=
3 sin 2x.
27.
3.r.
85
Sec.
Trigonometric Functions; Inverse Functions
1
55
Sketch each of the following graphs without constructing a table of values.
1
2
33. # = tan
x.
31. y = sin ^ a:.
32. y = cos 4z.
&
6
=
34. y
2 cos
35.
3.r.
?/
=
3 cos
36.
irx.
=
*/
r tan 3x.
4U
37. y
=
40.
!/
=
cos
43.
?/
=
sin (2x
46.
?/
=
sin
48. y
=
\/3 sin 20
50. y
= V2
8-5.
INVERSE FUNCTIONS
I
sin
3.
1
fa?
+
+
cos
2).
+
1).
cos
38. y
=
41. y
=
sin
44. y
=
cos (2wx
47.
0.
+ V5
30-3
|
?/
cos
-
(x
sin 6
*
^)
-
=
5 cos
42 * V
=
cos
(
45. y
= sin
f
39.
x.
|
-
cos
TT).
?/
sin 30.
~
3x
j
=
13
51.
=
2 sin
y
cos
^~7
-
2 )-
+ 7V
x
49. y
0.
cos 20.
(x + ~
sin
^
cos 20.
we defined a function by setting up a rule of
between
twa sets of numbers, X and 7, called the
correspondence
domain of definition of the function and the range set of the funcIn Section 2-3,
respectively. The function was called single-valued if just
one number y of the set Y is assigned to each number x of the set X.
If more than one number of Y corresponds to some value of x, the
function is multiple-valued.
tion,
pose a reverse problem. We
corresponding values of x.
Naturally, y is limited to lie in the range of the given function,
since otherwise no x exists. The function which makes correspond
to each such y all values of x for which y = f(x) is called the
inverse function corresponding to the given function.
We shall begin our discussion of inverse functions with an
is the set of all real numbers, Y is the set of
example in which
If
we know
assume y
that y
to be given
f(x),
we may
and ask for
all
X
all
non-negative real numbers, and the correspondence
is
deter-
mined by the relationship
y
=
x2
.
Ordinarily, we assign values to x in order to calculate values erf
x 2 In this case, we have a rule of correspondence that assigns just
one number y to ea$h chosen number x. Hence, y is a single-valued
function of x. The graph ofy = x 2 is shown in Fig. 8-8.
Assume now that y is given and that we wish to determine corr^sponding values of x. To do this we solve the given equation
x 2 = y for x, and obtain two numbers x = \/y and* x = ^/y cor.
s
responding to every, non-negative number
y.
Since the rule of
156
Trigonometric Functions; Inverse Functions
Sec.
8-5
correspondence assigns two values of x to each chosen number y,
see that # is a double-valued function of y. In this case, the
admissible values of y are restricted to zero and the positive real
numbers, while those of x comprise all real numbers, as was indi-
we
cated in the specification for the sets
X and
Y.
O
FIG. 8-9.
FIG. 8-8.
In the study of mathematics,
we
generally prefer to use the sym-
bol x to represent the independent variable and y to represent the
dependent variable. To be consistent with this preference, we shall
x 2 and y =
inverse of the other.
call
y
=
.^/x inverse functions, each being called the
The graph ofy = \/Hc is obtained by plotting that of x = y 2 as
shown in Fig. 8-9. We note that the roles of x and y are interchanged
in the two equations y = x 2 and x = y 2
Thus, we see that the
,
.
curve in Fig. 8-9 is actually the curve of Fig. 8-8 with the axes
interchanged and one of them reversed in direction.
8-6.
INVERSES OF THE TRIGONOMETRIC FUNCTIONS
The Inverse
Sine.
Let us consider the function y
= sin x and
attempt to apply a discussion similar to that in Section 8-5. Here,
as has been noted, the domain is the set of all real numbers, while
is the interval
1^7/^1. Referring to the graph of
x
in
and
recalling the periodic properties of this
y
Fig. 8-2,
graph, we see that, for every given number y such that 1 :f y 2S 1,
there are infinitely many values of x such that y = sin x. To designate the totality of all values of x such that y = sin x, we write
the range
=
sin
x
which
is
read x
fully that the
is
sin
sin" 1
or esc y.
t/
sin""
1
y,
the inverse sine of
symbol
which equals -
=
y.
The student should note
care-
l
y must be distinguished from (sin y)~
9
Sec.
8*6
Trigonometric Functions; Inverse Functions
Another notation that
function
is
57
frequently used to represent this inverse
is
=
x
which
1
arc sin y,
read x is the arc sine of y.
In order to conform to the preferred practice of considering y
as a function of x, we may designate the inverse of the sine func-
tion
is
by writing
=
y
sin" 1 x
or
y
=
arc sin x.
We
note the following properties of this inverse function.
# is a number such that \x\ > 1, then y does not exist. This
property follows from the fact that the sine function takes on only
the values from -1 to 1. Hence, the inverse sine function sin* 1 x is
defined only when
1 ^ x =i 1. For example, sin' 1 2 is not defined,
If
since there
If # is a
is no number or angle whose sine is 2.
number such that \x\ ^ 1, then y certainly
exists.
More-
over, because of the periodicity of the sine function, there are
*
x corresponding to every such
infinitely many values of y = sin
value of x. For example, it x = 1/2, then y = sin' 1 1/2 means that
1/2. Then y may be
y is any number or angle such that sin y
from
these by integral
taken as ?r/6, 5-77/6, or any value that differs
1
multiples of 27r. The totality of these values of sin- 1/2 may be
represented as
+
2n?r
+
and
2mr, where n
=
0,
2,
1,
.
convenient to plot the graph of y = sin' 1 x for a
further study of the inverse function. Since y = sin' 1 x and x = sin y
express exactly the same relation between x and y, the graph of
Fig. 8-2 may be used as a graph of the inverse function. We obtain
the graph ofy = sin* 1 x simply by interchanging the axes in Fig.
8-2 and reversing one of them in direction. The result is shown in
We
shall find
it
Fig. 8-10.
The question now
arises
into intervals within each of
ing to each x such that
ing a first interval
\x\
whether the t/-axis can be subdivided
which y has just one value correspond-
^
One way
1.
TT
^
^
of doing this is
by
select-
TT
-2 ***2"
Other intervals, such as -^
z
selected.
With the
^
y
^
-
z
and
~ w
O
z
g
y
entire y-axis thus subdivided,
the graph ofy = sin" x as consisting of
many single-valued functions or branches.
1
~
^
z
we may
>
are then
think of
the graphs of infinitely
Sec.
Trigonometric Functions; Inverse Functions
158
8-6
3TT/2
37T/2
7T/2
O
-i
i
O
-1
/I
-7T/2
"-7T/2
FIG. 8-11.
FIG. 8-10.
To avoid any ambiguity
in later applications as a result of this
multiple-valued property of the inverse sine, we shall often restrict
y so as to make the function single-valued. There will then be but one
value of y corresponding to each value of x such that sin y
shall
determine this value from the branch for which
^
y
x.
^
We
-
called the principal branch. The values of y chosen
this branch are called the principal values of the inverse-sine
This branch
from
--
=
is
function and are represented by the equation
y
Note that
=
Sin" 1 x
or
y
=
Arc
sin x.
in this case the initial letter of the
name
is
capitalized.
This restriction to the two quadrants containing the smallest
numerical values of y results in a single-valued function y = Sin- 1 x.
f or x in the interval
The values of y are such that ^ y ^
1
-1 g x ^ 1. For example, we have Sin- (-1) = -7r/2, Sin- 1 (-1/2) =
1
= 0, and Sin- 1 1 = ir/2.
-7T/6, Sin-
The Inverse Cosine. We shall next consider the function y = cos x.
With the help of Fig. 8-3, we see that the range of y = cos x is the
1 S y ^ 1, and that for every y in this interval there are
interval
values of x such that y = cos x. We are thus led to
the inverse cosine function, which makes correspond to y all the
values of x such that y = cos x. If again we interchange the symbols x and y, we may write the inverse cosine function as
infinitely
many
y
=
cos~ x x
or
y
=
arc cos x.
Sec.
8-6
159
Tr/0onomefr/c Function^; Inverse Functions
This inverse function has the following properties.
If # is a number such that \x\ > 1, then y does not exist, because
there is no value of y for which |cos y\> 1.
If x is a number such that \x\ ^ 1, then there are infinitely many
values of y designated by cos' 1
The graph ofy =
cos* 1 x is
x.
shown
in Fig. 8-11.
The method for
=
similar to that used for graphing y
sin' 1 x. We first
write x
cos y. We then plot a cosine curve by proceeding as for
Fig. 8-3, except that values of the independent variable y are laid
off on the ?/-axis.
plotting
it is
The principal branch of the curve in Fig. 8-11 is the portion of
the curve for which
g y ^ TT. It is represented by the principal
value of the function, which is denoted by
=
y
Cos" 1 #
or
y
=
Arc cos
x.
The Inverse Tangent. The inverse tangent function
=
y
1
tan" x
or
y
=
is
arc tan x.
7T/2
-2
-3
o
-1
-7T/2
-7T
FIG. 8-12.
represented by the graph in Fig. 8-12. We note that there are
infinitely many valu.es of y for every value of x.
The principal branch of the graph of y = tan- 1 x is the portion
It is
for which
~
<
y
y
<
~
= Tan"
This
1
x
is
or
represented by the equation
y
=
Arc tan
160
Trigonometric Functions; Inverse Functions
Sec.
8-6
The Inverse Cotangent, Secant, and Cosecant. The other inverse
trigonometric functions are
1
or
y = cot" x
y = arc cot x,
:
y
?/
=
=
sec" 1
a:
or
esc" 1
a:
or
y
y
=
=
arc sec x,
arc esc #.
\Y
FIG. 8-13.
show only the graph ofy = cot- 1 x. See Fig. 8-13. The
< y < TT.
principal branch of this curve is given by
Principal Values of the Inverse Cosecant and Secant. The selection
of principal values of ?/ = esc- 1 x and y = sec- 1 x is by no means uniform among all authors. Some writers adopt the range between
and 7T/2 for both functions when x is positive, and between TT and
7T/2 when x is negative. The authors prefer, however, to use the
We
shall
definitions
Csc- 1 x
=
Sin- 1
Sec" 1 x
=
Cos" 1
and
(-)
We
have, therefore, the following ranges of principal values of
the inverse trigonometric functions
:
7T
-o- SI
1
Sin" x
^
7T
>
^
Cos" 1 #
< Cor
-
^
Csc- 1 *
^
-J
1
(Csc^ x
*
0),
^
1
a;
Sec" 1 x
^
TT,
<
TT,
^
TT
1
(Sec"
x
^ y)
Sec.
8-6
161
Trigonometric Functions; Inverse Functions
34
(Ox
Sin- 1 =
o
Solution: Let Sin- 1
3
=
-=
Then
6.
5
is
is
used,
o
3\
1 Hence, cos (SinJ
/
=
=
Let Tan' 1
Solution:
and
(
-
x)
=
and cos
=
Since only the
5
a first-quadrant angle and cos
cannot equal
4
-
Example 8-4. Find the value of sin [Tan- 1
x being a positive number.
=
>
5
1
principal value of sin- =
4
=^
sin
=
(
-
x)],
Then tan
0.
6
between
w/2 and 0. If angle
constructed in standard position, as shown
x,
6 is
lies
/
in
Fig.
then sin 6
8-14,
1
Hence, sin [Tan"
(
-
x)]
is
found to be
=
FIG. 8-14.
Example 8-5. Find Arc cos
Solution: Let Arc cos (cot
(cot 60).
60)
=
Then
6.
=
cos 6
cot 60
=
.5774,
by Table
II.
Therefore, the value of 6 is Arc cos .5774. Since 6 is restricted to the interval from
to 180, 6 must, in this case, be a first-quadrant angle. Hence,
Arc cos .5774
5444'.
5444', and Arc cos (cot 60)
=
=
=
EXERCISE 8-2
In each
-
<
of the problems
^_ 5
K
O/y.
from
1
to
O-r
y
=
2x
8.
Tan- 1
12. Sin- 1
0.
(
16.
19.
9.
- J)
&/
Cos- 1
Cos- 1
(
1
+6
3
4
2'
22, find the value of the given expression.
1.
V
Tan-'
3
In each of the problems from 7 to
15.
y=rnx+b.
3
6
5.
Cot- 1
find the inverse of the given function.
1
1
3
11.
6,
_/*-!
-
13.
17.
Arc cos
Cot- 1
Tan-
-^
^
(
- V).
Tan-
Cos- 1 (-
1).
14. Csc-i 1.
o
1
20.
10.
1
[sin
18.
Arc tan (-
(-
ir/2)].
1).
21. Sin22. Tan- (-1.414).
^-0.414).
In each of the problems from 23 to 37, evaluate the given expression.
1
1
23. tan
26. cot
1
(sin- j|)
Sin- 1
24. sin
27. cos
1
(sin- 1)
1
(Cos' 1)
25. sin
281 tan (Sin- 1 .6450).
162
Sec.
Trigonometric Functions; Inverse Functions
30. tan (Cot- 1 2).
29. sin (Cos- 1 .9200).
32. sin
1
33. cot
(Cot7
V
4
1
1
34. sin
(Cot- 1)
37. Sin- 1
36. cos f Sin- 1 5
\
o
35. cc
1
Sin- i)
(
5/
\
31. sec
(Cos- 1)
8-6
(tan
^
In each of the problems from 38 to 56 simplify the given expression, taking u as
a positive number.
38. Cos- 1 (sin u).
In 38 to 40, 54, 56,
u O/2. In
44, 46, 48, 55, u <1.
39. Sin- 1 (- sin u).
40. Sin- 1 (cos u).
41. esc fsin- 1
\
42. sin (Sin- 1 u).
43. tan
(Tan-
i
u
1
u).
_
VI +
u
45. cot ("Tan- 1
-
V
U
47. tan ( Sin- 1
\
49. tan
48. esc (Sin50. cos
(Tan-
1
54. Sin"
56.
Cot-
v
u).
53. sec (Sin- 1
55.
(sin u).
u
1
+u
51. sec (Cos- 1 u).
52. cot (Sin- 1 u).
1
l
(Cos~
V
Cos-
1
14).
(cos
\
(tan
In each of the problems from 57 to 65, draw the graph of the given function by
changing from the inverse function to the direct function and using the period,
amplitude, and phase angle of the function to assist in plotting.
57. y
= --sin-
59. y
=
1
x.
58. y
=
a:.
60. y
= ^ sin-
Q
cos- 1
2 tan- 1 x.
1
-
sc
1.
Zi
61. y
= - cos-
63. y
=
65. y
= 4 tan-
x
1
o
(Cos-
+ 2.
+
1
62.
64. y
1).
69.
70.
71.
< u g 1.
1.
<u
Prove that cos (Tan- u) > 0, if u ^ 0.
Prove that tan (Cos- it)
< u 5 1.
0, if
Prove that Sin- u + Sin- (- ti) = 0, if - 1
Prove that Tan" w + Tan- ( - u) = for all
72. Prove that
=
tan- 1 z
4- 3.
3 sin- 1 (2x
+
^
0, if
0, if
1
1
u
1
1
1
1
Tan- 1 u
73. Prove that Cos-
74. Prove that
=4
1
66. Prove that sin (Cos- 1 u)
67. Prove that cos (Sin- 1 it)
68.
?/
1
Tan-
1
u
tt
+ Tan4-
-
1
(l/i*)
=
y
Cos- (-u)=ir,
1
Tan~*
v
= Tan-
>
if
1.
values of u.
w >
0.
if-lgul.
"
1
j**
J
'if
ti
>
and
v
>
0.
1)
-
2.
V linear Equations and Graphs
9-1.
SOLUTIONS OF SIMULTANEOUS EQUATIONS
Often problems arise that involve two or more unknowns and as
many equations. The solution of such a problem requires the determination of numbers which simultaneously satisfy the given equations.
Equations for which we seek
common
solutions are referred
to as a system of simultaneous equations. If the system has at least
one solution, it is said to be consistent; otherwise, it is called
inconsistent.
Let us investigate the following pair of simultaneous linear
equations
:
aix
+
biy
=
f
a,
(
a2x
+
b2 y
=
c2
(9-1)
.
x and y which satisfy both
and
equations, excluding from consideration the cases a\ = &i =
It is desired to find all pairs of values of
a2 =
&2
=
0.
We
proceed by multiplying both sides of the
and both sides of the second by bi, obtaining
b 2 biy
=
first
equation by & 2
c 2 bi.
Subtracting the second equation of this pair from the
obtain
(9-2)
If (aib 2
(aib 2
02&i)
^
0>
i
first,
we
a2 bi
we ^ nd that
_
we see that we have
which may exist.
Thus,
exactly one value for x in any solution
163
1
64
and Graphs
Linear Equations
Sec.
91
multiplying the original equations by a? and a lf respectively,
the first equation thus obtained from the second
subtracting
^nd
one, we obtain
By
-
(ai& 2
(9-3)
-
If (ai&2
a2 6i)
T* 0,
we
=
a 2 bi)y
a2 a.
find that
4i
CL 2C\
d\C2
~~ _____________
a\b 2
Again, we have
-
aic 2
CL 2
bl
exactly one value for y in any solution which
may
exist.
Hence, if (di& 2
&2&i) is not zero, we have at most one solution
of the pair of given equations, namely,
'
x
(9-4)
=
~
l
-
l
y
>
Substitution of these values for x and y in (9-1) shows that we
really have a solution. The reader should verify this solution by making the substitutions.
Consistent and Independent Equations. If (eii&2
e^&i) ^ 0, we
have exactly one solution of (9-1), which is expressed by (9-4),
and the given equations are called consistent and independent. We
may consider, as an example, the pair of equations
=
24,
5x-2y=
22.
2x
+
3y
I
\
Here a^b 2 - 0361 = (2) (-2) and y = 4 give a solution. In
(5) (3) = -19, and the values x = 6
other words, when these values are
substituted in the two given equations, both equations are satisfied.
Thus,
2(6)
+ 3(4) =
24,
-
22.
I
I 5(6)
-
0261
=
If di
^ 0,
division
Cases with ai& 2
di&2
&2&1
=
0.
2(4)
0.
,
02
=
Let us now consider cases in which
=
by
2
di gives
,
01.
01
We then
define k
=
>
and we have
ai
a2
(9-5)
=
Now
let ai
tion,
and since
k
=
-2.
,
we
0.
=
Then, since
a^bi
=
0,
fan,
61
62
=
fc&i.
cannot be zero by our initial assump0. In this case, if we define
we have a2 =
see that (9-5) again follows. Thus, ai& 2
- a2 bi =
has
Sec.
9-1
shown
been
and Graphs
Linear Equations
to
mean
proportional.
note also that
We
that
if
ai& 2
the
0261
members of
left
=
1
(9-1)
65
are
equations (9-2) and (9-3)
0,
become
f
X
=
Ci&2
(
y
=
aic 2
(9-6)
C2&1,
-
a2 ci.
It is clear that these equations cannot be satisfied by any pair of
values of x and y, unless both right members are also zero. Accordingly, if we wish to determine whether or not the given system has
a solution when ai& 2
&2&i = 0, it becomes necessary to take into
account the two cases of the right sides of (9-6) being zero or not
zero.
Consistent and Dependent Equations. In the
first
case referred
to in the preceding paragraph, a,ib 2
o^bi = 0, and Ci& 2
^2&i and
a2 Ci are also equal to zero. Hence, substituting the values
a,iC 2
of & 2 and 02 from (9-5), we have
2)61
(ci/b
=
c 2 )ai
(ci/fc
=
0.
Since a lf bi are not both zero, it follows that c 2 = kci. This means
that one of the original equations is a constant multiple of the
other, and any pair of values of x and y that satisfy one equation
Under
will also satisfy the other.
this condition, the equations are
and dependent.
as an illustrative example the equations
said to be consistent
We have
f
(
2x
+
3y
4z
+
Qy
=
24,
= 48.
(3) = 0.
Here ai& 2 - a2 b l = (2) (6) - (4)
Also, the coefficients of
the unknowns and the constant term in the second equation are
multiples of the coefficients of the unknowns and the constant term
in the first, and the multiplier is 2 for all three terms. Infinitely
many solutions exist. Some of them are x = 0, y = 8 x = 12, y =
:
and x
=
t,y
04 - 2t
o
,
where
t is
;
any number.
;
%
Inconsistent Equations. Finally, let us suppose that in the origand at least one of the numbers
a^bi =
equations aj)*
inal
Ci& 2
~
c 2 6i, aiC2
a 2 Ci
is
different
from
zero.
Hence, no solution of
(9-1) exists and the equations are inconsistent. This case is characterized by the condition that one of the numbers (cik
<%) 61,
(cik
02)0,1 is
not zero, in view of (9-5).
It follows that c2
^ kci*
1
66
and Graphs
Linear Equations
Hence, the multiplier for the
constant terms.
jto the
members
left
Sec.
91
of (9-1) does not apply
Consider, for example, the equations
2x
+
3y
=
24,
4x
+
Qy
=
7.
I
1
coefficients of the unknowns in the second equation are
of
those in the first equation, and the multiplier is 2 for
multiples
both terms; however, the multiplier for the constants is not the
Here the
same as that for the other terms.
Hence, these equations are
inconsistent.
9-2.
To
ALGEBRAIC SOLUTION OF LINEAR EQUATIONS IN
TWO UNKNOWNS
and independent system of two linear equatwo unknowns, we reduce the system to one equation in one
unknown by eliminating one of the unknowns. The following
example will illustrate two commonly used methods for eliminating
the unknowns.
solve a consistent
tions in
Example 9-1. Solve the equations
2x+3y= 24,
5x
Solution:
Since a\b 2
a 2 &i
=
-
2y
=
22.
(-2) -
(2)
(5)
(3)
= -
19, the
equations are
and independent, and there is but one solution. If we use the method of
elimination by addition or subtraction, the procedure is the same as that indicated
in obtaining the solution (9-4) from equations (9-1).
To eliminate y multiply the first equation by 2 and the second by 3, in order to
consistent
t
make
the coefficients of y numerically equal in both equations.
4z
+
Qy
= 48,
5x
-
6y
= 66.
Adding, we get
Solving for
Now,
x,
19$
=
114.
x
=
6.
we have
substitute 6 for x in the
3y
first
= 24
-
of the original equations.
2x
=
24
-
12
=
We
thus obtain
Then
12,
or
y=4.
Alternate Solution: If
by
solving the
first
we use the method
of elimination
equation for y in terms of
24
x.
-2x
by
substitution,
We thus get
we begin
Sec.
We
9-3
Linear Equations
Solving for
x,
-
2x
24
then substitute
~
for
and Graphs
1
67
y in the second equation and obtain
we have
Ux -
-
2 (24
2s)
= 66,
or
-
15x
+ 4s =
48
Hence,
19z
=
66.
114,
and
Substituting 6 for
x,
we
as before,
x
=6.
find that y
= 4.
A
Example 9-2.
grocer has some coffee selling at 80 cents per pound and some
at 90 cents per pound. How much of each must he use to get a mixture of 100
pounds worth 86 cents per pound?
Solution: Let x
= number of pounds of 80-cent coffee, and y = number of pounds
of 90-cent coffee.
Then
x
+y =
100,
and
+ 0.90y =
O.SOz
Simplifying,
x
+
y
z
+
9y
These equations have the single solution x
LINEAR EQUATIONS IN THREE
9-3.
0.86 (100).
we have
=
100,
= 860.
= 40, y =
60.
UNKNOWNS
In the solution of a system of three equations in three unknowns,
one method is to employ the following steps, which we do not justify
here
1.
Eliminate one of the unknowns from a pair of the equations;
then eliminate this same unknown from another pair of the
:
2.
original equations.
Solve the resulting
two equations for the two remaining
unknowns.
3.
Substitute the values found in step 2 in any one of the original equations to find the third
Example 9-3. Solve the system
2x
x
3x
unknown.
of equations
+ 3y - 2=5,
- 5y + 2z = 1,
+
y
-
42
= -
1.
168
and Graphs
Linear Equations
Solution: Eliminate z
from the
first
5x
Now
Sec.
and second equations by addition to obtain
+
y
=
11.
eliminate z from the second and third given equations
-
5z
9-3
90
=
by addition to obtain
1.
We then consider the equations
5x
+
y
=
11,
Solving these equations for x and y by subtraction, we have x
Substitution of these values in the first given equation gives z
2.
solution of the given system is z
2, y
1, 2
=
=
=2
= 2.
and y
=
1.
Hence, the
=
EXERCISE 9-1
In each of the problems from
Check
1.
(3x
I
1
to 30, solve the given system of equations.
all solutions.
x
-
2y
30
= 6,
= 4.
3x
2x
2x
3x
x
2x
2x
3x
- y = 7,
+ y = 8.
+ 30 + 1 = 0,
3.
6.
- 2/4-7=0.
+ 2y = 3,
+ 30 = 1.
+ 30 - 1 = 0,
+ 0+3=0.
9.
12.
15.
17.
- 3z = 6,
+ 20 = 3.
2x + 2y - 3 = 0,
5x + 30 + 4 = 0.
3z + 0+7=0,
4z + 80 + 9 = 0.
2x + 60 - 7 = 0,
3z - Sy + 9 = 0.
2x + 40 - 50 = 3,
- 7 = 2,
3z +
4x + 80 - 10z = 1.
x
+ 20 - 2=3,
+ 40 + 2s = 1,
+ - 30 = 2.
- 3x + 80 + 9s - 3 = 0,
2x + 30 + 2-4=0,
Bx - 20 - 22 - 4 = 0.
3z + 52 = 0,
- 42 = 2,
x
4z
32 + 1 = 0.
20
3x
40 + 22 = 3,
2x + y
=1,
5x - 3y 4- 42 + 5 = 0.
2x + 30 - 62 + 2 = 0,
- 2 = 0,
50 + 3
s
x
3z
-
19.
21.
23.
25.
a- V-21 f ~23 =0A
'
9-4
Sec.
Linear Equations
26.
By
2z
2z
=
169
27.
4,
-20 +3* =3,
+ 4s = 2.
3s
28.
+
and Graphs
(4
29.
13(3
1
1
_2
1
=
x
y
z
~~3
1
30.
(3(3 -f 0)
0)
-0) =5,
- 0) = 5.
-4(3 -0) =5,
-
3(3
+ 0) = 7.
*
Z
I'
-
4(3
3(3
4(3 -f 0)
(X
(X
+0) -
For each of the following systems of equations, determine whether it is consistent
and independent, consistent and dependent, or inconsistent.
31.
34.
37.
-
= - 5,
60 = - 3.
+ 40 = 3,
[2x
x + 20 = 6.
\
- 30 = 1,
/23
+ 6w = 2.
- x + 30 = 2,
2x = 60 + 14.
/23
43
1
30
32.
\
x
j
[210
38.
f
-
ay
9-4.
+
2x
35.
5
40.
-
(33
50
70
63
3x
+
+
+
63
9z
-
80
-
8=0,
-f
10
+
1
-
3
120
40
=
=
=
= 0.
= 0,
= 0.
3,
1.
a6.
33. J33
[63
36.
J45z
J153
43
39.
|
[150
- 20 = 8,
- 40 = 8.
- 270 = 21,
- 90 = 7.
- 30 = 6,
-203 +6 =0.
bx -{-ay
=
a6c.
GRAPHS OF LINEAR FUNCTIONS
The discussion of rectangular coordinates in Section 2-1 set the
stage for the pictorial representation of a function. By this representation of a function f(x), we mean the graph of the equation
y = f(x). It consists of all points, and only those points, whose
coordinates x and y satisfy the equation.
In the same section we considered the graphing of lines parallel
to the coordinate axes. The equation x = 3 was shown to represent
a vertical line, that is, a line parallel to the y-axis which intersects
the #-axis at the point (3, 0). This line thus includes all points 3
units to the right of the t/-axis. This example illustrates the fact
that a linear equation in x alone represents a line that is parallel
to the ^/-axis. Similarly, y = 2 was shown to be the equation of thfc
horizontal line whiph is parallel to and 2 units above the #-axis.
And this example illustrates the fact that an equation in y alone
represents a line parallel to the z-axis. Furthermore, it is proved
in analytic geometry that the graph of every first-degree, or linear,
equation in x and y is a straight line; and, conversely, that every
straight line
is
the crraDh of a linear equation.
170
Linear Equations
and Graphs
Sec.
9-4
We shall proceed by first preparing a table of corresponding
values in a given problem and then plotting the corresponding
'points on the coordinate system to obtain the graph of the equation,
The following illustrative examples will point the way toward an
understanding of the procedure in the graphing of linear equations.
Example 9-4. Graph the function 2x
+
3.
= 2x + 3. Then assign any values for or, substitute them in the
and obtain the corresponding values for ?/. The table and the graph are
Solution: Let y
equation,
shown
in Fig. 9-1.
Since a straight line is definitely determined when two points are known, only two
pairs of values of x and y are needed in graphing a linear equation. We can, however, use three points in order to check our work.
/U,5)
y
3
5
-1
FIG. 9-2.
FIG. 9-1.
Example 9-5. Graph the equation 2x
Solution:
The equation may be
table of corresponding values of
9-5.
3?/
=
6.
solved for y in terms of x.
y
A
+
= -
Then
+ 2.
|
# and ?/ and the graph are shown
in Fig. 9-2.
INTERCEPTS
In general, the points where a curve crosses the coordinate axes
are the easiest to obtain.
The ^-intercepts are the values of x at the points where the graph
crosses the x-axis. Since y =
on this axis, the ^-intercepts are the
values of x that correspond to y = 0. Similarly, the ^-intercepts are
the values of y at which the graph crosses the i/-axis. They are the
values of y that correspond to x
rule
=
0.
Hence,
we have
the following
:
To find the ^-intercepts, set y =
To find the ^-intercepts, set x =
and solve for
the equation and solve for
in the equation
in
x.
y.
Sec.
9-6
Linear Equations
Example 9-6. Find the
and Graphs
171
intercepts of the line
= 6.
Solution: To find the ^-intercept, let y = 0. Then 2x = 6, and x = 3.
To find the ^/-intercept, let x = 0. Then 3y = 6, and y = 2.
= 3 and = 2 found in this solution correspond
Note that the intercepts
2x
a;
the points (3, 0) and
coordinate axes.
(0, 2),
+
3y
?/
respectively,
where the
line in Fig.
GRAPHICAL SOLUTION OF LINEAR EQUATIONS IN
9-6.
to
9-2 crosses the
TWO UNKNOWNS
In the graphical solution of two linear equations in two unknowns,
the graphs of the two equations are drawn with reference to the same
coordinate axes. Since the solution of two equations in x and y
is a pair of values of x and y which satisf^ both equations, the
solution must represent graphically a point common to both lines
represented by the equations. Hence, the values of x and y which
satisfy both equations give the coordinates of the point of intersection of the lines. We find, therefore, that the two lines intersect
in a single point, are parallel, or are identical, according as the
equations are consistent and independent, inconsistent, or consistent
and dependent.
Example 9-7. Solve graphically
2x
+
3y
= 24,
5x
-
2y
= 22.
Solution: Tables of corresponding values for the two equations and also their
graphs are shown in Fig. 9-3.
It is seen from the graphs that the lines intersect at the point (6, 4). That x = 6,
y
=
4 gives the solution of the given equations
Y
2x
+
3y
=
24
5x
-
2y
=
22
y
-11
12
22
5
FIG. 9-3.
may
be checked by substitution.
172
I/near Equations
and Graphs
Sec.
9-6
EXERCISE 9-2
In each of problems from
x-
and
to
- 3y = 1.
= x -f 5.
= 3z + 4.
x
1.
4.
7.
1
y
9,
graph the given function. In each case give the
^-intercepts.
2.
y
5.
8.
y
= 2x - 8.
= 3.r.
= x - 3.
3.
9.
Solve each of the following systems of equations graphically.
12.
11. 130 - 2x = 0,
fa? - 3// = 1,
10.
\3x
13.
(2*/
I2x
2y = 0.
= + 3,
= 16 - 3y.
-
a;
x
+
/3x
-
I
14.
i
y
3x
=2.
=
=
4,
y
6.
(20
\2*
?/
=
=
-
a?
4.
#
3x
-
5.
= * -3,
= y + 3.
15.
-0
1.
=4.
- ~r
y)
16.
\2x
-
y
=4,
+2y =
12.
17.
+
4
-
4
= 50 - 3,
=4x +2.
18.
2*
-
30
=
12.
10
Deferminanfs
DETERMINANTS OF THE SECOND ORDER
10-1.
Let us consider the following system of two linear equations in
two unknowns:
.
,
(
aix
+
b\y
[
ClzX
+
b>2
(10-1)
The
y
nn
ci,
.
solution of these equations by the
given by
(10
-
o\
x
-2)
---
k-*!
,
""
t)lC2
p-
y
>
C2.
method of Section 9-1
-
is
-
2
0,201
It is
understood that a^b 2
We
~
a<2 bi
0.
more convenient way of writing the expression (tib 2
Onbi. The notation which we select for
this purpose will enable us to express also the numerators of (10-2)
by means of the same symbol with the proper changes in letters.
In choosing a symbol we shall select a form which will exhibit
the numbers a l9 &i, a2 &n in the same relative positions as in (10-1)
Thus, we write
shall at this point introduce a
.
,
02
62-
This arrangement of the four numbers in a square array, consisting of two rows and two columns, is then enclosed within vertical
bars, as follows
:
ai
hi
a2
This symbol represents a determinant of second order.
Thus, we start with a square array, or matrix, such as
174
Sec. 10-1
Determinants
We
&2 &i, called
then associate with this array a number ai& 2
determinant, which is denoted in the following manner
its
:
The numbers a u b lt a2
The numbers ai and & 2
,
b 2 are called elements of the determinant.
on the principal diagonal the numbers
on the secondary diagonal.
Note. We observe that the "expanded" value of the foregoing
determinant is equal to the product of the elements on the principal
diagonal minus the product of the elements on the secondary
a2 and
lie
;
&i lie
diagonal. It is interesting to note also that this value is the algebraic sum of all possible products obtainable by taking one and
row and one element from each column.
preceded by a plus sign or a minus sign, according
only one element from each
Each product
is
to a rule to be stated in Section 10-2.
Using the notation of determinants, we can write the solution
(10-2) of (10-1) in the form
X
(10-3)
=
y
=
0,2
We note that the value of each of the unknowns in (10-3) may be
written as a fraction whose denominator is the determinant of the
coefficients as they stand in (10-1), and whose numerator is the
determinant formed from that of the denominator by replacing
the column of coefficients of the unknown in question by the column
of constant terms.
Note. If a2 = kai and & 2
fc&i, where k is any number, then
=
0.
In this case, the equations of the system (10-1) are inconsistent
unless both numerators of the fractions in (10-2) are also equal to
zero, that is, unless
=
and
=
0.
Therefore, the equations of the system (10-1) represent distinct,
=
kc l9 or they represent coincident lines
parallel straight lines if c 2
if Co
=
kc\.
Sec.
10-2
175
Determinants
Example 10-1. Solve the system
of equations
-.x
\x
+
3y
=
1,
-
2y
=
22.
we have
Solution: Using determinants,
1
3
22
-2
-
2
-
66
-
68
4
3
-
8
-
9
-
17
3
-2
= 4.
Also,
10-2.
4
1
3
22
4
3
3
-2
88-3
_
-
8
85
-
9
=
-5.
17
DETERMINANTS OF THE THIRD ORDER
A determinant of the third order is a number designated by a
square array of nine elements arranged in three rows and three
columns and enclosed within vertical bars. An example is
D=
(10-4)
a2
62
The value, or expansion, of the determinant (10-4)
as the quantity
is
defined
&2
(10-5)
63
63
C3
C3
or as the quantity
(10 G)
D =
ai&3C2 ^
0162^3
Here the products such as ai& 2 c 3 ai& 3 c 2
the terms of the determinant.
,
,
and
are
known
as
Minors and Cofactors. In any determinant the minor of a given
element is the determinant of the array which remains after deleting all the elements that lie in the same row and in the same
column as the given element. Thus, in (10-4) the minors of a x 61,
,
Ci are, respectively,
C2
63
3
tt2
C2
62
&3
176
Determinants
The
column
Sec.
10-2
cofactor of an element which lies in the z'th row and fcth
is equal to the minor of that element if i + k is even, and is
equal to the negative of the minor
cofactor
=
if i
+
1)*+*
(
k
is
odd.
That
is,
minor.
Thus, in the determinant in (10-4), the cof actor of &i equals the
minor of ai, since 0,1 lies in the first row and in the first column
and i + fc = l + l = 2, which is even. Similarly, the cofactor of bi
is the negative of the minor of 61, since i4-fc = l + 2 = 3, which is
odd.
Often the following procedure may prove more convenient for
finding the sign corresponding to a given element. Beginning with
+ in the upper left-hand corner, change sign from place to place,
moving horizontally or vertically, until the position for the element
in question is reached. The schematic arrangement of signs corresponding to the elements of a third-order determinant is thus
as follows
:
Note that the sign for any position
is independent of the path
followed in arriving at that position.
We shall designate the value of the cofactor of an element by the
corresponding capital letter, and we shall use the subscript that
occurs with the element itself. Thus, the cofactors of a t 61, c x are,
,
respectively,
A,
(10-7)
C2
=
a.3
C'3
Hence, (10-5) for the expansion of the determinant
written as follows
may
also be
:
D =
(10-8)
aiAi
+
biBi
+ aCi.
This sum is called the expansion of the determinant according to
the elements of the first row.
We
observe at this point that the right member of (10-6) reprepossible products, here 3 in number, that can be formed
from the determinant in (10-4) by taking one and only one element
sents
all
!
from each row and each column.
of the determinant
is
It follows also that the
the same, regardless of the
value
row or column
Sec.
10-3
177
Deferm/ncmfs
according to which the expansion
the determinant as
D =
(10-9)
or as
D =
(10-10)
is
made. Thus, we
may
express
+ b 2 B2 +
03^.3
+
63^3
+
c^Cz.
These equations represent the expansion according to the elements
of the second column and according to the elements of the third
row, respectively.
Example 10-2. Expand the determinant
2-53
according to the elements of the
third column.
Solution:
1
7
4
2
1
1
7
4
first
The expansion according
2
6
row and according to the elements
to the elements of the first
6
1
-
1
4
+
1)
+ 3(42 +
-
6
2
1
7
row
of the
is
This reduces to
2(8
-
7)
+
5(24
Expanding according to the elements
()
2
-5
-1
7
2
-I-
-1
.
7
2
-5
2)
= 259.
of the third column,
=
3(42
+ 2) -
(14
-
we have
5)
+ 4(4 + 30) = 259.
PROPERTIES OF DETERMINANTS
From
the definition of the value of a determinant
we may deduce
the following important properties of determinants. These properties supply us with more convenient methods for evaluating a
determinant.
Note. For a more complete discussion of these properties, the student is referred to any one of the various treatises on determinants
or to texts on the theory of equations or on solid analytic geometry,
where he will also find proofs which apply to determinants of
any order.
The properties listed here will be employed in examples that follow, and their usefulness in simplifying determinants will be
illustrated.
178
Determinants
Property
The value of a determinant
1.
is
not changed
Sec.
10-3
if its
rows
and columns are interchanged.
Property
2. If all
the elements of a row, or of a column, are multi-
same number, the value of the determinant
plied by
multiplied by that number. For example,
the
Property
b\
ka,2
62
=
k
62
0,2
two rows, or two columns, of a determinant are
If
3.
ka\
is
identical or proportional, the value of the determinant is zero.
For example,
determinant
let
the first two columns be identical, as in the
CL\
D
Cii
C\
Q<2
C2
Cl'3
Then, expanding according to the elements of the third column,
we have
0,2
D =
02
=
0.
(73
Property 4. The value of a determinant is not changed if we add
to the elements of any column (row) any arbitrary multiple of the
elements of any other given column (row)
.
For example,
61
C2
(12
+
rib i
61
Ci
+
ribz
62
C2
3
The proof
first
column,
last
we
Expanding according
to the elements of the
find that
+ nbi
bi
0,2
+ nb2
b2
0,3
+ 7163
&3
L
The
follows.
ai
61
c\
2
0,2
b2
02
^3
dz
63
C.3
c
determinant vanishes, since two columns are identical.
Sec.
10-3
179
Defermmcrnfs
Example 10-3. Evaluate the determinant
43-1
D=
Solution:
By
5
1
2
4
adding 2 times the elements of the
first
row to the elements
of the
second row, we obtain
If
now we add 3
times the elements of the
first
row to the third row, the determinant
becomes
3-1
4
13
7
14
13
Expanding according to the elements
In this example, we
first
of the third column,
13
7
14
13
=
we have
-71.
converted the given determinant to one in which
one of the elements of the third column are zero. For the
final
but
expansion, the given
determinant was thus reduced to a determinant of the second order, and
was
all
its
value
easily found.
=
2 satisfies
and third rows are proportional.
Hence, the
Example 10-4. Without expanding the determinant, show that x
the equation
Solution: Substituting
is
satisfied
by x
x3
3
1
6
-4
-x
= 0.
2 for x in the determinant, we have
This equals zero, since the
equation
3x 2
first
= -
2.
12
-8
2
3
1
7
6
-4
1
180
Deferm/ncrnfs
10-3
Sec.
EXERCISE 10-1
In each of the problems from
to 12, evaluate the given determinant.
1
1.
5.
9.
*
In each of the problems from 13 to
equations by means
19, solve
of determinants.
Check
the given system of simultaneous
solutions by substitution in the
all
equations.
x
13.
2x
16. If
+ 30 =
- 4y =
14.
5,
7.
|2x
\3x
y=
iji)
Find the ^-intercept and the 0-intercopt of the
x
2
18. Solve graphically the following
f
3x
y
I
4
1
-3
1
+
-
D represents the determinant in Problem 10, show that D
of the straight line through the points On,
17.
15.
1,
+2y =4.
and
line
(j 2 ,
is
20
30
=
=
5,
5.
the equation
2/2).
whose equation
is
= 0.
system of equations:
= 0.
19.
Find the coordinates of the vertices of the triangle whose sides are the straight
x - y + 2 = 0,
2x + 90 + 15 = 0,
7x + 40 - 30 = 0.
and
lines
20.
Find the vertices of the parallelogram formed by the following
x
y
1
2
1
1
-3
2
1
x
= 0,
y
I
6-11
1
= 0,
x
y
I
1
3
1
2
4
1
=
0,
lines:
= 0.
Sec.
10-4
10-4.
Deferm/nanfe
181
SOLUTION OF THREE SIMULTANEOUS LINEAR EQUATIONS IN THREE
UNKNOWNS
Let us consider the following system of linear equations
ix
+
biy
+
(10-11)
The determinant
ciz
=
C2Z
=
di,
of the coefficients of the
i
D =
:
unknowns
is
61
C2
0,2
For the solution of such a system, we employ the following theorem,
which is known is Cramer's Rule.
If the determinant D of the coefficients of the system
not equal to zero, the system has just one solution. In this solution, the value of any unknown is equal to a fraction whose denominator is D and whose numerator is obtained from D by replacing
the column of coefficients of the unknown in question by the column
of constants di, d2 and c? 3
Proof. Let the numerators of the fractions for x, y, z be denoted
by DI, D 2 D 3 respectively. We proceed to show that if equations
(10-11) are to be satisfied, then
Theorem.
is
.
,
,
,
Dx =
(10-12)
Dy =
Di,
Specifically, the equation for
(10-13)
Dz =
D3
.
x will have the form
C2
C2
(12
#3
Z> 2 ,
63
63
C3
Similar equations may be written for y and z.
To find x the first equation of the system in (10-11) is multiplted by the cof actor At, the second by A 2 and the third by A 3
9
.
,
After adding and collecting terms,
(10-14)
(aiAi
we
obtain
+ a2^ 2 + a3A 3 )x + (biAi + b2A 2 + bzAz)y
+ (ciAi + c2^2 + c^A^z = diAi + ^2^2 +
^3^3.
x in (10-14) is the expanded value of D according to the elements of the first column. The coefficient of y is
The
coefficient of
biAi
+ b 2A 2 +
Defermmonfs
182
Sec.
10-4
this is equal to
b\
62
&3
&3
<?3
two columns are identical. Similarly, the
Hence, we have shown that the left side of
the expanded form of the left side of (10-13) and that
which equals
zero, since
coefficient of z is zero.
(10-14) is
the two are therefore the same.
The right
side of
is
(10-14)
cM-i
+ d2 A 2 +
d 3A 3 which
,
is
the
expansion of the determinant
61
i
i
This determinant may be obtained from D by replacing the coefficients of x by the column of constants that is, it is the expansion
of the determinant that we have called DI. Hence, the equation
Dx = DI in (10-12) is established. By a similar procedure we can
show that the equations Dy = D 2 and Dz = D 3 are also valid.
;
If
D T^ 0,
the value of x
is
given by the equation
d\
x
(10-15)
Similar values
may
bi
=
be found for y and
The proof of (10-12)
is
valid whether
z.
The proof
is
complete.
D=
0.
If
D^
or
D ^ 0,
the equations of the system (10-11) are consistent and have only
one solution, which is of the form (10-15) that is,
;
(10-16)
If
D = 0,
=
5"'
v
=
^'
and any one or more of the other determinants, D\, D2
not zero, the given system of equations has no solution and is
DZ, is
inconsistent.
,
Sec.
10-5
If
183
Defermmcrnfs
D = 0,
and
all
the other determinants are zero, the equations
may
be consistent or inconsistent. If they
are consistent, there are infinitely many solutions. This case will be
of the system (10-11)
treated in Section 10-5.
The following example
Example 10-5. Solve the system
D ^ 0.
of equations
-
x
x
2x
Solution:
which
will illustrate the case for
z
=
1,
2z
=
3,
y
+
+y
-
-
+ 3z = 4.
y
Here
=
D=
11;
= 2.
Hence, x
= -~ =
y
~
=
D ~5
^
=
-
If
O
j_J
we check by
substitution,
we
find that these values satisfy the given equations.
UNKNOWNS WHEN
SYSTEMS OF THREE LINEAR EQUATIONS IN THREE
10-5.
D
We
=
note that
when
D = 0,
the system (10-11) will not have a
any one of the other determinants DI,
from zero. Suppose that a solution is given by
solution if
x
=
y
r,
=
z
s,
=
D2 D 3
,
is
different
t.
Then the equations (10-12) become
r
=
It follows that Z?i
= 0,
s
Z?i,
Z> 2
=
0,
= D2
Z> 3 = 0.
,
*
= D3
.
The following example will illustrate the case of consistent equations where D =
and Z?i = D 2 = #3 = 0. The equations are said to
be dependent, and they have infinitely many solutions. The student
should construct an example to show that the equations (10-11)
may
be inconsistent
when
D = 0,
even though
Z>i
= D2 = D3 =
0.
184
Determinants
Example 10-6. Solve the system
Solution:
10-5
of equations
x
+
y
2x
-
y
4z
-
2?/
- z=3,
+ 30 =
+ 62 =
Here
Z)
Sec.
=
1
1
2
-1
4
-2
-
1,
2.
1
This equals zero because the second and third rows are proportional. But, we also
find that
13-1
1-1
1-1 3
2-2 6
3
D =
l
1
3
42
6
2
1
1
3
2-11
4-22
=
0.
In this case the given equations have a solution. In fact, the second and third
equations are proportional, and so either of those can be solved together with the
first equation for two of the unknowns in terms of the third. For example,
x
=4
-
5 -f 5z
2z
Thus, we have a single value of x and a single value of y for every value of
However, there are infinitely many values of
solutions of the given equations exist.
10-6.
and therefore
z,
infinitely
z.
many
HOMOGENEOUS EQUATIONS
The system (10-11)
is
homogeneous
if di
=
0,
d 2 = 0, and d 3 = 0.
= y = z = 0. When
Such a system always has the trivial solution x
= d2 = d3 = 0, it is seen that DI = D 2 = D& = 0, for each of these
determinants has zero for every element in one column. If D
0,
it follows from (10-16) that we can have but one solution, which
is given by
di
=
the given system is to have a solution besides the trivial
= 0, nonmust equal zero. It may be shown that, if
solution,
Hence,
if
D
D
trivial solutions
always
exist.
Example 10-7. Solve the system
x
of equations
-
2x
K/p
y
3y
_
2?7
+ 2=0,
+ 4z = 0,
"
-
9.
f)
10-7
See.
Solution:
Determinants
185
D is
The determinant
= 0.
Therefore, nontrivial solutions exist. To find these solutions, we proceed as follows:
Transpose z in each of the first two equations, and solve for x and y in terms of 2.
Then we have
-
1
D=
-
z
1
1
= -
2-3
-3
-42
1
-2
2
-42
2,
Hence,
=
I
-p
= 2,
j
and
i/
=
2
=
Substitution shows that those values also satisfy the third equation. The given
system therefore has infinitely many solutions, and the values of x, y, and z are related
by the equations x
=
z
=
and y
2z.
EXERCISE 10-2
In each of the problems from
to
1
equations by means of determinants.
6,
solve the given system of simultaneous
all solutions by substitution in the
Check
equations.
1.
+ y - 2 =
+3y - z =
+ y -3z =
3x
x
3
11.
I2x
2x
-
6x
+ 2y
20
10,
2x
5.
+ 2-2,
- 2z = 5.
x
3jc
-
y
-
For each of the following systems find at
that there is no nontrivial solution.
+ 22 = 0,
7.
2x
-
8.
+2=0,
10-7.
2x
x
- 2=0.
= -1,
= 0,
22
7z
-
3y
least
4-
2z
3
2.r
32
62
x
9.
0,
+3y - 2=0,
+ 3y +
2x
6.
-
+ 62 = 7,
+ 3y + 62 = 0,
+ 2y + 92 = 3.
- y = 3,
- 32 = - 1,
- y ~ 2.
?/
one nontrivial solution, or show
=
42
x
3.
-2y
- 4y + z=3.
- y - 32 = 7,
+ 2y - 2 = 10,
- 3# + 22 = - 7.
13,
5x
= -
4.
x
2.
11,
[
J2.c
= 0.
3x
-
+
2y
y
2/
+
32
= 0,
+42 =0,
- s = 0.
SUM AND PRODUCT OF DETERMINANTS
.Closely related to Property 4 of Section 10-3 is a theorem concerning the sum of determinants. We shall illustrate the theorem
for splitting the elements of a given column into two parts by
means
of the following equality for third-order determinants
+ 61
+ 62
03 + &3
ai
ci
0,3
02
0,2
2
Cs
as
03
d\
d\
62
:
186
Deferm/ncmfs
Sec.
10-7
This can be shown to be true by expanding the three determinants according to the elements in the first column and noting that
in the expansion the minors are the same for all three determinants.
Example 10-8. Show that
Solution:
Expanding each
of the determinants
according to the elements in the
This result
is
first
on the
left side of
the equation
column, we have
the expansion of the determinant on the right in the given equation
Computing the value of each determi-
according to the elements of the first column.
nant in the given equation, we see that each side reduces to 47.
A similar theorem for splitting the elements of a given row is illustrated by the
following example
:
ttl
+ Ci 61+6
CL2
(12
62
This can be shown to be valid by expanding according to the elements of the
first row.
Thus, consider the determinant
This
may
be written as a
2
sum
3
4
5
1
7
in various ways.
2
2
Examples arc
4
5
4
5
-2
2
3
5
and
1
We
7
7
1
product of two determinants of
equal to a determinant of like order in which the
element of the ith row and kth column is the sum of the products of the elements
shall state without proof the rule for the
the same order.
of the ith
The product
row of the
first
is
determinant and the corresponding elements of the kth
For example, for second-order determinants,
column
of the second determinant.
we may
write
+2-7 1-6 +2-8
1
2
5
6
1-5
3
4
7
8
3-5 +4*7
3*6 +4*8
Sec.
10-7
187
Deferm/ncrnfs
That the product of the values of the two determinants on the left equals the
value of the determinant on the right is checked by expanding. Thus, the desired
equality becomes
(4
-
(40
6)
-
42)
=
(950
-
946),
or
(-2). (-2)
To
is
illustrate the multiplication of
= 4.
two third-order determinants, we have that
equal to
3-1 +2(-3)
1-1
3-7+2-94- 5( -
+5-5
+(-!)( -3) +2-5
.4-1+ 6(-
3)
+0-5
l-7
4.
4)
+ (-l)9+2(-4)
7+6-9+ 0(- 4)
3-6+2-3+5-2
1
6
+ (-
1)3
+2
2
4-6+6-3 +0-2
This reduces to
22
which
30(-
is
equal to
21),
630.
which equals
Also,
-
19
34
14-10
7
82
42
14
computing the values of the given
factors,
we have
630.
EXERCISE 10-3
In each of the first three problems, combine the given determinants into a single
determinant, and evaluate the result.
1.
3.
188
Determinants
Sec.
10-7
In each of the following problems, find the product of the determinants without
evaluating the individual factors.
4.
8.
Complex Numbers
11
11-1.
THE COMPLEX NUMBER SYSTEM
There are many problems that cannot be solved by the use of
numbers alone. We observe, for example, that the equation
x2 + 1 =
has no real root, since x 2 can never be negative if # is a
real number. In order to provide solutions to such equations, a new
system of numbers, called the complex number system, was intro-
real
duced. Later in this book,
we
shall find
many
instances of solutions
involving complex numbers.
We
now define a complex number as an ordered pair of real
which
we denote by (a, b). If the numbers a and b are
numbers,
shall
regarded as the Cartesian coordinates of a point in a rectangular
coordinate system, we have a one-to-one correspondence between
the set of complex numbers and the set of points in a plane. The
plane is called the complex plane. Two complex numbers (a, b)
and (e, d) are equal if and only if they correspond to the same
point, that
is, if
and only
if
-
a
c
and
b
-
d.
Addition, subtraction, and multiplication of complex numbers
are defined as follows
:
(a,6)
(11-1)
(11-2)
(a, 6)
(11-3)
+
-
(a, 6)
=(a + c,6 + d)i
= (o - c, 6 - d),
= (ac - 6d, ad +
(c,d)
(c,
d)
(c, rf)
be).
For example,
(1, 3)
(1, 3)
+
(5, 2)
-
(5,
(1,3).
We
2)
(5, 2)
= (6, 5),
= l- 4, 1),
= (-1,17).
also define the following special
(11-4)
0-
(11-5)
1
(11-6)
i
=
(0,0),
(1, 0),
(0, 1).
189
complex numbers
:
1
Complex Numbers
90
Sec.
1
1-1
serves as a zero of the complex number
serves as a unit, in Accordance with the following
The complex number
system, while
properties
1
:
+
-
1
The
so-called
=
=
=
(a, 6)
(a, 6)
(a, 6)
(a, 6)
=
+
=
=
(a, 6)
1
(a, 6)
i=
imaginary unit
(a, 6),
0,
(a, 6).
will be discussed in
(0, 1)
more
detail in Section 11-2.
If
A;
is
a real number,
*
(11-7)
Also,
we
(a,
we define
= (fc, 0)
6)
=
(a, 6)
(fca,
kb).
define
-
(11-8)
=(-!) (a, 6) = (- a, - 6).
6) = (0, 0), the complex number
(a, 6)
Since (a, 6) + (a,
called the negative of (a, 6).
The Reciprocal
(x,
of (a,b).
^ 0,
If (a, &)
then
it
(a, b) is
has a reciprocal
y) such that
Furthermore, the reciprocal
(11-9)
=
(x, y)
(a, 6)
1.
given by
is
(x , y)
^
(11-3) and (11-5), the equation
written in the form
By
(ax
Since a + & ^ 0,
taneous equations
2
2
-
ay
by,
+
bx)
(&
=
-
6x
+
by
=
1,
ap
=
0.
'(#,?/)
=1 may
be
(1, 0).
we may determine x and
ax
&)
by solving the simul-
T/
j
1
The values of x and y can be foundry any of the methods taken up
in Section 9-1. The solution is
X
a
~a 2 +
We have, therefore, verified
_
~
y
b*'
(11-9)
Division of (a, b) by (cy d). If
as follows
b
a2
+
6a
<*
.
(c,
d)
=
0,
division can be defined
:
(a, 6)
where (u,v)
(11-10)
is
-5-
=
d)
(c,
(a, 6)
the reciprocal of (c,d).
(a, 6)
-!-
(c,
d)
=
(u, v),
By
(11-9),
(a, 6)
ac
(
+
bd
ad
+
bc
we have
Sec.
1 1
-2
Complex Numbers
For example,
(5,
13)
+
(3,
1
91
-,
-2) =
This result can also be verified by the method in Section 11-3.
THE STANDARD NOTATION FOR COMPLEX NUMBERS
11-2.
The
ing property
shall
i
-
(0, 1),
defined by (11-6), has the follow-
:
i*
(11-11)
We
number
special
=
(0, 1)
now show
(0, 1)
that
= (-
(a, 6)
1,
0)
= -
(1, 0)
= -
1.
and the binomial form a
4-
bi
are
equivalent, or that
(11-12)
in
(a, 6)
which a
Finally,
=
a
+ bi,
+ bi means al + bi. By (11-5), (11-6), and
al + bi = a(l, 0) + 6(0, 1) = (a, 0) + (0, 6).
(11-7),
by (11-1), we have
(a,
Hence, (11-12)
is
0)4- (0,6)
=
(a, 6).
established.
Real and Imaginary Parts of Complex Numbers. We call a the
and & the imaginary part of the complex number a + bi.
If a
and 6^0, a + bi reduces to bi, which is called a pure
imaginary number. If 6 = 0, the complex number a + bi, or al 4- bi f
reduces to the complex number al, which may be identified with
the real number a. The complex numbers, then, include both the
real part
numbers and the pure imaginary numbers as special cases.
Illustrations of various classes of numbers follow
Some real numbers are 2, 5, and \/3.
Some pure imaginary numbers are 3i, and x/Bt
Some complex numbers are 2 + 3i, and \/S i.
Note that the numbers 2, 5, \/3, 3f, and
\/5i may be put into
the standard form a 4- bi and written, respectively, as the complex
+ 3f, and - \/5f. Since
numbers -2 + Of, 5 4- Of, \/3 4- Oi,
= + Of, which may be written briefly as 0, we shall drop the use
real
:
.
of bold-face 0; similarly for bold-face
1.
Conjugate Complex Numbers. The conjugate of a complex num+ bi is defined as a bi. Likewise, a + bi is the conjugate of
a
2f
bi. Some pairs of conjugate complex numbers follow: 2f,
3 + 5f 3
own
is
A
its
x
5f
real
x
+
number
and
2yi
2yi,
ber a
;
,
conjugate.
;
.
192
Complex Numbers
Powers of
t.
= ft
6
i = i
2
.
= _^
= i2
i3
seen that
It is readily
Sec.
=
i
i, i*
= i2
i
2
11-2
=
1,
an(j so on
Therefore, successive positive integral powers of i have only four different values, namely,
these four values are repeated in regular order.
i,
i, and 1
1,
if
n
is
Hence,
any positive integer, we have in general
i
ft
9
;
These relationships afford a simple method for evaluating powers
3
i =
i
of i, as shown by the following illustrations i 7 = i* +3
38 =
2 =
103
4 26 1 =
=
=
i
&*+*
i
i
-1 and i
i.
:
'
;
-
;
11-3.
ON COMPLEX NUMBERS
OPERATIONS
STANDARD FORM
IN
From
the definitions given in Sections 11-1 and 11-2, it follows
that a 4- bi and c + di can be added, subtracted, multiplied, and
divided as if they were real binomials, except that, where i2 appears,
it is
replaced by
1.
Addition of two complex
numbers is effected by adding their real and imaginary parts separately and subtraction is performed by subtracting their real and
imaginary parts separately. Thus, in accordance with (11-1) and
Algebraic Addition and Subtraction.
;
(11-la)
(a
+
bi)
(a
+
bt)
+
(c
+
di)
=
(a
(c
+
di)
=
(a
+
c)
+
(b
c)
+
(b
+
d)
i,
d)
i.
and
(ll-2a)
-
-
-
For example,
+
(3
20
+
(4
- 50 =
(3
+ 4) +
(2
-
=
5)i
7
-
3t,
and
(3
+
20
-
(4
- 50 =
(3
-
4)
+
+
(2
5)i
= -
1
+
7i.
We
note that the sum of conjugate complex numbers is a real number, because (a + bi) + (a
bi) = 2a. Also, the difference of two
numbers
a pure imaginary number, because
is
conjugate complex
=
+
2bi.
(a
bi)
(a
bi)
Algebraic Multiplication. To find the product of two complex
numbers, multiply them according to the rules of algebra, and
2
replace i by -1 in the result. Thus
(a
Replacing
(ll-3a)
i
2
+
60
(c
+
di)
=
ac
by -1, we have,
(a
+
bi) (c
+ di)
+ adi +
in
=
bci
+
bdi 2
.
agreement with (11-3),
(ac
-
bd)
+
(ad
+
bc)i.
Sec.
11-3
Complex Numbers
193
respects the notation a + bi is more convenient than
In
(a, 6)
particular, the former notation makes it easier to remember how to multiply two complex numbers. For example,
In
many
.
+
(3
20
(4
-
5z)
=
(12
+
+
10)
+ 8)i =
15
(-
-
22
7t.
The student should note that the product of two conjugate complex
numbers is a non-negative real number, because (a 4- bi) (a bi) =
a2
+
b2
.
Algebraic Division. To obtain the quotient of two complex numbers, multiply the numerator and the denominator by the conjugate
of the denominator. Thus, if c + di ^ 0,
+ bi
c + di
a
+ bi
c + di
a
__
~~
c
_
+
adi
di __ ac
~~
di
c
c
(ac
+
bd)
c2
Therefore,
+
.
a
,.
bi __ ac
~
'c~+~di
The right member
is
of the
,
+ (be + d2
,
,
ad)i
,
+ bd ~^ be ad
2
2
c2 + d2
c + d
form A + Bi, and
,
bdi 2
bci
d 2i 2
2
.
lm
this equation agrees
with (11-10).
;
Example 11-1. Reduce -i
The conjugate
Solution:
to the form a
of i
is
i.
Example 11-2. Find the value
of 5
Solution: Represent the division as
the denominator by 3
5
+
3
-
3
13i
'
3
2i
+
2i.
We
+2i ~
__
+
+
bi.
Then
+
13i divided
-~
^
>
by 3
-
2i.
and multiply the numerator and
get
(15
-
26)
+
(39
+
9+4
2i
10)i ""
_
11
13
+
49
.
*'
13
EXERCISE 11-1
In each of the problems from 1 to 12, express the given quantity in the form
a*
and give its conjugate. In working these problems, note that
V
+ bi
a
l.
v^7^.
2.
4.
V-
5.
7. 3
10.
^'
2
.
V~ V- *
V2 + 3
\A 2
2.
2
-
8.
11.
V 17^.
-
3.
V4 - 3 VV15 V- 64a
3(xi 2 .
6.
3
6.
V
V-
9. 1
16.
4-
-
12. 3
a?
V+ V- 32a 6
+2
8.
2
3
.
Complex Numbers
194
In each of the problems from 13 to
13.
i.
26,
compute the value
.
15.
(-i) 13
-i 70
19.
37 183 .
i
23. i 14
14. i 12
17. i".
18.
21. i s i 29
22.
.
-
25. i 25
i
+
50
i
Sec.
-
i 75
(-i)
of the given expression.
- (-i) 17
16.
.
235 .
- (-
20.
i)
11-3
(-i)
24. i 10
18 .
+ i 30
i ao
+
.
602 .
.
26. i 10 -f i 100 -f i 1000 -f i 10000 .
100 .
i
In each of the problems from 27 to
36, find the values of
x and y which satisfy the
given equation.
=
=
27. (2z, 3y)
29. (to, 5y)
(18,
+
-
34. 2z
-
(y
40i
36.
x)f
=
=
6
-
-
2yi
3-ct
1
+2y,x +
=
5y)
(20
+ 3f.
+
8x
-
+ 3,
-
50
=
12
+
2x
7
+
+
6
5y
+
=
1)
(x, 1).
19).
35. 3x
2yi
(8, 9).
to
y),
-
33. 3z
2st.
+
+
30. (x
25).
31. (to
32. z
=
28. (2*, 3?)
(3, 1).
-
6
- 20)i =
- 30)i.
- 2)i.
(5
= (4
+ (x
0.
In each of the problems from 37 to 78, perform the indicated operations and
reduce to the form a + bi.
37.
(2,
40.
(1,
-
3)
-
43. (4,
46.(0,1)
7)
+
-
3)
38. (1,
(5, 6).
-
(7,
-
41. (2, 3)
3).
44. (0,
(4, 3).
-
+
1)
-
39. (1,
(3, 2).
(1, 1).
I)
3
.
4
V)
42. (1,
m
45.
0)
(1,
+
'
(0, 1)
-
~
(1, 1).
47
.
_
-
'
3
^~)
50.
V"17^ -
V^2
52.
(
-
54. (3
2i)
+ 2i)
+ 3i)
56. (6
58. (6
61. (5
1)
-f-
(
-
-
(6 -f 2i).
67. (3<
+ 4) +
-
8) (2i
(1
71. (3
72. (6
3
76
IU '
(2
+ i)
-
57.
-
59. (2
+
5i) (6
62. (3
-
3i) (4<
3t) (2
3i).
+2).
-
(
4i).
-f-
66. (3
69. (2
68. (2 + i) 2 + (5 - f).
2
- V3i)
(4 + 5i)
- v/3i) 2 + (3 + V30 2 - (4 - 3i) (i - 2i) (1 -f f) (1 - 3i) + (4 - 5i).
i).
70. (3
(6
55.
65. (5
4).
-
+ 3i) + (2 - 3t).
+ 9i) + (5 + 2i).
- 3 + 2i) - (5 - 2i).
- 5 V 17!).
(2i + 3) + (8
60. 2f (5 + 3i).
63. (1 + i V2) (5
2f).
51. (4
3f).
5i).
V^ + V^2 - V9.
53. (8
+ 5i).
17!).
(3i + V"
4- (3
-
+
a.
-
6
(3
-f
64. (5<
73. 1
-
- -
+ 3i)
2
49.
'
5
-s-
-4i
"'
(3-2i)
numbers
-
5i)
*-
3t)
-s-
- 5i).
(2
- 4<).
(5
W)
(2i
6).
75. i
1.
j^+j)^^^
(6
79. Prove that complex
-5-
f)
.
74. (6
5t).
-
''
7)
satisfy the associative
4-
(4
-
(4
- 7)
if
(a
+ bi}
(c
+ di) = 0,
then a
+ bi =
(4
+ Gi)
+
'
)
and commutative laws of
addition and multiplication and the distributive law.
80. Prove that
3i).
5i) (8
(3
-I-
or c
+ di = 0.
Sec.
11-4
Complex Numbers
195
GRAPHICAL REPRESENTATION
11-4.
As we have
P
point
~
seen, the complex number a + bi determines a definite
in the plane whose rectangular coordinates are x = a and
Conversely, to every point P in the plane corresponds a combi for which the values of a and b are the respective rectangular coordinates of P. See Fig. 11-1. In this system, the real numbers a + Oi are represented by points on the
#-axis, which is called the axis of reals. Pure imaginary numbers
+ bi are represented by points on the 2/-axis, which is called the
axis of imaginaries.
y
6.
number a +
plex
a
FIG. 11-1.
FIG. 11-2.
more convenient at times to represent the complex number
by the vector drawn from the origin to the point P. The
a~ + 6 2 and
length of the vector is given by the relationship r =
the direction is given by an angle
determined from the equations
a - r cos 6 and b
r sin 6.
In Fig. 11-2 is indicated the graphical addition of the two complex numbers a + bi and c + di, which may^ be represented either
by the points P l and P 2 or by the vectors OJ?i and OP^ Withjths
completion of the parallelogram OPiP-JP^ the sum of OP l and OP2
can be represented by the diagonal OP 3 Thus, either P 3 (a-f c,
It is
a+
bi
V
,
.
d) or OP represents the sum (a 4- c) + (6 + d)L Hence, the
vector which represents the sum of two complex numbers is the
sum of the vectors representing the given numbers.
6
+
To
and
l3
subtract c
c
+
di
from a +
bi graphically,
we merely add a +
bi
di.
Example 11-3.
Add
the complex numbers 2
+ 3f
+ 2t
let P 2 (6,
and 6
graphically.
Pi (2, 3) represent the number 2 + 3f and
2) represent
number 6 + 2i. Draw OPi and OP 2 in Fig. 11-3, and complete the parallelogram OPiPaPa. Then Ps^represcnts the sum 8 + 5i of the complex numbers
2 + 3i and 6 + 2i, and OP 3 represents the sum of the vectors OPi and OP 2
Solution: Let
the
.
Complex Numbers
196
Sec.
11-4
/'3 (8,5)
o
FIG. 11-3.
The difference of two complex numbers may be obtained
same manner if we apply the relationship
(a
+
-
to)
(c
+
di)
=
(a
+
bi)
+
(-
c
-
in the
di).
P
and Q represent the numbers a + bi and c + di, respectively,
complex plane, as shown in Fig. 11-4. Then c di is represented by Q', which is the reflection of Q through the origin.
Let
in the
P(a,b)
Q
fad)
O
FIG. 11-4.
Let us recall
how
the difference of two vectors
was represented graphically
OP, OQ,jand OQ' represent a 4-
Section 6-7 and
the vectors
respectively, then
representing the
11-5.
OR
is
number
was explained
in Fig. 6-15. If
bi, c
+
the desired vector and
(a +
(c
bi)
+
di)
di,
R
we
in
let
and -c -di,
is
the point
.
TRIGONOMETRIC REPRESENTATION
Let the complex number a
+
be represented by the radius
Then the distance
= r is called the modulus, or^the absolute value, of the complex
\OP\
number and the angle 0, which OP makes with the positive #-axis,
vector
drawn from the
bi
origin to the point P.
;
is called
an argument, amplitude, or angle of the complex number.
Sec.
11-5
From
Complex Numbers
clear that
is
Fig. 11-1, it
=
a
-
=
b
=
and -
cos 6
and
8
r cos
=
197
sin 0, or
r sin 6.
Hence,
a
+
bi
=
+
r cos
=
(r sin 0)i
+ i sin 0).
r(cos
This last expression is known as the polar form or the trigonometric form of the given complex number, as contrasted with the
standard or rectangular form a + bi.
To reduce a given complex number a + bi to the trigonometric
form r(cos 9 + i sin 0), we find r and
by means of the relation2
2 a =
=
b
6
r
r
and
r
cos
0.
sin 0. We have
+
ships
V^
,
a
Example
+
=
bi
rf -
+ - i) =
Represent the
11-4.
+
r (cos
i sin 0).
complex number
/5
9
+ ^r~
and change the given notation
i graphically,
V3/2
to the trigonometric form.
Solution:
/I
are [ \J
a
=
The
P whose
point
rectangular coordinates
1
.,
\/3\
^- /) represents the number Z
^
,
,
>
1/2 and 6
quadrant angle.
=
\/3/2 are both
We
from Fig.
see
=
7T/3 or
60. Hence, 1
J
\/3
zr
Here a
=
determined from cos 6
1
and
1
1-5 that
=
=
=
1.
^ and
Since
*
r
FIG. 11-5.
=
j^l- The
1/2 and sin 6
f =
+ ^1
^
6
1/2
i
l
(i
\J
So
sin
r
=
V2
=
=
^
+
Example 11-5. Express the complex number
Solution:
6<r
.
.
positive, ^ is a first-
determined from the equations cos 6
let 9
+
,
^
f)
i
\/2,
and
y=
=
C os
this case
60
/
1
--
\X/2. In
angle 6
is
we may
+ i sin 60.
in the trigonometric form.
is
We
a fourth-quadrant angle
thus have
-\/2
-4= - --Ut
=
\/2(cos 315
+
i
sin
315).
EXERCISE 11-2
In each of problems from
jugate graphically.
1. 3
2. 8
2f.
+
5. 3
-
5i.
6. 1
10.
1
to 12, represent the complex
+ 2i.
3.
3
-
7.
i.
1.
11.
+
.
number and
4.
its
2
-
con-
3t.
8. 1.
(1
-
12. 5
+ 12.
1
Complex Numbers
98
Sec.
1
1-5
In each of the problems from 13 to 24, perform the indicated operations graphThen check the result algebraically.
ically.
- 30 + (- 4 + 0- 60 - (4 + 30.
+ 20 - (5 - 0.
13. (7
15. (3
17. (3
21. 3
23. 7
+
+
19. (5
-
(1
-
(4
20. (3
-
16. (6
50.
- 40 - (2 + 0- 20 - (- 2 + i V).
(1
18. (3
+ 30 - (4 - 5i).
- 2i) + (6 + 20+ 40 - - 2 - 40.
14. (2
(
22. (v/2
24. (4
+
2i)
-
(5
+
+
30 -
-
40.
+ V&) - 7i.
+ 30 - (3 + 20-
(1
(2
In each of problems from 25 to 36, change the complex number to the trigonometric form and represent
25. 1
+
29.
|(1
QQ
*
Let
graphically.
-
26.
i.
+ V20.
,
d4 '
MULTIPLICATION
TI (cos
a+
i
- V30-
W ~
,,
AND
27.
5.
30. ~(1
1 -i
6 -4i
-2TT -3-+!'
11-6.
it
}
,.,
4l)
-
31. 5
Q(5
**
2
'
a
+ i sin a)
=
fi^Kcos
and r2 (cos
sin a)
/3
r% (cos
/3
a cos /3
+ i sin
sin
we have proved
32.
-?
Q,
3
1
+ i sin /?)
3 \/3i.
-2i
*
2
DIVISION IN TRIGONOMETRIC
+<
FORM
be any two com-
their product
is
given
/3)
a sin /3)
=
Thus,
12t.
6~+'
plex numbers in trigonometric form. Then,
by the relationship
TI (cos
+
-
28. 3
3t.
+ i(sin a cos +
cos
a sin
]8)]
]8)+ isin
(a+
j8)J.
/3
rir2[cos
(a+
that the absolute value of the product of two
the product of their absolute values, and an
angle of the product is the sum of their angles.
The result found for the product of two complex numbers can be
complex numbers
is
extended to the product of three or more complex numbers.
The quotient obtained by dividing the complex number
1*1 (cos a + i sin a)
by the complex number r2 (cos/J + isin/3) is
given by the relationship
ri(cosa
r2(cos
ft
+ fr'sinoQ _
+ i sin
~~
ft)
=
i sin
+ i sin a) cos
i
sin
cos
i sin
+
- + i sin (a (a
ri(cos
a
r2(cos
ft
[cos
ft)
|8
ft
ft
ft
ft)
It follows that the absolute value of the quotient of
numbers
quotient
is
is
ft)].
two complex
the quotient of the absolute values, and an angle of the
the difference of their angles.
11-7
Sec.
Complex Numbers
Example 11-6. Find the product
f sin
199
+
of 2(cos 30
30) and 3(cos 120
i sin
the rule for products in polar form, we have
i sin 120)
3(cos 120
By
Solution:
+ i sin 30)
2(cos 30
= 2-3 [COP
= 6 [cos
+
(30
+
150
120)
i sin
+
+ i sin
=
150]
6
(30
+
-
^~ +
(
120)]
-
3
- V3 +
(
)
1
i is
divided
+ -^A
by
Zi
From Examples 11-4 and 11-5
Solution:
-
1
=
i
and
i
-
1
1
+
2
_
i
\/3
+ i sin 315),
+
cos 60
+
ft (cos 315
cos 60
.
i sin
-f i sin
= -
[cos
\/2 (cos
-
11-7.
1)
+
i
Hb
^
(315
-
60)]
+ i sin 255]
75 + i sin 75).
we
[0.2588
+i
_
\~ (V3 -
i sin
255
Or, using exact values of cos 75 and sin 75
L.
60
- 00) +
[cos (315
- V2
- V2
315)
l
a table of trigonometric functions,
r
60.
i sin
2
= V2
= V2
From
=
i
f.
in Section 11-5,
\/2(cos 315
~
+
i
/^i
i
Example 11-7. Find the quotient when
find that the result is
(0.9659)].
previously found,
we obtain
-i
(VS +
4:
1)
_J
= - | [(V3 -
+ i (V3 +
1)
Z
1)].
DeMOIVRE'S THEOREM
we extend the law
n factors, we have
If
[ri(cos di
=
If
+
120).
+
of multiplication of the preceding section to
i sin 0i)] [r2(cos
02
+
2
rir 2
-
-
how we put
-
rn [cos(0i
TI
=
r2
=
+
+
i sin #2)]
-
-
+
n)
[rw (cos
+
n
+ i sin (0i +
=
= rn = r and
2
=
'
2
=
i sin
+
On
-
=
-
n )]
-
+ 0*)].
0, it fol-
lows that
[r(cos 6
+ i sin 0)] n =
r n (cos
n0
+ i sin n0).
known as De Moivre's Theorem.
Although we have derived De Moivre's Theorem
This result
values of n,
is
it
can be shown to hold for
erly interpreted.
all real
only for integral
values of n, if prop-
Complex Numbers
200
Example 11-8. Find the value
Since 1
Solution:
\/2
+ i sin
(cos 315
-
(1
=
i
A/2
i)
7=-
(
-p-i
The two
By De
+ i sin
1
i is
4
)]
-
=
=
+i sin 0) =
(cos
if
2,
315)]
by De Moivre's Theorem.
sin 26
we have
+
cos 2
2
sides are equal only
and
for cos 20
Moivre's Theorem for n
20
the polar form of
>
= [\/2 (cos 315 + i sin 315
= (V^) [cos (4 315) + i sin (4
= 4(cos 1260 + i sin 1260)
= 4(cos 180 + i sin 180) = - 4.
4
i)
Example 11-9. Derive formulas
cos 20
]
\\/2
V2 /
315). Hence, by De Moivre's Theorem,
4
Solution:
11-7
by De Moivre's Theorem.
4
of (1
Sec.
(2 cos
sin 0)
t
-
sin 2 0.
the corresponding real and imaginary parts are
equal. Hence,
-
cos 20
= cos
sin 20
= 2 sin
2
sin 2 0,
and
11-8.
cos
0.
ROOTS OF COMPLEX NUMBERS
Let p(cos + tsin$) be an nth root of the complex number
r (cos + i sin 0), where
^ ^ 360. Then
<
[p(cos
By De
+ i sin <)] n = r(cos 6 + i sin 0).
$
Moivre's theorem, this leads to
p
(11-13)
n
(cos
n<f)
+ i sin rup) =
Our problem now is to find
for which (11-13)
angles
all
is
<f>
+ i sin 0).
KCOS
non-negative numbers /> and all
satisfied.
Separating real and
imaginary parts, we have
n
p cos
(11-14)
ncf)
Squaring and adding,
p
2n
(cos
=
we
2
n<t>
r cos 0,
sin n<
=
r 2 (cos 2
6
r sin 0.
obtain
p
(11-15)
(11-14)
n
+ sin 2 w$) =
2
2*
2
Therefore, p = r , since cos a
is then given by the equation
From
p
+
=
we then have
cos n<j> = cos 0,
sin 2
a=
1.
+
sin
2
6).
The absolute value p
^
sin 0.
sin n<
from these equations that the angles n$ and 6 can
a
only by multiple of 2ir or 360. More precisely,
It is clear
(11-16)
where 'fc
n<f>
is
=
any
+k
integer.
360,
or
=
ft
7i
+
If
.
differ
11-8
Sec.
For
Complex Numbers
201
=
,
0, 1, 2,
(n-1) in (11-16), we obtain n distinct
values of the angle, all of which are non-negative and less than 2?r
or 360. Corresponding to these angles we obtain n distinct roots
fc
given by the formula
/b-360
cos
(11-17)
For example, for k
cipal root,
0,
we
1
fc
.
.
.
J-
n
sin
360
n
obtain one nth root, called the prina
with absolute value r 1/n and angle n
for k
;
1. The value
so on to
would yield the same root as k
n.860* = 6
cos
\
fl
.n
+
sin
(\Tb
k
k
n+
n
n
D
^
360)f =
sin
and
0,
-
and
Similarly,
it
same root as
This means that
1 yields the
and so
1,
only
cos -
+ 360)/ =
(\n
;
'
n
n
/
we have
360
n
k=n
k= n
1,
+
a second root, with absolute value r 1/n and angle
since
=
on.
distinct roots exist.
It is interesting to
note that the
points which represent the roots
are equally spaced on a circle
whose radius is <\/r and whose center
is
the origin. This fact
is illus-
trated in the following example
and Fig. 11-6.
FIG. 11-6.
Example 11-10. Find the three cube
+ i sin 60).
Solution:
The
60
three cube roots are found
+ i sin
60)
60
=
2[cos(20
As
just shown, the substitution of
Hence, for k = 0, we have 2(cos 20 +
140) and
for
k
=
by
evaluating (11-17). Thus,
+k
360-
3
=
t sin
roots of 8(cos 60
the root
is
+k
= 0,
i sin
120)
1,
20)
;
.
.
+
60
+ i sin ,
+ i sin
(20
i
we have
360>
3
+k
120)].
2 yields the three required roots.
for k
1, the root is 2(cos 140
=
+ i sin 260).
+
The
roots are repre-
2,
2(cos_260
sented by the equally spaced vectors OPi, OP 2, and OPs, terminating on the circle
whose radius is 2 and making angles of 20, 140 J, and 260, respectively, with the
;
positive z-axis.
Complex Numbers
202
Sec.
11-8
EXERCISE 11-3
In each of the problems from
expressing the complex
form.
1.
3.
(1
+ i)
(1
-
number
1
to 18, perform the indicated operations by first
Express the answer in rectangular
in polar form.
2.
V3i).
(-1 + V3i)
(\/
+
(l
+ i) (V + i).
1
~
l
4.
0.
fci'-VS.
1
+*
V3 +
7. -
3f
.
V3 + i
11
"'
"
2
13.,,
(1
+*
1
....
+.-..
(3 + 4i)
(2
+
1) (3
(2
+
3t)
+ 1)
14.
)
+ *)
(V3
.7.
i(l
-
_
-1
100
/
20.
V2i)J
[1
1
1
_ \70
(-i + iV
In each of the problems from 22 to 29, find roots as directed and represent them
graphically.
+ i sin
22.
Find two distinct square roots of 9 (cos 50
23.
Find four distinct fourth roots of 16(cos 36
+ i sin
24.
Find three distinct cube roots of 27(cos 165
+ i sin
-
25.
Find
26.
Find the three cube roots of
27.
Find the four fourth roots of
1.
28.
Find the two square roots of
i.
29.
Find the three cube roots of
-^
five distinct fifth roots of
32.
1.
(1
+
V3i).
50).
36).
165).
Sec.
11-8
Complex Numbers
203
In each of the problems from 30 to 34, the complex numbers E,
voltage, current,
and impedance,
E when I =
5
=
4
respectively,
and
E=
7,
and
31.
+ 4i amperes and Z = 30 - Si ohms.
Compute / when E = 110 + 3(K volts and Z = 20 - I5i ohms.
32.
Compute Z when
33.
When two
30.
Compute
7
+ 3i amperes and E =
impedances Z\ and
--=+
A
L%
2
115 volts.
are connected in parallel, the equation
determines an equivalent impedance Z.
ju\
=
5
34. If 2
+ 4f ohms and
and
z are
#2
=8
6i
Z designate
IZ.
ohms.
conjugate complex numbers, prove that
Compute
Z when
Z\
Equations in
Quadratic Form
12
12-1.
QUADRATIC EQUATIONS
IN
ONE UNKNOWN
This chapter provides an extension of the work on linear equations to second-degree, or quadratic, equations. Consider a quadratic equation in one unknown written in the form
ax 2
(12-1)
where
a,
6,
and
c
+
bx
+c=
(a
are given real numbers.
and
called the general quadratic equation in x,
^
0),
This equation is
is said to be in
standard form.
If 6 T^ 0, (12-1) is called a complete quadratic equation; if b = 0,
it is called a pure quadratic equation. Thus, 3x~
a; + 4 =
is a
4; and
complete quadratic equation in which a = 3, b
1, and c
=
x2
2
1 and c
2.
is a pure quadratic with a
In Section 12-4 we shall prove that every quadratic equation has
two and only two solutions or roots. The roots may be equal or
unequal, and they may be real or complex. Their natures depend
on the values of
a, 6,
and
c.
We
eral use for finding these roots,
well to the solution of equations
shall consider the
in gen-
and we shall then apply them as
which are not quadratic in x but
which can be written as quadratic equations
ing the unknown.
12-2.
methods
in expressions involv-
SOLUTION OF QUADRATIC EQUATIONS BY FACTORING
form can be
the
solution of the equation depends on the following
factored,
If the left side of a quadratic equation in standard
important principle
The product of two or more numbers equals zero
at least one of the factors is equal to zero.
:
That
is,
A B=
if
and only
if
A=
or
B = 0.
if
and only
if
Sec.
122
Quadratic Form
in
Equations
205
In practice, we apply this principle by equating to zero each
linear factor of the left side of the given quadratic equation, and
solving the resulting linear equations. The following examples will
illustrate its application.
Example 12-1. Solve 2x* - 7x
To
Solution:
find the values of
+
=
6
by
factoring.
x which satisfy the equation 2x 2
7x
+
6
= 0,
write the left side in the factored form
-
(x
This product equals zero
and only
if
-
x
2) (2x
=
2
2 or x = 3/2. Moreover,
Hence, x
3/2 satisfy 2x* - 7x + 6 = 0. Thus,
_
= 0.
3)
either
if
-
2s
or
=
2(2)2
-
= 0.
3
2 and 3/2 are solutions, because both 2 and
+6=8- 14 +6=0,
7(2)
and
2(3/2)
7(3/2)
+6=9/2-
2
a) Solve the equation 2 sin
Example 12-2.
all
-
2
non-negative angles x
-
x
+6=0.
21/2
sin
-
x
=
1
for sin x. b)
Find
than 360 which satisfy this equation.
less
Solution: a) Factor the given equation to obtain
(sin
Since sin x
1
=
or 2 sin x
x
-
+
1
1) (2 sin
=
0, it
+
x
=
1)
0.
follows that sin x
1
or sin #
=
1/2.
Check:
2(1)
2
-(1)
-1=2-1-1=0,
and
b)
x
=
When
sin x
=
1,
=
x
90; when
= -
sin x
1/2,
x
= 210
or
330. Therefore,
90 or 210 or 330.
Check: 2 sin 2 90
-
sin
2 sin 210
-
2 sin 2 330
-
2
-
90
1
210
-
sin 330
-
sin
=
1
2
Note that
/
=
1
2
-
2^
0,
-
^)
2(
=
-
-
1\ 2 -
/
^J
(
-
1
= 0,
-
1
= 0.
^)
(
-
1\
)
which the unknown is a trigonometric
thus have only two values of sin x which satisfy the
this equation is a quadratic in
function of the angle x.
equation.
1
-
=
and
1
-
We
The determination
solution of the quadratic,
and
of the angle, however, goes
it
may
beyond the algebraic
happen that there are more than two values
of x which satisfy the equation. For this reason,
it is
recommended that all solutions
be checked by substituting in the original equation.
206
Equations
In
Quadratic Form
Example 12-3. Solve the equation 3
-
sec x
=
write
-
values of x less than 27r radians.
=
Solution: Since sec x
and
of fractions, transpose,
-
2 cos 2 x
3/2 or cos x
x for which cos x
number
we can
>
-
2 cos x
3
1 for all
= 2 cos x
cos x
12-2
non-negative
We then clear
1.
factor, to obtain
cos x
=
Hence, cos x
real
1
cos x
Sec.
-
3
=
(2 cos
j;
= - 1. When
= 3/2.
-
z
3) (cos
= -
cos x
+
1)
1,
x
= 0.
=
There
TT.
no
is
Cfcecfc:
3 sec
=
TT
2 cos
TT
-
1,
or
3(-
= 2(-
1)
-
1)
1.
should be noted that factoring provides a method of solving any pure
2
c = 0.
Thus, the equation is equivalent to
quadratic equation ax
It
+
which gives
+
x
/i
- =0
x
or
/i/
-=
d
r
or x
=
This result agrees with that given by writing x 2
-
/!/
r
(Z
and then simply extracting square
x
=
\/
-
>
0.
"~~
/
zb
^
when -
a
0,
roots
both
of
Note that the roots are real when -
^
&
obtain
to
sides
=
and pure imaginary
EXERCISE 12-1
6. In each of the problems from 9
than 360 which satisfy the given equation.
Solve each of the following equations for x or
to 20, find
Check
all
all
non-negative angles
solutions.
1.
x2
3.
2x(x
5.
x2
= 0.
+ 5) = - x = 6.
o;3
-
7.
+
+
9. sin 2 6
7x
27
-
(sec 6
cot
cot
19. 3 sin 2
1
-
3.
+3) =0.
= 0.
=
+
13. sec
17
3x(x
sin
11. sin
15. (cot
less
2 esc
+
6)
l).(cot
+4 ~
+6
2.
z2
-
10x
4.
z2
+
6.r
6.
6x 2
8.
16x
+21 =0.
- 27 = 0.
- 4x - 192 = 0.
- a 2 + 2a6 - 6 2 =
0.
=
16.
+ 2) = 4.
+
3 tan
14. 3 cos 2
-
8 sin
2 esc
0=4 sin*
-
2
= 0.
= 0.
4
12
1
16.
6
3
2+3 cos
1ft
2
3 cos
20.
*
A
4
cos 2^ a
2 esc
esc
8
*
+ 6)
0-10 sin 0.
0.
esc 0.
12. 2 tan 2
143
(cot
=
10. sin 6
+3
"*"
+4
-I=
3 cos
" Q-^-2
8 cos
"^
_
-
2
COMPLETING THE SQUARE
The method developed here is based on the fact that we can make
2
any binomial of the form x + kx into a perfect square if we add to
12-3.
Sec.
it
let
1
2-3
in
Equations
207
Quadratic Form
the square of one-half the coefficient of x. To make this clear,
us recall from Section 1-18 the formula for a perfect-square
trinomial.
The formula
is
+ a) 2 = x + 2ax + a 2
2
(x
Since the coefficient of x in x 2
2
^J
(fc\
Thus, the
left
A*
or
kx
4-
is k,
.
the square of one-half of
2
Adding
-j
this to x 2
+^ =
+
fc#,
we have
+ ~)
x2
+
kx
member
is
a perfect square, namely, the square of
(x
Applicability of the procedure to a variety of processes, including
solution of quadratic equations, is illustrated in the following
examples.
Example 12-4. Solve x 2
-
2x
=
4
by completing the
square.
We first transpose the constant term, so that the left side will be of
+ kx. Hence, the equation becomes
= 4.
x - 2x
= 1 is added to the left side to make it a perfect square.
Now the quantity
I)
Solution:
the form x 2
2
2
(
To obtain an equivalent equation,
The result is
the same quantity
-
x2
2x
+
=
1
is
added to the right
side also.
5,
or
-
(x
Taking square roots
2
=
5.
we have
of both sides,
So the desired solutions are x
I)
x
-
1
1
+
\/5 and x
=
=
\/5.
=
1
V5.
Check:
j (1
and
+ V5) 2 -
(1
-
_
\/5)
2
-
2(1
2(1
+ V5) -4 = 1+2 \/5 +5-2-2
_
_
- V5) - 4 = 1 - 2 V5 + 5 - 2 + 2
2
Example 12-5. Solve 2x
-
+3 =
5x
Solution: Transpose the constant
2x 2
Since the coefficient of x
2
is
not
1,
_
\/5
-4=0,
- 4 = 0.
by completing the square.
term to obtain
-5x
we make
= it 1
5
3.
by dividing both
we have
o
>/5
3
sides
by
2.
Then
208
Quadratic Form
In
Equations
5\\ 2
/
(12 \
both
sides, thus
making the
~
*2
4.=-*
+
^ 16 ~ ~ 2 +
25
.
or
/
X
take square roots of both sides,
we obtain
~
and
2(1)
-
&)
2
=
2
-5(1)
Reduce x 2
Example 12-6.
(y
'
5
1
=
j
=t
T
Solving for
z,
1
l
-:
-r'
4
4
or
#
is,
=
+
is
16
we have x
K
o
=
X
That
16
5\ 2 _J_
~
16
4/
\
When we
"
Tr
~
,
2*
12-3
25
=
/ /
a perfect square. The result
_ 3 25 _
left side
5
Sec.
=
1.
+3=2-5+3=0.
+ y* -
4x
+
Qy
+4 =
the form
to
(x
-
2
ft)
r2 .
The solution of this problem requires that we complete the square of
the terms containing y as well as the square of those containing x. Hence, for
convenience, we write the equation in the form
Solution:
-
(x
When we
4z
)
+
(y
+ 6y
2
)
= -
4.
complete the squares in the parentheses, the equation becomes
- 4s + 4) + (0* + 6// + 9) = - 4 + 4 + 9.
(x*
Thus, the solution
is
-
(x
+
2
2)
(y
+ 3) 2 =
9.
Example 12-7. Reduce 9z 2 - 4y 2 - l&c - 16#
A(x -ft) 2 - B(y -k)*
- 43 =
=C.
- 2x
-
Solution: Write the equation in the form 9(x 2
Complete the squares in the parentheses to obtain
9(x
2
-
2s
+
1)
-
2
4(i/
+ 4y +
4)
)
= 43 +
9
to the form
2
4(?/
-
+ 4t/
)
= 43.
16.
Note that the numbers and 4, which are added within the parentheses to complete
the squares, must be multiplied by the coefficients 9 and - 4, respectively, to
determine the numbers that are added to the right side.
The reduced form is, therefore,
1
9(s
-
2
Example 12-8. Reduce \/3z
Solution:
I)
2
-
4(0
+ 4z -
+ 2) 2 = 36.
4 to the form y/a((x
For convenience, work with the quantity 3z
2
radical sign until the final result is obtained. Hence, write
3* 2
+ 4* - 4 =
-
+ 4#
h)*
-
-
A;
2
).
4 without the
Sec.
12-4
Complete the square
Now,
of the terms in x
209
Quadratic Form
in
Equations
and simplify
to obtain
write this result under the radical sign to obtain
Comparing
choose a
=
h
3,
= -
2/3,
=
and k
vX(#
form
this with the required
h)
2
+
/
l/3((
/c
2
),
we
<r)
)
see that
we may
d= 4/3.
EXERCISE 12-2
In each of problems from 1 to 15, solve the given equation by completing the
less
In each of the problems from 9 to 15, find all non-negative angles
square.
than 300 which satisfy the given equation. Check
2. x 2 + lOx = 40.
1. x 9 - 8x = 20.
4. X 2 + x + i = o.
5. z 2 + x + 2 = 0.
7.
z2
12. 1
14. 3
+z -
5
=
0.
8.
2x 2
= 3z + 9.
20.
22.
24.
x2
6x 2
9.
tan 2
esc 2
-
15. sec
0.
+
2 tan
1.
2.
cos
=
2.
to 25, reduce the equation to the form
1(5
17. x 2
4x +
+ 32^ - 18?y + 37 = 0.
2 2
- 32 = 0.
9x + 4?y
Sy
x 2 - 9// 2 - 4x 4- 36;// - 41 = 0.
2 4r 2 + 50# -f 32x + 41 = 0.
5//
%
=
^-4
01 =
13.
2
2
- 7x = 30.
- 5x - 1 =
3.
6.
+ tan-' 6 = sec 0+3.
cot
+ cot = 3 - 4 cot 0.
In each of the problems from
A(x -h) 2 +li(y -A-) 2 =C.
16. x 2 - 4?/ 2 - 2x + 1 = 0.
18.
all solutions.
2
2
19.
x
21.
4.c 2
+ 4i/ 2 + 4// -
23. 4x 2
25. 3* 2
2
-f 9^/
-
9//
y
2
2
2
6x
+
16y
+ 21 =
0.
- 40y -h 109 = 0.
- Ifxc - ISy -11=0.
lOx
+ 32o; -h 36y -f 64 = 0.
+ 2Qx - 2y + 11 = 0.
In each of the following problems, express the quantity inside the radical or
- h) 2
k 2 ).
parentheses in the form a((x
26.
V&
&r
-
27.
40.
\/x
2
-
V(9^ 2
12-4.
Gx
+
-
48,-c
-
v/2x
30. (x
*
29.
32.
~
-
2
16 J
6.r 4-
+
34)
41.
28.
2/3 .
31. (2x
20?/
-
+
28.c
+
34)~
-
14a:
+
II)
2
-
2
(4//
76)
3 '2
.
1/2 .
7
+ 23) 3
.
33. (9z 2
+ 24x + 25)~ 1/3
.
34. (7x*
SOLUTION OF QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA
applying the method of completing the square to the general
quadratic equation (12-1), we can obtain a formula for the roots,,
either real or complex, of any quadratic equation whatever. The
general equation is
By
(12-1)
ax 2
Transpose the constant term
ax
2
+ bx + c = 0.
c,
and obtain
+ bx
=
c.
21
in
Equations
Dividing by
a,
Quadratic Form
Sec.
1
2-4
we have
c
6
+ ax = --a
x *2|
2
6\ 2
- = & ^2 to both
)
(1
4a
2 o/
sides to obtain
.
*
o
+
.
b
62
.
:C+
=
-
which becomes
2
ft
_ 4ac
&2
Extracting square roots of both sides gives
Solving for x,
Vb 2 - 4ac
b
__
+ 25~
*
,
25
we have
6
=
*
-\/&
2
4ac
&
Hence, to solve a quadratic equation, put
form ax 2 +
60?
+c=
0,
and substitute the
it
into the standard
coefficients a, &,
and
c in
the formula just derived to obtain the roots
,10 ON
(12-2) xi
=
- & + Vb2 -
4ac
%
,
x2
and
-
=
b
- Vb 2 -
4ac
^
That these numbers #1 and #2 are solutions of the given quadratic equation is shown by substituting each of them in (12-1).
The details of this substitution for x l follow
-:
/b 2
-
26
b2
Vb 2 -
4ac
+
2
(6
+c.
- 4ac)
+ b Vb 2 - 4ac
2a
b2
-
2ac
- 6 Vb 2 -
-
-
4ac
62
+ 6 \b 2 - 4ac
.
A
2
Hence, the number x satisfies the equation ax + bx + c = 0.
similar computation shows that #2 is also a solution of ax 2 + bx + c
= 0. Consequently, the two expressions given for x in (12-2) are
actually roots of (12-1). In Section 12-7 we shall study the expressions in (12-2) further and shall determine when the roots are dis-
tinct
and when they are
real.
Sec.
2-4
1
Example 12-9. Solve 5z 2
= 5,
Here a
formula, we obtain
Solution:
=
fr
=
=
8
6#
c
6,
21
Quadratic Form
In
Equations
1
by the quadratic formula.
Substituting these values in the
8.
_
_6
~
\/36
db
6
160
-f
Therefore, x\
j^
These values are seen to
=
_
*~~
Hence,
si
=
1,
-r
-
+2 =
= 1, c = 2,
Vr~^~8 _1
6
1
= g
when
satisfy the original equation
2
Example 12-10. Solve x
Solution: Since a
10
10
= 2 and z 2 =
=
6 db 14
\/196
db
10
x
by the
substituted for x.
formula.
we have
\^J "_
2
1 db
2
1+V?
~
j
and
z2
1
=
Ctecfc:
Similarly, the second root
may
Example 12-11. Solve 2
all
non-negative angles x
We make
Solution:
sin 2
less
be checked.
x
-f
3 cos x
-
3
=
by the formula, determining
than 360.
+ cos
use of the identity sin 2 x
2
=
x
1
to transform the
given equation into one involving a single trigonometric function of x.
2
cos 2 x we have
Replacing sin x by 1
t
2(1
-
+
cos 2 x)
-
3 cos x
= 0.
3
we obtain
Simplifying,
2 cos 2 x
-
3 cos x
= 0.
-f 1
--
Solving for cos x by the formula, we find
cos x
Hence, cos x
The
=
solutions
=3
V9 -
=t
A
mining
=3
1
A
4
=
or 1/2. Therefore, x
or 60 or 300.
may be checked by substitution in the original equation.
1
Example 12-12. Solve cos x tan x
all
8
4
-f sin
2
x
=
sin x
1
by the formula,
non-negative values of x less than 360.
Solution:
Making
6 use
of the identity
tan x
-
cos x
-sin
x
cos
a:
.
2
h sin x
,
=
cos x
=
>
we have
.
1
sin x.
deter-
212
Equations
Quadratic Form
in
Transposing and simplifying, we get
sin 2 x + 2 sin x
Sec.
1
2-4
= 0.
1
x by the quadratic formula, we obtain
Solving for sin
sin z
-
=
2
2
= -
-
/o
V2.
1
1
one solution. However, sin x =
\/2must
be excluded, since tho sine of an angle cannot be numerically greater than 1.
The value
=
3
sin
Check: For sin x
+ V2
1
=
cos x
\/2
=
is
we
1,
-
\/2 V2
find that
and
2
=
tan 3
~
+ (V2 -
V2 -
1
+3 2
From
-
=
2
I)
- (V2 -
1
1).
2
This reduces to
or
2
we have
Substituting these values in the original equation,
V2V2-2-
V2 -1
_
V2V2 -
_
2\/2
x/2
=
2
the table of trigonometric functions,
=
-
1
V
_
+
1
- V2.
=
we have x
2428' or 1 5532'.
EXERCISE 12-3
Solve each of the following equations for x or 8 by the quadratic formula. In each
of the problems from 14 to 24, find all non-negative angles 6 less than 360 which
satisfy the equation. Check all solutions.
1.
s*
2.
x2
-
3.
x2
4.
x2
+
5.
2
6.
7x2
7.
+ 6* - 7 = 0.
+ 2x - 5 = 0.
z + 2z + 1 = 0.
2z 2 4- 3* + 2 = 0.
9. (2x
-
I)
-
2
2(2x
13.
3+2
17.
-
8
= 0.
2
21. (sec
23. (esc
12-5.
10. (a 2
2X
+
3
3
*
x
+
;
'
3
"*"
cos
0+1)
+ 2)
= 0.
= 0.
_ 9 = o.
- 5 = 0.
20
+
I
_ gj.
- 7x
- 6 2 )z 2 -
ft
14 - ww
cot 2
1
+
^
"
-
cot *
2 cot
(sec
22. (sin
2
24.
-
-
3
3
-
3
(a
+
2
-
62)
= 0.
1
4
:
tan
=
-
+
3
=
"
3
"
18. 16 sec 2
on
-
'
3
16. esc
2 cos 2
+ 2) = sec + 3.
+ esc = 1.
-
4abx
8
6
= 0~
v
+r 3x
o-Vr
+5
-- = 3.
tan
cot
- 2 sin = sin + 3.
sin 2
^ __ ___ _
- +
^
3
15.
1)
-
5
11
11.
-
8. 9.r 2
-
x
x
2
sin 0.
8 sec
+
2 cot 6
_
3
+ 3)
1
-
= 0.
1 ~~ o
cot
(3
-
sin 0)
=
T +|
= 3(sin + 3).
-
EQUATIONS INVOLVING RADICALS
Sometimes an equation in which the unknown appears under a
radical sign can be reduced to a quadratic by raising both sides to
Sec.
1
2-5
Equations
Quadratic Form
in
213
a power sufficient to remove the radical. The process must be
repeated until the unknown no longer occurs under a radical.
The operation of raising both sides of an equation to a power may
lead to an equation redundant with respect to the original that is,
the final equation may possess roots that are not roots of the original equation. Such roots are called extraneous roots. For this
reason, every root obtained must be checked by substitution.
;
+ 4 = 2.
- 3x + 4 = 8.
= 0.
Example 12-13. Solve the equation \/x* - 3x
Cube both
Solution:
sides to obtain x 2
-
x2
Factoring and solving for
x,
we
4
Transpose, and get
find that
=
x
-
3x
x
or
1
= 4.
Check:
>X(-1) 2 -3(-l)
and
+4 = ^1 +3 +4=^8=2,
_
Hence, both
1
and 4 are
+4 =
3(4)
^/g
=
1
\/x
2.
roots.
Example 12-14. Solve the equation \/2x
-
+3 =
1.
Solution: Transpose one radical to obtain
When
=
1
both sides are squared, the result
is
2x
like terms,
Combining
-
=
1
+ \A~T3.
_
+ 2>A +3 + x +
we obtain
-
5
=
2 V-c
square both sides to get
x 2 - Wx
+
25
=
- Ux +
13
=
0.
x
=
13.
x
Now we
1
1
3.
+ 3.
4(s
+
3).
Transposing and combining gives
;r*
By
factoring
and
solving,
we
find that
_
x
,
and
Hence, 13
is
=
1
V2(l) -
1
V2(13) -
1
a root, but
1 is
or
_
- VI +3 = 1 - 2 ?* 1,
- V13 +3=5-4 = 1.
not.
EXERCISE 12-4
Solve each of the following equations. In each case check for extraneous roots
3.
V% - 2 = 4.
Vz + 5 = 1.
5.
\/z
1.
2
-
16
=
2.
4.
2x l
.
6.
V3z +4 =
2.
^3z -1=7.
>A 2 - 2 = V2.c +
6.
214
-
7.
x
9.
\/2x
11.
V^T =
-
3
+
x
\/5
3
=4 -
4-
3x.
IN
=
5.
Sec.
*
8.
0.
\X4oTT~5
EQUATIONS
12-6.
Quadratic Form
in
Equations
-
5* 1
10.
\/z
12.
\/% x
-
'
1
2
12-5
+6=0.
+ Vz - 3 = 2.
QUADRATIC FORM
Frequently, an equation which
unknown may be considered
not quadratic in the given
is
as a quadratic in
- 3or 2 +
some expression
and 2(x 2 - 2x) 2
2 =
Thus, or
6
(x
2x)
may be treated as quadratic equations in the
2
2
or
and
expressions
(x
2x), respectively. This type of situation
was met earlier in Examples 12-11 and 12-12. The following
4
involving the unknown.
=
2
examples will further
in quadratic form.
methods used
illustrate
4
Example 12-15. Solve the equation or
Solution: Let x~ 2
= y,
-
3or 2
in solving equations
+2=0.
so that the given equation becomes
_
7/2
Factor, to obtain
+
3y
-
=
2
(y
-
1) (y
y
=
l
or
y
or 2
=
1
or
z~ 2
=
2)
Therefore,
0.
0.
= 2,
or
= 2.
Hence, the solutions are
x
Example 12-16. Solve
Solution: Let x 2
2
(z
=
-
2
= y,
and
1
2)
-
2
y
-
2
Factor, to obtain
2)
7y
get
10
=
2
+
-
1
=
-
7(.r
we
so that
a:
7=
+
= 0.
-
2) (y
y
=
2
or
y
2
=2
or
.T
2
-
2
x2
=4
or
z2
=
7,
Hence,
= 0.
0.
(y
5)
10
=
5,
or
x2
-
= 5.
Then,
and the
roots are
&
=
2
and
=
\/7.
Sec.
127
Equations
2
Example 12-17. Solve x
+
-
x
i'n
V^ + z +
2
2
215
Quadratic Form
= 0.
3
V# 2 + * + 3 = Then we can write
+ 3) - 2 -\A + z + 3 - 3 = 0, or y - 2y - 3 = 0.
(x +
Therefore, y = - 1 or 3, and \/z + z+3=-lor Vz 2 + x + 3 = 3.
Solution: Let
2
?/.
2
2
re
2
definition of the radical,
By
\A +05+3=
2
1
\/a
is
a non-negative number.
Hence, although
consistent with the original equation, there are no values
is
of x which satisfy this equation.
Consider, then,
V# 2 + # + 3 = 3.
z 2 + x + 3 = 9,
This leads to
x 2 + a?
or
Hence,
x
=
2
or
a?
= -
-
6
= 0.
3.
Substitution shows that each of these values of x satisfies the original equation.
EXERCISE 12-5
Solve each of the following equations. Check
1.
3.
5.
7.
9.
^+
12 = 0.
- 4 = 0.
(x + 2) + 3(z + 2)
x 4 - 13x + 36 = 0.
(3* - 4) + 6(3x - 4) + 13 = 0.
f x + 1V +
- 48 = 0.
+
X2
_
2
2
2
4jr*
2.
6.
x4
2
8.
(x
10.
(a;
-)
-
-
4. r*
2
2(2;
all solutions.
6z
1
-
liar*
= 0.
3
+ 8 = 0.
= 0.
2
2
+ 3z) 2 -
2
-
2
a:)
-
14(z
20(x
2
2
+ 3x) + 45 = 0.
- x) + 36 = 0.
THE DISCRIMINANT
12-7.
be recalled that the two roots of the general quadratic
are
equation ax- + bx + c =
It will
-
,
and
The expression
62
4ac
2a
which appears under the radical sign, is
the discriminant of the quadratic polynomial ax 2 + bx + c, or
called
4ac,
the discriminant of equation (12-1).
In what follows
shall
make
we
shall
assume that
a, 6,
and
the solutions Xi and
1.
If 6
,
2a
are real, and
we
use of the discriminant to determine the character of
the roots without actually solving the equation.
2
c
4ac
=
x>2 ,
0,
we
By
inspection of
reach the following conclusions
each of the two roots #1 and x 2
and both roots are thus
real.
is
:
equal to
216
in
Equations
If 6 2
2.
real,
- 4ac
and they are
If b 2
3.
4ac
V& 2 ~~ 4ac
both roots are
is real,
distinct.
is
negative, then
V& 2 ~~ 4ac
be summarized as follows
may
imaginary, and the
is
numbers of the form a +
roots are distinct complex
These results
Sec.
then
is positive,
12-7
Quadratic Form
fti
and a
/3i.
:
4ac is a perfect
Furthermore, if a, 6, and c are rational, and ftrational square, then the roots are rational; otherwise, they are
irrational.
The following examples will illustrate
character of the roots.
Example 12-18. Determine the character
+
=
7x
49
+
2x*
Solution:
Here a
=
2, b
62
The
discriminant
discriminant
is
is
-
=
7, c
4ac
positive,
-
2kx
-f
4
and
15.
to determine the
of the roots of
15
=
0.
Hence,
120
=
169
=
(13)
so the roots are real
2.
and unequal. Since the
a perfect square, the roots arc also rational.
Example 12-19. Determine
kx 2
=
-
how
=
all
values of k for which the roots of the equation
are equal.
Solution: The discriminant must equal zero for the equation to have equal roots.
2
2
or k - 4. When
Hence, 6 - 4ac = 4& - 16& = 4k(k - 4) = 0, and so k =
=
k
0, the equation is not quadratic. Therefore, the roots arc equal only when k = 4.
EXERCISE 12-6
Determine the character of the roots of each of the following equations by means
of the discriminant.
1.
4.
7.
10.
13.
+ 3s + 4 = 0.
6z - 7x + 3 = 0.
5z 2 + 7x + 2 = 0.
4* 2 - 8z + 2 = 0.
4x2 _ i 2x - 9 = o.
ar
a
2
+ &c - 9 = 0.
x + IQx -f 2 = 0.
4z - 12z + 9 = 0.
2x2 _ x + 3 = Q.
x 4 4z - 18 = 0.
2.
x2
3. x*
5.
2
2
8.
11.
14.
2
2
6.
x
9.
x2
12.
+
+
3x
-
2x
5z 2
15. x
2
IQx
+z
-
4z
-
+2
+1
+5
+7
= 0.
= 0.
= 0.
= 0.
= 0.
6
Sec.
12-8
217
Quadratic Form
in
Equations
SUM AND PRODUCT OF THE ROOTS
Adding the two roots of the general quadratic equation
12-8.
ax 2
we
+
-
xi+x*=
b
+
-
-
2
\/b
4ac
XlX2
=
0,
6
-
Vb
4ac
26
6
-_=--.
=
Ya
we have
+ Vb
- -ac
2
6
-
2
- Vb 2 -
b
4ac
2
b
__
""
we
=
c
,
2
-
-
+
Also, multiplying these roots,
Hence,
+
bx
by using (12-2),
obtain,
2
2
(b
4ac)
__
""
4a 2
4ac
c
__
""
4a 2
a
sum and product
have, for the
xi
(12-3)
x2
= ~
xix a
=-
+
and
(12-4)
-
of the roots,
>
tZ
These formulas are used in various ways, for example, in checking roots of a quadratic equation, and in forming an equation if its
roots are known.
To find the factored form of the quadratic polynomial ax 2 + bx
+ c, let us make use of the sum and product formulas just found.
for 6, and solving XiX 2 = - for c, we have
Solving Xi + x.> =
b
a(xi
Therefore,
ax 2
a
a
=
+
bx
-f
+
x<2 )
and
c
ax 2
+
+c=
ax 2
=
=
a(x
if
c
=
x\
bx
and
0, its
may
a(x
- (xi
2
- xi) (x - X2).
xi) (x
=
#2)
0.
write the equation equivalently as
(X
its
2
we have
+ x<)x + a(x\X2)
+ x )x + (ziofe))
are the roots of the quadratic equation
factored form can be written
^ 0, we may
This form or
a(x\
x>2
a(x
Since a
Hence, by substitution,
a(x^x^).
-
Xi) (X
=
3fe)
expansion,
X2
(#1 + X 2 )X
0.
+ XiX2 =
0,
be used in writing an equation whose roots are known.
Example 12-21
indicates the procedure.
Example 12-20. Without solving the equation, find the
5.c
=0.
of the roots of the equation 3x 2
sum and
the product
+2
Solution: In this case, a
=ft
o
,
=
and the product
3, b
-
5,
of the roots is
and
c
=^
do
-
=
2.
Then
the
sum
of the roots is
218
Example 12-21. Write a quadratic equation
roots are (1
\/3
Xl
=
1
X2
=
(1
sia*
=
+
+
\/3
+
(1
12-8
form (12-1), given that the
and x 2
=
+
(1
~~
(1
- V3
l>)
\/3 i)
2
is x
2r
have
t,
^
4
-f
V3 we
=2
f)
1
'
i)
=
=
0.
1
-
3
i
a
=
4.
Writing the desired equation in factored form, we have
Alternate Solution:
(*
-
+ VSO) (*-(!- V3 <)) =
(1
0.
we obtain
-
(z
or
((*
-
-
1
\/3
-
(x
i)
+ V3
1
- V3 i) ((* - 1) +
- 1) + 3 = 0,
(3
1)
Hence,
and,
i
^
+
Therefore a suitable equation
Simplifying,
in the
Sec.
i).
Solution: Since x\
and
Form
Quacfrof/c
in
Equations
finally,
a;
2
-
+4 =
2x
= 0,
i)
x/3
f)
=
0.
0.
EXERCISE 12-7
In each of the problems from
of the given equation.
1. z*
2* - 1
0.
=
+
6x
4.
2
7. 5x*
-
2.c
+
3
-
Qx
-
1
Form an
10. 1,
1/t
14
3
-
to
9,
2. 3.r 2
-
x
5
sum and
find the
+
=
2
8. 5
2
4- 6.c
+
1
6. 5.D
-
=
+
3. a*
0.
-ir-4*+i=
the product of the roots
9.
0.
2
=
2
-
lOOx 2
0.
6z
+
-
40x
= 0.
1
+17-0.
equation with each of the following pairs of roots.
3.
11. 0,
1
15. 2
'2'3'
18.
= 0.
= 0.
1
V3, V5.
19. 0,
+
2.
12.
iV
16.
-
13. 3, 6.
1, 1.
17.
i.
|(1
V3 -
20. a
6f
21.
.
V3
GRAPHS OF QUADRATIC FUNCTIONS
12-9.
To graph any quadratic function
the form ax4-
bx
+ c and
+
bx
+
c,
we
set
of
ax 2
y
construct a table of values
of y corresponding to assigned values of
x. The graph is of the type shown in
and is called a parabola.
As found by the quadratic formula,
the two solutions of the general quad-
Fig. 12-1
ratic equation (12-1) are given
(12-2)
Xl
=
X2
=
4ac
2a
and
FIG. 12-1.
6
\/b 2
4ac
by
Sec.
1
2-9
Equations
Since (12-1) states that y
solutions Xi
Quadratic Form
in
=
in the equation
219
y
=
ax 2
+
bx
and x 2 are ^-intercepts of the curve.
let A and C be the ^-intercept points, and
In Fig. 12-1,
the mid-point of AC.
or
We note
Now
B is
the point
(
^-
=
ax-
+
=
ax-
bx
+
+
bx
k
c
-f c
=
Q.
be
which represents a straight
fc,
may
This line
parabola, depending on the value of
and y
B
let
V
>
consider the equation y
line parallel to the x-axis.
the
7
.
Therefore,
c,
that
OB = OA + AB,
""
Xl
X2 - Xl +X2 -
-
+
or
may
not intersect the
Solving the equations y
k.
simultaneously, by elimination of
The roots of
?/
we
=
k
obtain
this resulting equation are
and
6
\/6
2
4.a (c
k intersects the curve in two
If the value of k is such that y
distinct points, the discriminant in
and the roots
will be real
with abscissa x
section,
= -
^ci
is
/c)
2a
2a
and
(12-5)
distinct.
is
greater than zero,
The point on the
line
y
k
then equidistant from the points of inter-
whose abscissas are x and x z
.
=
k does not
intersect the curve, the discriminant in (12-5)
is less
than zero,
and we have a pair of conjugate complex
In this case,
If,
is
on the other hand, the value of k
is
such that y
roots,
z>a
the real part of these roots.
The points with abscissa x =
and arbitrary ordinates
lie
0.
on a vertical
line, called
of the curve; the curve
the axis of symmetry, or simply the axis
is
said to be symmetric with respect to
the axis.
The point
of intersection of the axis and the parabola
is called
the vertex of the parabola. If the coefficient a of the second-degree
term of y = ax- + bx + c is positive, the vertex is the loivest point,
and the curve is said to be concave upward. If a is negative, the
220
Equations
vertex
is
in
Quadratic Form
the highest point, and the curve
Sec.
12-9
said to be concave
is
doivmvard.
To
find the coordinates of the vertex,
the equation x
=
b
=
,
and
(12-6)
y
That
=
ax 2
+
bx
+
c
a& 2
-
-r-o
b2
b2
,
-^
4o
in another way. Let us
demand
=
k intersect the parabola in two coincident
us insist that the two roots x\ and xs in (12-5)
that a horizontal line y
coincide.
=
of the vertex are thus found to be
The vertex may be characterized
points.
solve simultaneously
of the axis and the equation y
The coordinates
of the parabola.
we
is, let
Then the discriminant
in
(12-5)
is
equal to
0,
and
=
^-- The value of y corresponding to this value of x is
Za
the ordinate of the vertex, as found in (12-6) We say that the line
#1
x2
.
y
is
= -
-
b2
4ac
4a
tangent to the parabola at the vertex.
-
2
Example 12-22. Graph x
Solution:
Let y
ponding values of
The
=x -
?/,
2
6x
as in the
-f 4, assign
values to
accompanying
table.
x,
and compute the corresis shown in Fig. 12-2.
The graph
coordinates of the vertex are
=!(-i.il)
(2,
+ 4.
Gx
-4)
and
b2
2/=c
^-=4
9=
-5.
(7,11)
Hence, the axis
is
the line x
positive, the vertex (3,
=
3.
Since a
is
5) is the lowest point
on the curve, and the curve is concave upward.
Sec.
12-10
Equations
in
Quadratic Form
2
Example 12-23. Graph y = x - Qx + 9 and ?/
same coordinate system as was used for the graph
The
s2
-
Qx
= x2
of y
+
14 relative to the
6#
Example 12-22 but applying
shown in Fig. 12-3.
Solution: Tables similar to that in
curves are constructed.
=
221
-f 4.
to the
first
two
three curves are
(C)
y
-
j-2
_
Ox
+
14
FIG. 12-3.
Curve (A)
of
.r
2
Gx
H-
4
roots of x 2
6.c
because x 2
Gx
0.
+
+ 14 =
1
has imaginary roots.
reader should relate the discriminants of the quadratics to a study of these
1
The
two points, corresponding to the roots 3 =fc \/5
Curve (#) is tangent to the x-axis at (3, 0), because both
9 =
an equal to 3. Curve (C) does not intersect the z-axis,
crosses the x-axis at
graphs.
12-10.
QUADRATIC EQUATIONS
TWO UNKNOWNS
IN
The general equation of the second degree
ax
(12-7)
2
+
bxy
+ cy +
2
dx
+
ey
in x
+f =
and y
is
0,
and / are given real numbers. An equation of
this form, in which at least one of the coefficients a, 6, and c is
different from zero, is called a quadratic equation in x and y.
By a solution of such an equation, we mean a pair of real or
complex numbers which, when substituted for x and y in (12-7),
where
a, b, c, d, e,
will reduce the left side of the equation to zero.
infinitely
many
pairs of
numbers which
Usually there are
satisfy the equation.
222
Equations
in
Quadratic Form
Sec.
12-10
If c T^ 0, (12-7) may be solved for y in terms of x by means of
the quadratic formula. Corresponding to each real value assigned
to x f we then obtain, in general, two values of y. We then have
pairs of numbers (x, y) which, if real, may be plotted in a rectangular-coordinate system. It is shown in analytic geometry that
the graph so obtained will be one of a class of curves called conic
sections, inasmuch as they may be obtained as curves of intersection of a plane and a right circular cone. At this time we shall
confine ourselves to merely listing the curves which comprise this
and indicating briefly the form of the quadratic that corremore adequate discussion of this
sponds to each of the graphs.
class
A
subject is given in analytic geometry. However, typical examples
of these curves are given here.
1. Parabola. When A ^ 0, the equations y = Ax- + Bx + C and
=
x
Ay- 4- By + C represent parabolas with vertical and horizontal
axes of symmetry, respectively.
the equation x-+ y 2 = C represents
a circle whose center is at the origin and whose radius is \A?3a. Ellipse.
When the constants are positive, the equation
Ax- + By- = C represents a curve called an ellipse. If A = B, the
2.
Circle.
When C
is positive,
ellipse is a circle.
Point Ellipse. If A and B are positive and C
0, the equaAx- + By 2 = C is satisfied by only one point, namely, the origin.
The graph is then said to be a point ellipse.
3c. Imaginary Ellipse. If A and B are positive and C < 0, there
are no (real) points on the graph, and we say that the equation
Ax 2 + By 2 C represents an imaginary ellipse.
4a. Hyperbola. When A, B, and C are positive, the equations
Ax- By 2 = C and Ay 2 Bx 2 = C represent hyperbolas.
3b.
tion
the equation xy = C represents a
curve called an equilateral hyperbola.
5. Pair of Straight Lines. The equation Ax 2 + Bxy + Cy~ + Dx
f Ey + F =
represents two straight lines, which may be either
distinct or coincident, if the left side can be expressed as the product of two real linear factors.
We use the quantity b 2 4ac, which is usually called the characteristic of the equation ax 2 -f- bxy + cy 2 -f dx + ey -f / = 0, to determine the nature of the conic corresponding to a particular form of
the general quadratic equation. In analytic geometry the following
statements are shown to be true
1.
If b 2
4ac = 0, the conic is a parabola or two real or imagi4b.
Hyperbola.
When C ^
0,
:
nary parallel
2.
If
2
b'
4ac
imaginary
<
lines.
0,
the conic
ellipse.
is
an
ellipse or
a point ellipse or an
Sec.
3.
12-10
If b 2
4ac
>
0,
the conic
223
Quadratic Form
In
Equations
is
a hyperbola or two intersecting
lines.
In Section 12-9 the graph of the parabola y = ax 2 + bx + c was
discussed. In the following illustrative examples, the procedures
for graphs of other quadratic equations are considered.
2
Example 12-24. Graph x
+ y2 =
9.
=
=
to obtain the ^-intercepts, which are
Solution: Set y
3; and set x
to obtain the ^-intercepts, which are it 3. Solve the equation for y, obtaining
y
V9 -
=
x2
.
construct a table of other corresponding values pf x and y. To yield real
y, the numerical value of x cannot exceed 3. As shown in Fig. 12-4, the
resulting graph is a circle with center at the origin and radius 3.
Then
values of
IY
-3
FIG. 12-4.
2
Example 12-25. Graph 4z
+
9?/
2
=
36.
to obtain the ^-intercepts, which are
Solution: Set y =
3; and set x
2. Solve for y and obtain
obtain the ^-intercepts, which are
y
=
o
V9 -
x2
to
.
Construct a table and draw the curve, as shown in Fig. 12-5.
an
=
This illustrates
ellipse.
X
-3
-
2
1.49
1
1.89
2
1.89
1.49
FIG. 12-5.
Note that the numerical value
yield real values of y.
of
x must be equal to or
less
than 3 in order to
224
2
Example 12-26. Graph 4z
-
2
9s/
Quadratic Form
in
Equations
Sec.
12-10
= 36.
3. Setting x = 0,
Solution: Setting y = 0, we find that the ^-intercepts are
2
- 4. Hence the curve has no ^/-intercepts.
however, results in the equation y =
Solving the given equation for
we have
?/,
=
y
The accompanying
table
is
V* 2
|
9-
constructed.
FIG. 12-6.
Note that
This
x must be equal to or greater than 3 in
The graph of the given equation is shown in Fig. 12-6.
in this case the numerical value of
order to give real values of y.
illustrates
a hyperbola.
EXERCISE 12-8
Identify and graph each of the following.
1.
x2
4.
4z 2
7.
y
10.
12.
+y =
2
-
9*/
2
25.
2.
=
5.
36.
+
-
?/
2
2
?y
17.
19.
4*/
= = x 2 - 3x + 2.
8.
2
2
2z +
4y =4.
3z 2 - 4xy + 2^/ - 6x + 3?/ = 7.
14. 4* - 4ry +
2 2x - 2 = 0.
xy +
y
2x - xy - 28y 2 = 0.
9z 2 - 24xy + 162/ 2 + 3x - 4?/ = 6.
2
15.
9z 2
2/
2
12-11.
GRAPHICAL SOLUTIONS
QUADRATICS
2
= 36.
= 16.
4.
3. 4,r 2
6.
x2
9.
5^ 2
-
+
= 0.
=
0.
9?y
2
=
28?/
ftcy
-f-
2
9?/
2
.
= 2x +
13. x + xy - 2y* + 3y - 1 = 0.
2x + 4?y - 12 = 0.
16. z 2 - 3x - 3y* + I8y - 27.
18. 4x + 4xy - 3i/ 2 + 4x + lOy = 3.
20. 4x 2 + 3xy + 4i/ 2 - Sx - 8y = 24.
11.
5ar?y
?/.
2
2
OF SYSTEMS OF EQUATIONS
INVOLVING
In Chapter 9 we solved systems of two or more linear equations
both algebraically and graphically. Frequently, however, simultaneous systems include one or more equations of the second or
higher degree. We have, therefore, to consider the problem of finding systems of values of the unknowns x and y that satisfy two equa-
Sec.
12-11
Equations
one of which
tions,
is
in
225
Quadratic Form
quadratic and the other of which
is
linear or
quadratic.
We
shall begin
types of systems.
by illustrating some graphical solutions of several
The graphical method yields only the real solu-
tions of a system, but
may
prove advantageous in suggesting
solutions and interpreting results. In general, this method yields
at best only approximate solutions. The graphs should be drawn as
it
accurately as possible.
Example 12-27. Solve graphically the system
'
x2
x
- 8* + 3y = 0,
- Zy + 6 = 0.
*
Solution: Solving each equation for y in terms of x,
y=
8x
-
x2
,
and
x
we have
+6
=-3
Construct tables of values, and draw both graphs, using the same coordinate
system, as shown in Fig. 12-7.
7 123456
+ 3^-0*
+*x
7 8\
(A): J?2 -8.1:
(B):
*-
FIG. 12-7.
The
line
and the parabola are seen to
It follows that these points represent
intersect at the points (1, 7/3)
common
roal solutions, possibly
and
(6, 4).
only approxi-
mate, of the system. A check by substitution shows that the real solutions are,
in fact, x = 1, y = 7/3; and x = 6, y = 4.
The
Example 12-27 suggests a procedure for finding
graphical solutions of quadratic equations in one unknown.
solution of
226
Equations
in
Quadratic Form
-x
Example 12-28. Solve the equation x*
Since x
Solution:
=
2
x
+
Thus, the original equation
2,
is
2
=
Sec.
12-11
graphically.
both sides of this equation
may
be set equal to
y.
by the system
y = x\
replaced
f
(
y=x+2.
y
=x +2
2
From
d=l
1
2
4
3
9
4
16
-2
the accompanying tables of values of x and ?/, the graphs shown in Fig.
The graphs show that the line and the parabola intersect at
12-8 are constructed.
the points
1,
(
their abscissas
1)
must
and
(2, 4).
Since these points are
x2 = x
+
=
2.
'(-2,0)
FIG. 12-8.
Example 12-29. Solve graphically the system
t
(
Solution:
From
the
first
and
From
and
to both graphs,
2.
2
Hence, the required roots of the original equation x
and x
common
satisfy the equation
the second equation,
3x
2
-
2?/
given equation,
2
=
6.
x
2
=
are x
=
1
12-12
Sec.
Equations
The necessary
The
ellipse (A)
in
227
Quadratic Form
and the graphs are shown
tables are given here,
and the hyperbola (B) are seen to
in Fig. 12-9.
intersect in the following four
distinct points:
(2,
V3),
(2,
-
x/3),
(
-
2,
V3),
(
-
2,
-
>/3).
These values of x and y already appear in the tables used for constructing the
and need not be checked by substitution in the original equations. However,
such checking is usually desirable.
graphs,
EXERCISE 12-9
Solve each of the following systems of equations graphically.
1.
4.
7.
10.
13.
12-12.
As
may
ALGEBRAIC SOLUTIONS OF SYSTEMS INVOLVING QUADRATICS
in the case with linear equations discussed in Section 9-1, it
happen that in a system of equations involving quadratics
228
in
Equations
Quadratic Form
Sec.
12-12
part of the graph of one equation coincides with part of the graph
of the other. Such a condition gives rise to infinitely many soluUsually, however, there are only a finite number of points
of intersection of the graphs corresponding to the given equations,
tions.
and the algebraic problem consists of finding the pairs of numbers
(x, y) which satisfy both equations. We can say in this case that
two simultaneous equations in x and ?/, of degrees m and n, respectively, can have at most mn solutions. Thus, a system of one linear
and one quadratic equation can have at most two solutions, and a
system of two quadratics can have at most four solutions.
When a system consists of two quadratic equations, the algebraic
solution usually leads to a fourth-degree equation in one of the
unknowns. Since we have not presented a general method of solving
a fourth-degree equation, we shall consider here only systems whose
solutions can be effected by the theory of quadratic equations. The
methods of procedure in some of the more important types are
shown
:
One Linear and One Quadratic Equation. A system of
type can always be solved by the method of elimination by
Case
this
in the following three cases
1.
substitution.
Example 12-30. Solve the system
x
x2
- 3?/ + 6 = 0,
- Sx + Zy = 0.
Solution: Solve the linear equation for y in terms of x, obtaining
x
+
6
Substitution for y in the quadratic equation yields
,,_
8a;+
3(?L6)=0.
Collecting terms gives
x2
=
-
Ix
+ 6 = 0.
and x = 6. Substituting these values in the
linear equation, we obtain y = 7/3 and y = 4. Hence, the solutions are
x = 1, y = 7/3;
and x = 6, y = 4.
These values can readily be verified as solutions of the given system. Note that
The
roots of this equation are x
1
they correspond to the coordinates of the points of intersection in Fig. 12-7.
Example 12-31. Solve the system
+ 2y + 4 = 0,
x* + 4y* - 2x x
3
= 0.
Sec.
12-12
229
Quadratic Form
In
Equations
Solution: Solve the linear equation for 2y, to obtain
= -
+ 4)2
+ 4).
We then have
- 2z - 3 = 0,
+
+
2y
(x
Substitute in the quadratic and collect terms.
*
+
z2
(x
or
2x 2
One
solution
*=
The other
6x
= 0.
13
is
solution
-
3
+ i V17
y=
'
2
5
+ t \/17
5
-
4
is
*
=
-
3
- tVl7
y=
'
2
i
yT7
'
Since these values arc imaginary, the graphs of the two given equations do not
intersect.
by = c. When the
consists
of
two
system
equations containing only squared terms in
each unknown, it can be solved for x 2 and y 2 by the methods used
Case
Two Equations
2.
Form ax 2 +
of the
2
for linear systems in Section 9-2.
Example 12-32. Solve the system
+
x*
2x 2
Solution:
To
2y*
-
2
2/
=
=
17,
14.
eliminate y 2 multiply the second equation
,
by 2 and add the two
equations, as follows:
+ 2y* = 17
- 2y = 28
x2
2
4z 2
5.r 2
Solving for
x,
i
Now
=
45.
we have
substitute 9 for x 2 in the
first
2
2?/
=
=
17
y
=
or
3.
of the original equations.
-
9
=
Then
8,
2.
Hence, we have the following four solutions:
(3,2),
These
may
be written
(3,
-2),
(3, =b 2),
(
-
(-3,2),
(-3, -2).
3, d= 2).
2
2
of the Form ax + bxy + cy = d. If the
system is of this type, the solution is effected by elimination of the
constant term. The resulting equation is then solved for one
Case
3.
Two Equations
terms of the other. This procedure gives us two linear
and y which may be combined with either of the
given quadratic equations to form two systems of the type con-
unknown
in
equations in x
sidered in case
1.
230
Equations
Quadratic Form
in
Sec.
12-12
Example 12-33. Solve the system
_
x*
2x 2
xy + 2y = 1,
2
2xy + Sy = 3.
2
Solution: Multiply the first equation
from the new equation, as follows:
3z 2
2x 2
-
by 3 and subtract the second given equation
xy
+ 6?/ = 3
+ Sy = 3
- 2y = 0.
(x
-
2
3xy
2xy
x2
2
2
Factor, to obtain
+ y)
(x
2y)
=
0.
Hence,
x - 2y = 0.
x + y =
or
combine each of these two equations with the
form the following two systems:
f
x 2 - xy + 2y 2 = 1,
x + y = 0;
and
f
x 2 - xy + 2y 2 = 1,
- 2y = 0.
We may
We
first
given equation to
then proceed by the method for case 1.
solutions of the given system consist of the solutions of these two systems.
The
Hence, we have
s = - 1/2, y = 1/2;
x = 1/2, y = - 1/2;
* = l,y = l/2;
-1,0 = -1/2.
in connection with systems described
another
method
is
effective
Occasionally,
under case 3. The following example illustrates this method.
*=
Example 12-34. Solve the system
f
+
x2
= 37,
2
9?/
xy=2.
I
Multiply the second equation by
equation, to obtain
x 2 + 6xy + 9y 2
Solution:
the
first
Also subtract Qxy
=
12 from the
6,
to obtain 6xy
=
12.
Add
this to*
= 49.
equation to obtain
x 2 - Qxy + 9?/ 2 = 25.
The left side of each of these new equations is a perfect square. We chose the
multiplier of xy, which is 6 in this case, so as to obtain perfect squares. We now
first
have the system
/
I
(x
(x
+ 3y) 2 = 49,
- 3y) 2 = 25.
i
Hence,
x
+ 3y =
7
or
x
+ 3y = -
7.
Also,
= 5.
x
x
or
3y = 5
3?y
We now solve the following four systems of linear equations:
3 +
+ 3y = 7,
+ 3y = 7,
+ 3y = ~ 7,
-3y = 5; \a? - 3y = - 5;
\a?-3y=5;
\x
fa;
fa?
f
Sec.
12-13
Equations
in
Quadratic Form
231
An equation in x and y is said to be symmetric in x and y if the
equation is unchanged when x and y are interchanged. When a
system consists of two quadratic equations both of which are symmetric in x and y, the substitutions x = u + v,
v will give
an equivalent system which may in some cases be solved by previous methods.
y-u
EXERCISE 12-10
In each of the problems from
1
to 21, solve the given system of equations
algebraically.
22.
Complete the solution of the system in Example 12-34.
In each of the problems from 23 to 26, solve the given system by the method of
Example 12-34.
23.
x2
xy
25.
X2
xy
+ =
= 25.
+ V2 = 12.
7/2
24.
50,
xy
26.
25,
+ y2 = 4;
xy + x + y = 10.
z /?/ +
A = 56,
X2
29.
2
2
28.
30.
3+0=2.
12-13.
An
EXPONENTIAL
*2
a?y
Solve each of the following symmetric systems.
27.
z2
+ 42/2 = 13,
= 3.
+ = 144,
= 56.
7/2
+ 02 - x - = 2,
xy + 3x + 30 = 2.
x 2 + 02 = i 3>
3x 2 + 2x0 + 30 2 = 42.
x2
AND LOGARITHMIC EQUATIONS
equation in which the unknown occurs in an exponent is
an exponential equation. Such an equation is usually solved
by taking the logarithm of each side and solving the resulting
equation. When this latter equation is in linear or quadratic form,
it may be solved by preceding methods.
called
232
Example 12-35. Solve
Quadratic Form
Equations
in
for x: 5* +3
= 625.
Sec.
12-13
Solution: Write the equation in the form
5*+ 3
5*.
This equation
and only
is satisfied if
x
if
=
+ 3 = 4,
=
3 ** 1
Example 12-36. Solve the equation 2
Solution:
that
is,
=
x
3 4 *.
of each side to the base 10,
Taking the logarithm
log (2**+*)
=
1.
we get
4
log (3
*),
or
(3z
+
1) log
_ _
Therefore,
-
4 log 3
From Table
III, log
= 0.3010
2
= 4z log 3.
2
3 log 2
=
and log 3
0.4771. Substituting these values,
we have
_
~
X
__
0.3010
4
0.4771
-
+
1)
0.3010
3
'
or
Example 12-37. Solve
for x: log (x
=
1
-
log
(3a?
+
Collecting terms containing logarithms on one side
as a single logarithm, we have
Solution:
member
2).
log (x
+
+ log
1)
(3x
+ 2) =
and writing that
1,
or
+
log (x
+ 2) =
1) (3x
1.
Writing the members in exponential form, we have
(x
+
1) (3*
+
2)
+
5z
-
This equation reduces to
3z 2
Solving for
x,
we
find that
a;
=
1
or
a;
=
10'
=
8
= 0.
10.
g
=
'
o
Checking, we find that x
=
1 satisfies
the original equation, whereas x
gives rise to logarithms of negative numbers. Since negative
real logarithms, this latter value of x is not to be used.
Example 12-38. Solve y
= log,
(x
+
\/l
+#
2
)
for
y.
Solution: Write the given equation in exponential form, obtaining
e*v
Solving this equation for
x,
-
we have
2xe
-1=0.
is
o
o
numbers do not have
x in terms of
Transpose and square to remove the radical. The result
=
Sec.
12-14
233
Quadratic Form
in
Equations
EXERCISE 12-11
Solve each of the following equations for the
1. 2*
= 64.
7. 5*
=
15.
10. (3*) (2*)
= 36.
5. 4*
= 24.
8. 3*
=
11.
5"*
17.
21.
23. e 2 *
3)
=
1000.
= 2.403.
22. Iog 2
=
1) t= 1
-
x
25.
log (to
+ 2).
+ 3) = 3.
4e*+* = 7.
(-!)+ Iog
2
(x
16.
29.,=^.
12-14.
= 32.
= 3.
15. (1.5)*
18. Iog 2
6*.
= 0.4136.
8
12. log*
= 4.83.
2
26. e* + 2 *- 2
=
9. 3(2*)
3
z 10***
= 243.
3 3 ** 1
3.
= 0.6932.
14. s l4 =0.04681.
17. 3*+ 2 =
5.03 = (3.17)i'<*-i>.
20. log (4* 19. log (3x - 5) = 3 - log 7.
13. log* 2
16.
x.
= 256.
2. 4** 1
= 0.0001.
4. 10*
unknown
31. e 4 *
-
-
e 2*
10
= 0.
GRAPHS OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
The graph ofy =
log x is
shown
in Fig. 12-10.
By
assigning
values to x, one finds corresponding values of y from Table III.
few pairs of values are shown
in the
A
accompanying tabulation.
FIG. 12-10.
The graph
=
be obtained from a table of exponential
Here, however, we shall proceed as follows: Take the
logarithm of each side to the base 10, to obtain
of #
e*
may
functions.
log
y
=
x log
e.
234
Equations
Prepare the accompanying
Quadratic Form
in
table,
Sec.
12-14
and construct the graph in Fig.
12-11.
EXERCISE 12-12
Graph each
1.
y
= Iog
7
of the following.
2.x
x.
= Iog
7
3.
y.
T/
= 3-.
*-"*-?
x
4.y =
.
7.
y
=3+4(2-).
10.
y
=
100e- 06 *.
2
y
=
y
=
e"'.
11. v
=
7 3 *- 1
2
10-*
5.
8.
e2 .
6.
9.
.
12.
y
=
16(5-).
y
=
log c x.
=
3.9 2 *- 3 .
?/
13
13-1.
Theory of Equations
INTRODUCTORY REMARKS
With the ever-increasing importance of mathematics
neering and the physical
in engi-
problems are constantly occurring that involve the solution of equations. Often these equations
are of the simple algebraic or trigonometric types which we have
already learned to solve. There are many other problems, however,
which require the solution of equations of higher degree than the
second and of some types of somewhat more complicated transcendental equations. Equations of the third and fourth degree can
be solved by methods analogous to those which we used for quadratic equations.
Because of their complexity, however, these
methods are seldom used. It has been proved that no such procedures exist for equations of degree higher than the fourth.
sciences,
In this chapter we shall consider various properties of polynomial equations in general. Some of these properties will be of
considerable use in later studies of mathematics, while others are
considered here merely for the aid they give us in determining roots
of equations.
13-2.
A
SYNTHETIC DIVISION
simplification of the ordinary
method of long
division, called
synthetic division, will be presented here. This abbreviated method
not only enables us to quickly find the quotient and remainder when
a polynomial f(x) = a Q x n + a^ x n 1 H
h a n is divided by a binomial
of the form x
r, but also affords a simple process for substituting
values of the variable into a polynomial.
We shall divide 2x 3 - 9x- + I3x + 5 by x - 3 to illustrate the pro'
cedure in synthetic division as compared with that of long division
considered in Section 1-19.
235
236
Sec.
Theory of Equations
By
long division
2z 3
we have
-
Qx 2
+
13-2
:
a-
13x
2x
2
3
-
3x
+4
I3x
Qx
4z
4a
+
5
-
12
+
17
3x + 4, and the remainder is 17.
Thus, the quotient is 2x
Since like powers of x are written in the same vertical column,
2
the
work may be shortened by writing only the
coefficients, as in
the following schematic arrangement:
1-3
2-3+4
+
Next,
we
note that the
first
17
term
in the divisor x
written, since the divisor is always linear,
in it is
term
and the
r need not be
coefficient of
x
always unity. Moreover, it is not necessary to write the first
row that is to be subtracted, since its coefficient is
in each
always the same as that of the term directly above.
Also, only
the first term of each partial remainder needs to be written down,
for the second term
first
row.
is
the
same as the term
directly above
it
in the
Finally, the coefficients in the quotient need not be
written, since these are precisely the leading coefficient in the
dividend and the remaining partial remainders, excepting the
Hence, we may indicate the process in the following way
:
2-9+13+5
-6
+
9
-
12
+
17
|
-3
last.
Sec.
13-2
237
Theory of Equations
This scheme can be written compactly as follows
-
2
9
13
-
5
6
9
12
2-3
4
17
|
:
3
If we replace -3 by +3 in the divisor and add the partial products in the second row instead of subtracting them, we obtain the
same
The synthetic
result.
division then takes the following
2-9
13
5
6
-9
12
2-3
4
17
form
:
3
|
Here the numbers 2, -3, and 4 in the third row are the coefficients
of the quotient, and the last number 17 is the remainder.
We can now outline the procedure for synthetic division. Note
that in every step of the procedure, immediate reference is made
to the illustrative example.
To divide f(x) = a x n + a^ x n ~* H
h a n by x
r, first arrange
in
of
for
zero
the coefficient of
f(x)
descending powers
x, writing
any missing power of
x.
process in three rows, as
( >,
its
in the
:
row, write the coefficients in f(x) in order, as
!,
right, put the constant term of the divisor with
sign changed. We have then
Step
a
Then arrange the numbers involved
shown in the following steps
In the
1.
an
.
first
At the
Example
a
Step
2.
ai
a2
an
Bring down the
2
r
-
|
first coefficient
place of the third row. Thus,
9
13
5
3
|
a of f(x) into the
first
we have
Example
i
an
a2
\_r_
/
2
9
13
5
|
3
i;
r, and write the product a r in the second,
Bring down the sum of ai and a r into the third row.
Thus, we now have
Step
3.
row under
Multiply a by
ai.
Example
a
an
ai a 2
|_r_
2
-
9
6
apr
+
i)
23
13
5
.
238
Sec.
Theory of Equations
13-2
In the example, multiply 2 by 3 and write the product 6 in the
second row below 9. Then add 9 and 6, writing the sum 3 in
the third row.
Step
4.
Multiply o r
row under a2 and
,
4- di by r, place the product in the second
add. Continue this process until finally a product
has been added to a n
The complete
.
example follows
solution of the illustrative
2-9 13
6-9
2-3 4
5
|
:
3
12
17
9
In the last operations,
3 is multiplied by 3, and the product
9 equals 4. Finally the
is written below 13. The sum of 13 and
5 to
product of 4 and 3, or 12, is written below 5 and added to a n
give 17.
Step
5.
When
the process
is
completed, the last
number
in the
directly below a n is the remainder. The other numbers
in this row, read from left to right, are the coefficients of powers of
third
row
x in the quotient arranged in descending order.
The entire synthetic process of dividing a polynomial f(x) by
x
r, although it is somewhat complex notationally, can be conveniently exhibited as follows
:
an
a n _i
a2
ao ai
r
[
b n ~2r
bpr b\r
bo
b n -\r
R
b n -i
62
hi
bn
of the
Here the expressions for the coefficients 6 &i> b 2
=
=
a
r
6
+
6
a
of
are
x in the quotient
Oo r2
&i
2
Oi,
powers
-1
2
=
b n -i
+ a! r- +
+ a n -i. Hence, the quotient
+ ai r + 02
do r
,
,
,
_i
()
,
11
,
may
be written
q (x)
(13-1)
=
b x*- 1
+
Also, the remainder assumes the
R =
(13-2)
a
r
n
The expression for R is
inf(x). In other words
+ air"-
1
Bi
xn
~2
+
+
& n -i-
h a n ^r
+ an
form
H
.
precisely the result of substituting r for x
B=/(r).
(13-3)
we
r and q(x) from f(x),
subtract the product of x
Finally,
we obtain the remainder R = f(r). Or, if we transpose the product,
we have the usual statement found in discussions of division. That is,
if
(13-4)
/(*)
= (*-r).j(*
Sec.
13-2
239
Theory of Equations
The following examples will further illustrate the process of
synthetic division.
Example 13-1. Divide 3x*
Solution: Since x
of the missing
4z 2
+x -
3
,
we have the
+ 2.
2 by x
= x + 2, we have r =
r
power x
-
Writing zero for the coefficient
2.
following result:
30-4
-
1-2
6
12-16
30
3-6
8-15
28
|-2
Note that the first -coefficient 3 is brought down into the first place of the third
Next 3 is multiplied by
6, is written in
2, and the product, which is
the second row under 0. The sum of and - 6, or - 6, is written in the third row
row.
directly below 0. Proceeding in this way,
is 28.
+ 8x
fix 2
we find that the quotient is 3x 3
15
and the remainder
Example 13-2. Given /(x)
=z 3
2x 2
+ 5x -
4, find
By (13-3), /(r) equals the remainder obtained in the division of f(x)
Hence, we have the following results
Solution:
by x
r.
:
a)
Therefore,
/(-
Therefore, /(I)
= -
1)
12.
= 0.
c)
Therefore, /(3)
= 20.
EXERCISE 13-1
In each of the problems from
find the quotient
1.
3.
5.
7.
9.
11.
1
to 15, divide the
first
and the remainder, by using synthetic
- Sx + 7, x - 1.
- 5x + 6, x - 4.
z 4 - 3x 2 - 6x + 6, i + 3.
2x 4 - x 3 - 6z 2 + 4z - 8, x - 2.
3
+ 3x 2 - 2,r - 5, x - 2.
4
z 4- z 3 - 59s 2 - 69z + 030, z 2 x2
2. x*
x2
3
,
.c
divide successively
by each
factor.)
a?
-
42.
4.
z
6.
x3
function
by the second, to
division.
-
-
x2
Sx
2
Sx
+
+
fix
- 3.
- 3.
6, x
- 4.
x
24,
6,
-
8.
2x 3
3x 2 + Or
+ Sx 2 + 4, x
10.
2x 5
-
x
+
2.
z
+ 2.
(Hint: Factor, the divisor
and
3z 3
+
2z
4- 1,
240
Theory of Equations
12. x*
13.
14.
z5
aw
-
+
15. x*
3x 3 + So; 2
2x 4 - 21x
1.
1, o
x
y*
y.
-
Sec.
13-2
+ 2, x - 3x + 2.
+ 18, 2 + 2x - 3.
2
3x
a;
y
For each of the following polynomial
method of synthetic division
functions, find the indicated values
by the
:
= 3x - 7x2 _ 5x + 6 Find /(I) and/(- 4).
= - 2x + 2x - 5x + 2. Find/( - 2) and/(0).
= x - 4x 3 - 4x + 24x - 9. Find/(3) and/(9).
= x* - 3x3 _ 13^2 + 21 X + 18. Find /(I) and/(3).
= 7x 4 + 37x3 _ X _ 14a + 4. Find /(I) and/(- 5).
3
16. f(x)
.
17. /(x)
2
3
a;*
2
4
18. /(a?)
19. /(x)
2
20. /(x)
.
THE REMAINDER THEOREM
13-3.
In Section 13-2, it was shown that the remainder in the division
of a polynomial by a binomial x
r can be found without actually
performing the division. Thus, in establishing (13-3), we have
proved the following theorem.
Remainder Theorem. If a polynomial f(x) is divided by x r,
the remainder is the value of f(x) for x = r; that is, the remainder
from this theorem that f(x) is exactly
and only if /(r) = 0. Hence, we have proved
It follows
x
r
if
divisible
by
also the fol-
lowing theorem.
Factor Theorem. If f(r) =
nomial f(x), and conversely.
Example 13-3.
3
157. Therefore, since /(
therefore, that
x
is
a factor of the poly-
2
2
According to the factor theorem, /(
of /(x). But /(-3) = (-3) 4 - 2
Example 13-4.
r
+ 3 a factor of x 4 - 2x + 3x - 5?
= x 4 - 2x 3 + 3x - 5. Also, x - r = x + 3, and r = -
Is x
Solution: Here/(x)
=
then x
0,
-
3) 7* 0,
x
3)
must equal zero
if
x
-f
3
is
3.
to be a factor
+3 (-3) -5 =81 +54 +27 -5
+ 3 cannot be a factor of x 4 - 2x 3 + 3x 2 - 5.
2
(-3) 3
Given /(x) = x 3 - 3x 2
2 is a factor of /(x).
+ 5x -
6.
Show
that /(2)
=
and,
= x 3 - 3x 2 + 5x - 6, we have by substitution
3 3 (2) 2 + 5 (2) - 6 = 0.
/(2) = (2)
From the fact that/(x) equals zero when x = 2, it follows from the factor theorem
Solution: Since /(x)
that x
-
division
2
is
and
a factor of /(x).
The student should check
this result
find that
/(x)
= x3 -
3x 2
+
5x
-
6
=
(x
-
2) (x
2
-
x
+ 3).
by synthetic
Sec.
n
13-4
241
Theory of Equations
Example 13-5. Find under what condition
an integer and a ^ 0.
(x -f a) is
a factor of x*
+ an
,
where
is
=r +
Solution: In this case, f(x)
equal zero only
if (
a)
=
n
-
an and/(
a)
,
an that
,
= (
only when n
is,
a)
+ an
w
This
.
sum can
odd.
is
EXERCISE 13-2
In each of the problems from 1 to IS, determine
of the first. If it is a factor, find another factor.
1.
2z 3
3.
z3
5.
x3
7.
xs
9.
x4
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
+ x + 6, x - 2.
- 3z - 1, x - 1.
- z 2 - llx + 15, x - 3.
- 256?/ 16 z - 2?y 2
- 4z 3 - x 2 + 16.c - 12, z +
-
+
.
,
the second function
2.
z3
+ 4z 2
4.
z7
-
6.
z5
+ 243,
8.
2z 3
Gx 2
2z2
if
1,
-f
z
3z 2
-f
.r
-
a factor
+ 6. x + x + 2.
2
bx
-
is
1.
+
3.
-
9z
111,
z
-
1.
1.
+ 5s 4 + 20z + GO* 2 + 120z + 120, x + 1.
+ aft (5 - 6r?& )z - 2a 6 2 (5 + 3a6 2 + 12a 5 x - ab.
2x* - 3x* - 3z - 2, x + 2.
i - 10.r + IS* - 24.1* + 75, x - 2.
2x 4 - 31x + 21x - I7x + 10, x + 1.
- 4(Xr - x 2 + lllj - 90, 2x - 3.
12.x
12z - 22.r 2 - 34z + GO, 3x + 5.
24* - 122x + 159x - Ix - GO, 3z - 4.
24x^ _ 74^.4 __ 85.r 3 + 311x 2 - 74x - 120, 2x + 1.
Show that the equation x + 2.r 3 - 9r - 2x + 8 = has the roots 1,2, - 1, - 4.
Find all roots of the equation 12.r - 4Qx* - x + lllx - 90 = 0, given that
.r
5
3
5z 3
2
2
2
4
)a:
4
,
8
3
4
fc
2
3
3
4
3
2
4
2
2
4
a double
that
is, that (x
a divisor of a n
3/2
root,
b is
Prove that a
22. Prove that a + b is a divisor of a n
23. Prove that a + 6 is a divisor of an
24. Prove that neither a + b nor a even integer.
is
21.
13-4.
We
3/2)
2
is
a factor of the
left side.
b n for
-
bn
-f-
bn
b is
if
every positive integral value of n.
n irf a positive even integer.
if
n
is
a positive odd integer.
a divisor of a n
+ 6n
if
n
is
a positive
THE FUNDAMENTAL THEOREM OF ALGEBRA
usually assume that every algebraic equation with real or
coefficients has at least one real or complex root. Although
complex
the existence of such a root is not to be taken for granted, a proof
is beyond the scope of this book. Accordingly, we shall accept the
following theorem as true.
Fundamental Theorem of Algebra. Let f(x) be a polynomial of
degree n with complex
f(x) =0 has at least one
The first complete and
was given by Gauss in
Since that time, many
coefficients.
complex
Then the algebraic equation
root.
rigorous proof of the fundamental theorem
the beginning of the nineteenth century.
proofs have appeared, but most require
knowledge of the theory of complex functions.
242
Sec.
Theory of Equations
13-4
repeated application of the fundamental theorem, it can be
roots of any polynomial equation with
real or complex coefficients is equal to the degree of the polynomial. We shall state the following theorem and give its proof.
By a
shown that the number of
Theorem. If f(x) is a polynomial of degree n, the equation
f(x) =0 has exactly n roots.
Proof. Let f(x) be a polynomial of degree n with real or complex coefficients. By the fundamental theorem, there is a number TI
such that f(ri) =0. Then, by the factor theorem,
=
f(x)
where q\(x)
-
(x
n)
qi(x),
a polynomial of degree n
is
1
whose leading
coeffi-
cient is OQ.
has a root by the fundamental theorem.
Likewise, q\(x) =
root
denote this
by r2 then QI (r 2 ) = 0. Also,
we
If
,
qi(x)
=
(x
-
r2 )
92(3),
2 with leading coefficient a
is of degree n
tf 2 (#)
can continue in this way
also has a root.
Similarly, # 2 (#) =
until we come to a polynomial of the first degree with root rn Then
where
.
We
.
(13-5)
f(x)
=
a (z
-
n)
(a
-
r2 )
(a
-
r n ),
where a
is the coefficient of x in our final first-degree polynomial.
Since f(x) equals zero when we substitute for x any one of the n
has at
numbers r lf r2
it follows that the equation f(x) =
, r,
rn
least the roots r a r 2
Moreover, there are no other roots. For, suppose that r is some
rn Substitution of r for x in (13-5) yields
root other than n,
,
,
.
,
/(r)
.
,
,
=
a
(r
-
ri) (r
-
r2)
(r
-
r n ).
The right
side of this equation cannot equal zero, since no one of the
has no
factors can equal zero. Therefore, /(r) ^0, and f(x) =
more than the n roots found before.
It may happen that a certain root appears more than once among
the numbers r t r2
r. In that case it will be counted as many
,
,
,
times as the corresponding equal factors appear in (13-5). If a
certain factor (x - r) appears
times in (13-5), then r is said to
be a root of multiplicity m. A root is called a simple root, a double
root, a triple root, and so on, the proper name depending on how
many times the same factor appears. Hence, combining the statements that "f(x) = has at least n roots' 1 and "f(x) = has no
more than n roots," we can conclude that a polynomial equation of
the nth degree has exactly n roots, a root of multiplicity
being
m
m
counted as
m roots.
Sec.
13-5
243
Theory of Equations
It is to be noted that a rigorous proof of this theorem requires the
use of induction. (See Chapter 16.)
13-5.
OF COMPLEX ROOTS OF AN EQUATION
PAIRS
We wish to remind the student that while an equation of the nth
degree has n complex roots, the number of real roots may be less
than the degree of the equation. For example, x 2 + 1 =
has no
real roots. Determination of the number of real roots may be simplified by use of the following theorem.
Theorem.
if
If all the coefficients of f(x) =
are real numbers, and
number a + bi is a root of f(x) = 0, then the conju-
the complex
gate a
bi is also
a root. It
understood that a and b are real and
is
6^0.
Let x in f(x)
Then we have
Proof.
a
+
bi.
f(a
+
bi}
=
a (a
=
a xn
+
bi)
n
+
+
a x xn~ l
ai(a
+
H
h
bi)*-
1
+
a n be replaced by
+
.
If we expand the powers of a + bi by the binomial theorem and
simplify the resulting expression, then all terms which contain even
powers of i will be real, while all terms which contain odd powers
of i will be pure imaginary. Denote by P the aggregate real part,
and by Q the aggregate imaginary part. Then we have
= P + Qi = 0.
with (11-4) P = Q = 0.
/(a
Hence, in accordance
Now, replace x by a
+
bi)
,
0. In those terms of f(a
bi)
f(x)
raised to an even power, the result will remain the
bi in
in which
bi is
same as in f(a + bi) However, all terms in f(a bi) in which
is raised to an odd power will have their signs changed. Hence,
.
/(a
But, since
we have shown
-
bi)
that
= P-
P-Qi
In other words, a
bi is also a root of
Example 13-6. Solve
x*
-
x3
-
Qi.
P=Q=
=
2x 2
+
6s
bi
0,
we
conclude that
0.
f(x)
-
4
=
=
0.
0,
one root being
1
+
i.
1 + i and 1
i are roots of the
sum
for
the
and
and
(12-4)
product of two
given equation. Using Eqs. (12-3)
2
are
roots
a:
2x +2 = 0.
these
numbers
of
we
find
that
roots,
conjugate complex
Dividing the original polynomial by this quadratic function, we get the quotient
Solution: According to the last theorem, both
x2
+
x
roots, x
-
2.
= -
Solution of the equation x 2
I.
2 and x
=
+
x
-
2
=
yields the remaining desired
;
244
Theory of Equations
Sec.
13-5
-
vX
EXERCISE 13-3
1.
Solve X s
3
2.
Solve x
3.
Solve z 4
4.
Solve x*
5.
Solve
a;
5
+ x 2 .- 2x -f 12 = 0, one root being 1 + V3f.
+ 3z + I2x - 16, one root being -2 - 2 V~ 3.
- 2x - 7x + ISx - 18 = 0, one root
being 1
-f 3z -f 7z -f 6z + 4 = 0, one root being - 1 - Vi.
- 8z -f 27z - 46z + 38x - 12 = 0, one root
being
2
3
2
3
2
4
3
and one root being
6.
Solve x*
7.
Find a
Find a
8.
13-6.
f.
2
+x + 1 =
2
by considering the equation to be a quadratic equation in x 2
two
whose roots are 2 and 1 -f 2i.
equation of lowest degree having the roots i and 1 -
real cubic equation,
real
2
1.
THE GRAPH OF
.
of
A POLYNOMIAL FOR LARGE VALUES OF
i.
x.
In graphing a polynomial function, it is helpful to know the
location of points on the curve for numerically large values of x. It
can be shown that, when x is numerically sufficiently large, the term
a x n of highest degree is numerically larger than the sum of all the
other terms combined. Therefore, the sign of this term determines
the sign of the entire polynomial.
Let us consider the values of the function f(x) = a; 3 + 5x 2 - Ix
13 as x assumes various values from left to right along the #-axis,
that is, for increasing values of x. The results may be tabulated
conveniently as follows
:
When x is negative, but numerically sufficiently large, it is seen
that f(x) is negative. Thus, when x = - 10, x* = - 1000, while the
sum of other terms, or the value of 5x 2 -Ix - 13, is only 557 and
;
=
-443. For points far enough to the right of the origin,
=
for
x
say
10, f(x) is positive. For example, /(10) = 1000 + 417 =
1417. Hence, the graph of y = f(x) would be located below the
#-axis on the left but would rise above the #-axis as x gets
larger
/(-10)
and
larger.
As a second
x4
+
7x 2
-8x +
=
illustration, let us consider the function f(x)
10. Here f(x) is positive for large numerical values
of x, regardless of the sign of
x.
Hence, in this case the graph
Sec.
1
3-8
245
Theory of Equations
would be above the
right and left of the
numerical values of x to both
re-axis for large
origin.
The following helpful conclusion can be drawn from the preceding discussion involving large values of x.
When x is sufficiently large and positive, f(x) has the same sign
as the leading coefficient a. When x is negative and sufficiently
large numerically, f(x) has the same sign as a when n is even, and
has the opposite sign when n is odd.
The following symbols are sometimes found in discussions of the
values of functions for numerically large values of x. When the
symbolic statement /(-foo)>0is used, what is meant is that, for
all sufficiently
the statement
tive values of
large positive values of x, f(x) ig positive. Similarly,
means that for all sufficiently large posioo) <
means
oo) >
x, f(x) is negative; the statement /(
/(+
which are sufficiently large numerco
and
means that, for all negative
ically, / (x)
positive
/(
) <
of
x
values
which are sufficiently large numerically, f(x) is negative.
2
Thus, if f(x) =x* + 5x -7x- 13,/(- oo) <0 while /(+ oo) > 0.
=
x* + 7x 2 - Sx + 10, /(- oo)>Oand/(+ co)>0.
However, if f(x)
that, for all negative values of x
is
13-7.
;
ROOTS BETWEEN a AND
b
IF f(o)
AND
Another helpful theorem relating
equation
is
f(J>)
HAVE OPPOSITE SIGNS
to the roots of a polynomial
the following.
Theorem. If the coefficients of a polynomial f(x) are real, and if
a and b are real numbers such that f(a) and f(b) have opposite
has at least one real root between
signs, then the equation f(x) =
a and b.
We shall not give a proof of this statement here, but shall merely
mention the following geometric considerations. The graph of a
polynomial is a continuous curve; that is, it has no "breaks." Therefore, if the points (a, /(a)) and (b,f(b)) lie on opposite sides of
the #-axis, the graph apparently has to cross the #-axis at least
once between these points.
13-8.
RATIONAL ROOTS
The following theorem
is
fundamental for the solution of equa-
tions having integral coefficients.
Theorem.
If the equation
f(x)
with integral
-
a xn
coefficients
+
aix
n~ l
H
h an
=
has the rational root
-,
a
>
where
c
and d
246
Theory of Equations
Sec.
13-8
are integers having no common factor > 1, then c is a divisor of
the constant term an , and d is a divisor of the leading coefficient a
.
We shall make
use of a principle from the theory of numProof.
bers: If an integer c divides the product of two integers a and 6
and if 6 and c have no common divisor other than
1, then c is
a divisor of a.
XI
Let
-3
a
common
a
Multiplication
w
Since c divides
now
^
c
"+
'
a "~ l
l
+ a" = a
d
gives
+ aicn~
all
and d are integers with no
c
Then
1.
cn
^ + ai d^r +
n
by d
aoc
where
0,
divisor other than
cn
term. If
=
be a root of f(x)
~l
+ a ndn =
n
h a n -icd
d H
terms before the
0.
final one, c also divides that
c is factored into primes,
none of these primes
is
a
divisor of d, and therefore of d n . Thus each prime divides a n , and so
c itself divides an
In a similar fashion, it may be shown that d
.
divides a
.
-
rational roots of 3z 3
Example 13-7. Find the
I7x 2
+
I5x
+
7
= 0.
st
Solution:
The
possible rational roots are of the
form
-r
>
where
c is
a divisor of
ct
the constant 7 and d
is
a divisor of the coefficient
1/3,
7,
1,
Using synthetic division to check
-
3
3
By
the remainder
17
15
7
3
20
-20
35
-35
-28
theorem, /( -
=
We
see, however, that /(O)
there must be a root between x
1)
= -
=
1
hope that a rational root
sibility is
1/3.
The check by
3
3
-
28. Hence,
by
following result:
1
-
1 is
not a root.
the theorem in Section 13-7,
and x = 0. Examining our list of possible
1 and 0, we see that a posbetween
lies
synthetic division follows:
17
15
18
21
7
-16-7
-
|-
4-7. Therefore,
roots, in the
possible roots are
7/3.
we obtain the
1,
The only
3.
|- 1/3
=
1/3 is a root. After dividing the given polynomial by x + 1/3, and
2
18z 4- 21 to 0, we obtain the quadratic x 2
6x + 7 = 0.
equating the quotient 3x
Hence, x
This equation has the roots 3 =h \/2, which are not rational numbers. Therefore,
= 1/3.
the only rational root is
In this example, as is sometimes the case, it has been possible to find all the roots
of the given equation.
Sec.
13-8
247
Theory of Equations
EXERCISE 13-4
1.
Show
that l&r 3
-
33z 2
+
5
=
=
0.
+
-
1 and 0,
between and 1, and between and 2. Find these three roots.
Find the rational roots of 2x 3 - 9z 2 + 3x + 4 = 0.
3x - 5 = has at least one positive root.
Prove that x 3 + 2x 2
x 3 4- x 2 + x
Prove that x*
3 =
has at least one positive root and at
least one negative root.
Prove the following corollary of the theorem in Section 13-8. If f(x) =
x n + aix*- 1 +
+ a n = has integral coefficients and has an integral
2x
has real roots between
1
2.
3.
4.
5.
root
6.
r,
then
r is
a divisor of a n
Find the integral roots of x*
.
-
1
+ x + 1 = has no rational roots.
= 0.
Find all the integral roots of 8 + z a + x + I = 0.
5
3
Show that 3 is a root of x 3 - ^ x - 2x + - = 0. Why does this
z
&
Show
2
that the equation x
1
8. Solve the equation x 3
7.
9.
10.
a:
2
diet the
not contra-
theorem in Section 13-8 or that in Problem 5?
In each of the problems from 11 to 20, find all roots of the given equation.
12. 4x 3 - 16z 2 - 9z + 36 = 0.
11. x 4 - Sx 2 + 16 = 0.
13. 3x*
3
15.
2z
17.
5z 3
19. 2x*
+ x 2 + x - 2 = 0.
- x 2 + 2x - 1 = 0.
- 13z + 16s - 6 = 0.
- x 2 - 4z + 2 = 0.
2
+ 3z 2 - 6x - 9 = 0.
x
llz 2 + 37x - 35 = 0.
x* - Sx 3 + 37z 2 - 50z = 0.
3z 3 - 13z 2 + 13z - 3 = 0.
14. 2x*
16.
18.
20.
s
14
Inequalities
INTRODUCTION
14-1.
In previous chapters we have explained some methods of determining the roots of an equation. By applying these methods, one
can find the values of an unknown for which a certain function of
the unknown equals zero. Often, however, it is necessary to solve an
inequality, that is, to discover for what values of the unknown a
certain function is less than or greater than another function.
The present chapter is concerned primarily with the solution of
inequalities, and the following discussion is essentially an extension
of the study of the order relation undertaken in Section 1-8. We,
therefore, recommend that the student thoroughly review Section
1-8 before starting the study of the present chapter. Since the solution of inequalities often involves the use of absolute values, a thorough mastery of Sections 1-9 and 1-10 is also a requirement.
The classification of inequalities corresponds to that of equalities
or equations. As in the study of equations, there are two kinds of
inequalities involving unknowns, namely, absolute inequalities and
conditional inequalities.
An absolute inequality is an inequality that is satisfied by all
values of the variable or variables for which the functions appearing are defined.
A conditional inequality is one that is true only for certain values
of the variable or variables. Thus, x 2 + 1 >
(where x is real) is
an absolute inequality, because it is true for every real value of x
but x
1 >
is a conditional inequality, because it is valid only
;
when x >
1.
PROPERTIES OF INEQUALITIES
14-2.
The
gous
have
rules for dealing with inequalities are to some extent analoto those for equations. In transforming inequalities, we shall
occasion to use the following elementary principles which
248
Sec.
14-3
249
Inequalities
follow at once from the fundamental properties proved in Section
1-8.
Here are three
From
From
From
12
8
>
<
12
<
a
If
1.
Principle
then a
&,
illustrations
c
<
b
c.
:
+ 3 > 8 + 3.
- 2 < 12 - 2.
12 - 8 > 0.
follows that 12
8, it
12, it follows that 8
>
8, it
Principle 2. If a
follows that
<
b
c>
and
<
then ac
0,
be and -
<
c
Two illustrations are given here
From 3 < 5, it follows that 3 2 < 5
Q
A
From 8 < 10, it follows that < ~
c
:
2.
1
,
Principle
Here
14-3.
an
is
From
3. If
3
<
a
<
b
and
illustration
4, it
c
<
then ac
0,
>
be and c
>
c
:
-3 >
follows that
-4.
SOLUTION OF CONDITIONAL INEQUALITIES
The process of
solution of a conditional inequality consists in
values
of
the variable which satisfy the inequality.
finding
solution consists of a set of values of the variable, rather than one or
more isolated values as is usual in the case of a conditional equation.
A
all
The discussion in this section is limited to inequalities involving
rational functions in only one variable. If this variable is x, the
or f(x) < 0. For
inequality can be written in the form f(x) >
instance, suppose
we want
to solve the inequality
x2
We may
-
>
x
2.
then obtain the following equivalent inequality:
/(a)
=
x2
-
x
-
2
>
0.
seen that this transformed inequality has the same solution set as the original inequality.
In solving this transformed inequality, we find first the values
of x, if there are any, for which f(x) changes sign as x increases
in magnitude. If f(x) is a polynomial, such a change of sign occurs
It is easily
when f(x) =0.
In the example under consideration, changes of
sign are obtained only at points where
2
/GO = x
-x-2=Q.
we must find the roots of x 2 - x - 2 = 0. These are 1 and
and
2,
they determine on the #-axis three intervals throughout each
of which f(x) retains the same sign. In other words, to find the
Hence,
250
Sec.
Inequalities
14-3
> 0, we find the interval or intervals within which
the
has
f(x)
sign indicated in the given inequality. In the example
under consideration this sign is positive, because f(x) is to be > 0.
solution of f(x)
The method
is applied in Example 14-1.
In general, the solution of an inequality is obtained by equating
the function f(x) to zero and solving the resulting equation. If the
is
inequality
of the
i?i
form
^
0,
ps)
where p(x) and q(x) are poly-
2
nomials, it may be cleared of fractions by multiplication by [q (x) ]
Since the square of any non-zero real number is positive, the sense
of the inequality is not changed by multiplication by this factor.
This leads to the form f(x) ^ 0, where f(x) is a polynomial. The
.
values of x for which f(x) changes sign are called critical values.
When the critical values are arranged in increasing order, they
determine on the #-axis intervals, throughout each of which f(x)
cannot change sign. Consequently, the required solution is represented by the set of values of x for which f(x) has the same sign
as that indicated in the given inequality.
Note. In general, it can be shown that
if a factor of f(x) appears
an odd power (that is, if f(x) = has roots of odd multiplicity),
the function will change sign at values of x for which this factor
vanishes. If a factor of f(x) appears to an even power (that is, if
to
=0
has roots of even multiplicity), the function will not
change sign at values of x for which this factor vanishes. Therefore, it is sufficient to test only one value of x in one interval. This
test gives the sign off(x) in that interval. The sign of f(x) in each
of the other intervals can be quickly and easily determined from
the multiplicity of the critical values. Substituting into f(x) a value
of x in each interval then provides a check of the solution.
f(x)
Example 14-1. Solve the inequality
x
>
2.
= x2 -
x
-
x2
Solution:
An
equivalent inequality
f(x)
From
the preceding discussion
it
These roots are
As shown in
1
and
2
>
0.
follows that the critical values are the roots of
x2
-
is
-
x
-
2
= 0.
2.
Fig. 14-1, the points
1
and 2 determine on the z-axis the following
three intervals
(a)
x
< -
1;
(b)
-
1
<
x
<
2;
(c)
x
>
2.
Sec.
14-3
251
Inequalities
Throughout each of these intervals, f(x) retains the same sign. This condition
2
x
2 shown in Fig. 14-2. Here
may also be seen from the graph of y = x
we note that in each of these intervals the curve lies either entirely above the
a>axis or entirely below it.
The
= (-
>
solution of f(x)
- (-
can
now be found by examining
Thus, for a value such as x
of these intervals.
-2=4. Hence, f(x) is
=x
x
2 shown in
1.
x-axis to the left of x =
For the value x = in the interval (6),
we have /(O) = - 2. Hence, f(x) is nega2)
2
2)
2
In the graph of y
= -
the sign of /(a?) in each
2 in the interval (a), /( - 2)
positive throughout the interval (a).
above the
Fig. 14-2, the curve lies
throughout this interval, and the graph
below the g-axis.
For the value x = 3 in the interval (c),
tive
lies
=
/(3)
tive,
(3)
2
-
(3)
-
= 4.
2
and the graph again
So/(.r)
lies
is
posi-
above the
.r-axis.
These same
may
results
also be obtained
following much shorter procedure:
Select a value of x in the interval (6) for
by the
which /(x)
easily evaluated.
is
=
isz=O.Since/(0)
-2,/(x)
this interval. Therefore, f(x)
vals (a)
and
(c),
changes at the
Such a value
<0 throughout
>
for inter-
because the sign of f(x)
values x
critical
=
1
_
and
* A
x=2.
We see,
therefore, that in the intervals (a) and (c), f(x) has the sign indicated by
the given inequality. Since f(x) must be greater than zero, the solution set of the
- 1 and x > 2.
given inequality is described by x <
Example 14-2. Determine the values
Solution:
We shall
of x for
which \/x 3
2x 2
-
3x
is real.
solve the equivalent problem
f(x)
3
Solving the equation x
2x 2
= z 3 - 2z - 3x ^ 0.
3# = 0, we find that the critical values are
2
1, 0, 3.
(d)
As shown
in Fig. 14-3, these critical values determine
on the #-axis the following
four intervals:
(a)
(c)
x
< - 1;
< x < 3;
(6)
(d)
x
Throughout each interval f(x) has the same sign.
In this example we may select z = 1 in the interval
-
3(1)
= -
4, f(x)
<
throughout this interval.
1
>
(c).
<x < 0;
3.
Since /(I)
=
(I)
3
2(1)
2
252
Sec.
Inequalities
As we proceed
into the interval
>
Therefore, f(x)
and becomes <
(6),
we
14-3
when x = 0.
when x = - 1
find that/(z) changes sign
in the interval (6).
Again f(x) changes sign
from the interval (c)
we find that/(#) changes sign when x = 3 and is > in the
in the interval (a). Proceeding to the right
into the interval
(d),
interval (d).
Thus, we have the following results:
in (a), /(a)
in (),/(*)
<
0;
in (&),/(*)
<0;
in(d),/(x)
Hence, the inequality /(x)
2x 2
3x =
condition x 3
1,
=x
2x 2
3x
^
are
which the original expression
is satisfied
is
1
solution.
^ x ^
V#
Example 14-3. What values
in intervals (b)
and
0;
0.
(d).
And, since the
also allowed in the original problem, the values
the
3 are included in
0,
3
>
>
>
3
2# 2
3x
solutions
3x 2
^- > 0?
X
t
Solution: Clear the given inequality of fractions by multiplying by (x
- 2) (& 8 - 3rc 2 ) > 0. Solving the equation z 2 (,r - 2) (x
(x
we find that the critical values are 0, 0, 2, 3.
obtain f(x)
=
Hence, we have
(a) x < 0;
the,
for f(x)
These are the values of x for
3.
is real.
x3
of x satisfy
the
Therefore,
and x ^
following four intervals, as
< x < 2;
(c) 2
(b)
shown
in Fig. 14-4:
<
3;
x
<
(d)
Z
>
2)
-
2
3)
and
= 0,
3.
Throughout each of these intervals /(x) has the same sign.
z = 1 in the interval (b). Since /(I) = (1 - 2) [(I) 3
=
1)
2)
3(1)
2, it follows that/Or) >
throughout this interval.
We see that /Or) does not change sign for the critical value x = 0, because x
is a double root of f(x) = 0. Hence, f(x) >
in the interval (a).
Let us
-
2
]
initially test f(x) for
(
= (
As we proceed to the right from the interval (b) the function f(x) changes sign for
in the interval (c),
each of the critical values x = 2 and x = 3. Hence, f(x) <
and f(x) > in the interval (d).
Therefore, f(x) is positive in the intervals (a), (6), and (c?). That is, the solution
set of the original inequality is described by x < 2, excluding the value x
0,
=
and x
>
3.
Example 14-4. Solve the inequality x 2
=
2
2x
+3 >
0.
=
+
x
Solution: Let f(x)
2x
3. Since the roots of f(x)
are imaginary,
there are no critical values. Hence, the graph of y
f(x) lies either entirely above
the x-axis or entirely below that axis.
=
Testing for x
above the
= 0, we find that/(0) = 3.
z-axis.
This result indicates that the graph lies
is" satisfied for all real values of x.
Consequently, the inequality
14-3
Sec.
253
Inequalities
-
2x
1
1
Example 14-5. Solve the inequality
By
Solution:
Section
b-a<x<b+a.
r
3~
Hence,
be written as follows
may
the inequality
1-10,
<
I
which
1,
<
1
<
We may
2x
1
and
5
O
2x -
-
or
1
i
1
a
equivalent to
is
equivalent to
2z
|
1
<
1
3,
< 2x <
2
4.
5
O
<
and determining the common
1
3<2x
we have
O
described
is
by
proceed by solving individually the two inequalities
or
1,
1
- 1 < 3, or x < 2. The common solutions
< 2. So we again have - 1 < x < 2.
2x
x
1
<
|
that the solution set of the original inequality
Alternate Solution:
2x
is
6
:
l-3<2x<l+3
We thus find
- 1 < x < 2.
-
x
|
<
satisfy the
2x
1
< 1, we have
- 1 < x and
inequalities
For
x.
For
solutions;
r
o
EXERCISE 14-1
In each of the following problems, solve the given conditional inequality or
inequalities.
1.
x
-
3
4.
x
-
1
7.
6r
<
2.
x
0.
5.
4x
8.
4x
11.
-
14.
-
+ 3 < 0.
10. 3
-
13.
<
4x
+
0.
<
2x
+
2-i + i
1.
<
1.
<
0.
-
16
<
-
8
>
3x
-
3
<
6x
<
3.
1
<
^? + ? <
1
0.
10.
x
6.
4x
-
16
>
0.
9.
4x
< -
3x
-
<
3.
12.
2x 2
22. 4z 2
25.
3ic
20.
32.
+
2
5x
< -
1.
3#
<
4.
27. (x
+ !)(*+ 2)
29. (x
-
31.
Vz 2
33.
x
x
35.
1) (x
+2
+1
25
-
(x
2) (x
is real.
>
o.
23.
^
^
X
2
I
<
3
<
0.
_
1
X
+ 3) <
0.
28. (x
-
0.
30. (2x
3)
>
-
-
32. \/x*
34.
1) (x
-
5z
3
+2)
x2
<
|
21. x(3x
<
4) (x
|
144.
24.
6a?
0.
>
-
26. 3# 2 4-
1
-
2*
|
.
-
x
|
IB.
1.
18.
19.
+5 >
3,
2
4.
<
< -- -
^
X
X
7.
1.
1.
9.
+ 5)
(x
+ 2)
+
6
-
6)
>
(3x 4- 1)
is real.
0.
^
0.
254
Sec.
Inequalities
14-4
ABSOLUTE INEQUALITIES
To prove the truth of an absolute inequality, one must use the
known properties of the order relation. When none of these seems
readily applicable to the given inequality, it may be helpful to
replace this inequality by an equivalent one which may be more
Repeated replacements may have to be made.
In carrying out a sequence of replacements, one need not verify
equivalence at each stage, provided that the final inequality can be
shown to imply the original one.
The methods for proving absolute inequalities may be used also
to prove theorems involving inequalities, as in Example 14-7.
easily treated.
+ 62 ^
2
Example 14-6. Prove that a
The given
Solution:
inequality
a2
that
is
2ab for
all real
equivalent,
-
2ab
+
(a
-
2
b2
by
numbers a and
Principle
1,
b.
Section 14-2, to
0,
to
is,
This last inequality
6)
0.
because the square of every real number
is true,
negative. Therefore, the original inequality
is
is
(I
Example 14-7.
a
.a
.
,
prove that
-r
Solution:
o
<
distinct positive real
r
<
inequality T
,
,
is
is
by
equivalent,
is
true because
Similarly, the inequality
,
<
ad
this inequality,
by
Principle
-3
is
-f ed
1, is
<
<
ab
1,
CL
I
the inequality
a
T
o
/
+ be.
be.
<
,
,
is
a
-f c
,
+a
o
.
is
-5
by
true also.
equivalent,
<
<
by
Principle 2, to
be -f ed,
<
be.
equivalent to the given condition
T
c
<3
Hence,
/*
<^
,
<
>
to
a
,...
C
-5
equivalent to
ad
This last inequality again
inequalities
<
Principle 2, Section 14-2, to
equivalent to the given condition T
it is
,
by
equivalent,
Principle
Principle 2. Therefore, the inequality T
and
T
d,
ad
This inequality
if
,
.
ab -h ad
This inequality
numbers, and
+c <c
3
a
+
b
The
and d are
If a, 6, c,
non-
true also.
,
<
is
true also.
c
j are true,
a
Therefore,
it
follows that the original
14-4
Sec.
255
/nequcrfiffes
EXERCISE 14-2
1.
Prove that a 2
2.
Prove that
+1
2
a
>
Prove that a 2
4.
Prove that a
>
5.
Prove that T
+ -a >
Prove
a2
o
a2
7. If
that
+ b2
a
if
^^ ^ ^TT
+
3.
6.
2a
2t
a
a
if
a3
if
2
if
a
6
>
1
<
a
if
>
and
<
>
6
and that a 2
;
+
Note that a 2
is real.
<
1.
0.
a
<
if
<
a
1.
1.
a and b are positive and a
+ b + c ^ a& +
+ c 2 ^ 26c, and
2
= 2a only when a =
1
2
fcc
2
2a6, 6
c2
9* b.
+ ca. (Hint: From Example 14-6,
+ a 2 ^ 2ca. Add these inequalities.)
-
a and 6 are two positive real numbers, the quantities
and
\^ab,
>
~-r
are called, respectively, the arithmetic mean (A), the geometric mean (Cr), and the
harmonic wean (H) of a and b. Prove that
< A, except when a 6.
=
H <G
(In this case,
we have
8.
Prove that a 3
+ 63 >
9.
Prove that ab
+ cd ^
C2
10.
=
-I- rf2
Prove that a
3
z\
%2
13. If 21, z 2
14. If z is
,
1
1, if a, 6, c,
a and 6 are positive and a
and d are
^
as a
3
,
positive,
and
if
a2
>
6.
+ &2 =
1
and
=
\
|
21
6, if a, 6,
and
c are positive.
6.
a complex number, the modulus or absolute value of z is
\/x 2 4- y 2 Show that \z + z 2
\zi\ +\z*\ and that
is
z
|
$
+ c o according
6 + ab < a 4 + 6 4 if a ^
= x + yi
denoted by
I
6), if
3ab(a
^T
Prove that 7
12. If 2
//.)
1.
b
11.
A =G=
|
.
|
22
|
,
for all
l
\
complex numbers
and z 3 are complex numbers, prove that
a complex number, prove that
2
|
|
|
|
Zi
and
z\
+z 2
+|
\
22.
\z\
+23
<2||.
1
-h
|2
-2 3
|.
15
Progressions
SEQUENCES AND
15-1.
SERIES
Sequences. An infinite sequence, called more simply a sequence
or sometimes a progression, is a single-valued function whose
domain of definition is the set of positive integers. A finite sequence
is
1,
a single-valued function whose domain consists of the integers
for some positive integer m.
2,
,
m
In specifying a sequence,
it is
necessary to give a definite rule of
correspondence which assigns to each integer n a single definite
number, or term, of the sequence. This term may be denoted by a n
In particular, there is a first term ai corresponding to the integer 1,
a second term a2 corresponding to the integer 2, and so on. The
sequence may be specified by the array of numbers
.
(15-1)
aij
This sequence
a,2,
as,
,
an
,
.
denoted briefly by {a n }. (As usual, throughout
of dots
stands for numbers assumed to be
present but not written.) A finite sequence has a "last" term and
am or simply by {an }.
may be designated by a i9 o-2
The nth term, or general term, of a sequence is denoted by a n
From the rule specifying the nth term for each n we obtain the
is
this discussion, a
row
,
.
f
first,
for
second, third, and other terms of the sequence by substituting
1, 2, 3, and so on in turn. For example, if an = 1/n,
n the values
the sequence
is
111
lig'a'i""'
We
should note that {an } is the symbol for the sequence or function as a whole, whereas an is the symbol for the nth term or value
of the function corresponding to the integer n.
256
Sec. 15-1
257
Progressions
There are many methods for specifying the function in the
Two of these methods follow
tion of a sequence.
of
Explicit Formula. In one method, the nth
n itself by means of an explicit formula.
Here are a few
If
T,
If
an
an
=
n, the
Tl
T*
If
an
=
sequence
~T~
'
L
is
given in terms
.
.123-^
ti16 se(l uenc e is
~
>
^
the sequence
'
1
=
>
9
O
Zi
n
jr
^71
is 1, 2, 3,
.,
2 _j_
term
illustrations.
n
=
defini-
:
.,23
^
is 1,
i
.
1U
>
>
.
O
o
Recursion Rule. In another method, one or more of the first several values of a n are given explicitly, and a rule is then given
whereby a n can be calculated from some or all of its predecessors.
A few illustrations are given here.
Let a n+ i
=
an 2
+
I
with
a2
as
a4
Let a n+i
=
(n
+
Let a n+2
in this
=
=
=
example a n
o n with
= <Wi +
as
a4
05
=
=
=
0.
Then
+ 1 = O2 +
2
a2 + 1 = I 2 +
a3 2 + 1 = 2 2 +
ai
2
l)a n with a\
a2
as
a4
Note that
=
a\
=
=
=
=
2ai
3a 2
4a 3
=
nl
ai
=
+ ai
+ 02
3
4 +
a3
1
1
=
=
=
1,
2,
5,
Then
1.
=
=
=
fife
1
2
1
3
2
4
6
=
=
=
and 02
=
=
=
+
+
+
1
1
2
2,
6,
24,
=
1
1
1.
=
=
=
Then
1,
2,
3,
Let {an } be a given sequence of terms ai, 03,
, a*,
Form a new sequence {s n }, where s n is obtained by adding the first
n terms of {an }. The sequence of partial sums is then given by
Series.
.
Si
sn
=
=
=
Oi,
52
ai
+ 02 +
Oi
+ 02,
+ an
The sequence {s n } formed in this
based on the given sequence {an }.
The
'
>
(n
,
way
is called
=
1, 2, 3,
)
the (infinite) series
series as just defined is usually written in the following
abbreviated form:
(15-2)
'
ai
+ a2 +
258
Sec. 15-1
Progressions
Two
illustrations of series follow.
Illustration
partial
Let an
1.
= - and
=
sn
n
+ a^ H
ai
series
may
si
=
ai
=
1,
2
=
Oi
+
a2
=
1
+o =
1
,
1
3
9
>
11
1
=
Let an
2
4.
=
an(^ $n
+
+ #2 +
&i
<&n-
Then the
sums are given by
=
The
Then the
.
then be written
Illustration 2.
partial
an
sums are given by
1
The
I-
ai
+
=
a2
112
+ -
g
g
3
series is
2
n2
12
6
+
'
n
Limit of a Sequence. One of the most important questions relating to sequences is whether or not a given sequence an has a limit
as n increases indefinitely. If such a limit
exists, the sequence
is said to be convergent or to converge to the limit <. Symbolically
this statement may be expressed as follows
{
}
:
=
lim a n
w-oo
This notation
is
is
.
read, "the limit of a n as
n
increases indefinitely
."
In order to help
make
the concept of limit clear,
the sequence given by a n
= - The terms
we
shall consider
are
71
dl
=
1,
d2
=
1
H
Z
'
a3
1
=
7j
O
>'>
0n
=-1
fl
When we examine
n
is,
these terms, we notice that the larger the number
the smaller the term becomes. In other words, as n increases,
the closer to zero
as
we
please, if
is
the
number -
we merely
n
In fact, - can be
n
made
as small
choose n sufficiently large. For example,
Sec.
15-1
259
Progressions
- <
n
1000
for every n larger than 1000. We may conclude that for an arbitrary real positive number d, we can find a value of n, say any
- <
n
-
N
integer
- <
n
d.
n
for every
100
The
^
such that for
>
^
a
We
larger than 100.
all
see also that
n > N,
integers
is
it
true that
and the sequence <-> are therefore related as
limit
(nl
indicated by the following statement.
The sequence
<
-
1
(n)
converges to the limit
=
<
to each arbi-
0, if
trary positive number d there corresponds an integer
that
0-d<-<0
+ dfor
n
In general, the limit
Definition.
number d >
<
d
{an converges to the
0, there corresponds
+ d for
<
an
every n
> N.
:
sequence a n converges to the limit <,
there exists a positive integer
such that
\
{
:
}
number d >
also be put as follows
A
be defined as follows
A sequence
N such that
may
may
of a sequence
,
a positive integer
such
> N.
every n
Definition of Limit of Sequence.
if to each arbitrary real
limit
This definition
N>
if
for each
N
n
|
<
|
Convergence of a Series.
d for every n
We
>
N.
shall again consider the infinite
series
ai
(15-2)
which
is
+
a2
+
+ an +
-
,
sums
the sequence of partial
Slj 52,
-
*
'
'
,
Sn
'
'
'
,
of the sequence {a n }. By the following definition the series
vergent if the sequence of partial sums is convergent.
is
con-
sequence of partial sums of the infinite series
(15-2) converges to a limit, and if lim S n = Sj then the series is
Definition.
If the
n-
said to converge to the limit S,
and S
CD
is called
the
sum
of the infinite
series.
The new use of the word "sum" for the value S of an
infinite
perhaps unfortunate, for it seems meaningless to talk about
adding up the terms of an infinite series. Actually, S is not a sum,
but it is rather the limit of a sequence of partial sums of the series.
If a series does not converge to a limit as n becomes infinite, we
say that it is divergent, or that it diverges.
series is
260
Sec.
Progressions
151
EXERCISE 15-1
1.
2.
Given
=
a\
1
and a+i
=
Given a l =4, a 2
the sequence { a n }
3.
Given
ai
=
1,
a2
of the sequence
4.
5.
Show
Show
of
6.
five
terms of the sequence {a n }.
an
Find the
= 2a n +i
and a n + 2
=
an }
2,
=
and a n+ 2
.
first six
terms of
+
(n
na n Find the
l)a n+ i
.
first six
terms
.
= 3 n satisfies the recursion rule an+ = fln+1 + 6an
w
=
2) also satisfies the recursion relationship
2
that the sequence a
Problem
Find the
.
.
that the sequence a n
.
(
4.
A and B are any real numbers whatsoever, show that the
= A3 n + #( 2) n satisfies the relationship of Problem 4.
= 2a n + 1 with ai = 2. Let sn = + 2 +
+ an Find
Assuming that
sequence a n
7.
{
3,
= n + an
Given an +i
ai, 02,
,
8.
Show
9.
Prove that
10. If s n
+
that
=n
(2n
15-2.
-
.
i
05 and
if s
cr n
2
,
1)
s lf s 2)
ai
=
sn
+
a%
s.
,
+
sw -i for
+
n >
1,
= 2n
show that a n
= n2
,
then s n+ i
sn
= a n +i
= SL
and therefore that
given ai
1
1
+3 +5 +
.
ARITHMETIC PROGRESSIONS
An arithmetic progression is a sequence of numbers in which
each term after the first is obtained from the preceding one by
adding to it a fixed number; this number is called the common
difference.
Note that the common difference may be found by subtracting
any term of the sequence from the one that follows. Thus, 1, 5/2,
4, 11/2, is an arithmetic progression with the common difference
3/2, since 11/2 -4 = 4-5/2^5/2-1 = 3/2. Also, 5, 1, -3, -7 is
an arithmetic progression with the common difference 4, since
-7 -(-3)
=-3-l = l-5 = -4.
necessary and sufficient condition
that three numbers A, B, and C form an arithmetic progression is
It follows, therefore, that a
C-B^B-A.
15-3.
THE GENERAL TERM OF
AN
ARITHMETIC PROGRESSION
Let a denote the first term of an arithmetic progression, and let d
denote the common difference. Then, by definition, an arithmetic
progression with n terms may be written as follows
:
a,
Hence
(15-3)
if ln
a
+ d, a + 2d
;
a
+
3d,
,
a
+
(n
represents the value of the nth term,
ln
=
a
+
(n
-
l)d.
-
l)d
15-4
Sec.
Also,
we may
n terms
write an arithmetic progression of
following manner
a,
We
261
Progressions
in the
:
a
+
d,
a
shall be concerned
+
-
2d,
with
,
-
ln
2d,
Zn
-
five quantities in
d, t.
connection with an
arithmetic progression. These are the first term a, the
terms n, the nth term ln , the difference d, and the sum
number of
of the n
Sn
terms.
SUM OF THE
15-4.
TERMS OF
FIRST n
AN
ARITHMETIC PROGRESSION
Sn
represent the sum of the first n terms of an arithmetic
progression. If we write the indicated sum in both direct and
Let
reverse orders,
Sn =
a
+
In
+
(a
we have
+
d)
+
+
+ 2d) +
(a
(l n
-
2d)
+
-
(l n
d)
+
Zn ,
and
S =
(l n
-
Adding the right
d)
+
(Zn
sides,
-
2d)
+
+
(a
+ 2d) +
we have n terms each
(a
+ d) + a.
of which
is
a
-I-
ln .
Thus,
2S n =
(a
+
+
Zn )
(a
+
Zn)
+
+
(a
+
+
^).
k)
+
(a
+
Zn)
=
n(a
+
t).
Therefore,
Sn =
(15-4)
(a
- l)d from (15-3) for
(n
have another useful form for the sum. This is
If
we
substitute a
+
Sn =
(15-5)
Example 15-1.
[2a
+
(n
Z
in (15-4),
we
-
Determine which of the following sequences are arithmetic
progressions:
a) 3, 7, 10;
6) 6, 1,
-
4;
c)
3z
-
y,
4x
+ y,
5x
+ 3y.
10-7 and 7 3 are not equal, the sequence
not an arithmetic progression.
5. Since these differences are
6 =
b) In the second sequence,
- 4 is an arithmetic progression.
equal, the sequence 6, 1,
Solution: a) Since the differences
3, 7,
c)
=
x
10
is
-4-1=1
We find that (5x + 3y) - (4x + y) = x + 2y and (4s + y) - (3& - y)
+ 2y. Since there is a common difference, the given sequence is an arithmetic
progression.
Example 15-2. Find the twelfth term, and
of the arithmetic progression 4,
7, 10,
.
also the
sum
of the first 12 terms,
262
Solution:
We have a = 4, n = 12, and d = 3.
- l)d = 4 +
12 = a + (n
Z
Also,
Sec.
Progressions
by
Then, by (15-3),
-
(12
1)3
=
37.
(15-4),
Si.
=
+ Z.) = y (4 + 37) = 246.
(a
\
Example 15-3. The
sixth term
is
Solution:
third term of an arithmetic progression
Find
the twenty-second term.
3/2.
By
(15-3),
h =
a
2d and
-f
f
a
{a
By
solving these
l
(15-3),
15-4
n =
1/4
two
+ 2d
+ 5d
= a -f
= 3/4,
= 3/2.
we
linear equations,
+ 21(1/4) =
Z6
5d.
is
3/4,
and the
Thus, we have
=
find that a
1/4 and d
=
By
1/4.
11/2.
Example 15-4. Find each value of x for which the three quantities 3#
5,
x
+ 4,
2 form an arithmetic progression.
3x
C B =B
A, we have
+ 4) = (x + 4) - (3x - 5).
Solution: Applying the condition
(3x
o,
-
i
Solving,
15-5.
i. A
we obtain
ar
-
2)
15
=~
-
(x
m,
The
-*u
.
i-
arithmetic progression
.
is
25
-j-
37
31
>
-7-
>
-j-
ARITHMETIC MEANS
The terms of an arithmetic progression between any two given
terms are called arithmetic means between the given terms.
If we let the given terms be a and Z n any number, say fc, of means
,
be inserted between a and ln by using the formula
(n
l)d with n = k + 2. As soon as we have found d,
insert the required means.
may
Example 15-5. Insert three arithmetic means between
Solution: Let a
=
0, I*
=
and n
1,
113
Hence, the three means are j
>
^
>
=
3
+ 2 = 5.
Then
1
ln
=a+
we can
and
1.
=
+ 4d, and d =
j
If only a single arithmetic
mean
is
to be inserted
between two
given numbers, then the inserted value is called the arithmetic
mean of the given numbers. Thus, if a, x, b form an arithmetic
progression, x is called the arithmetic mean of a and 6. Since
x = x a,
6
Thus, the single arithmetic
half their sum.
mean
of
two numbers
is
equal to one-
Sec.
5-5
1
263
Progressions
EXERCISE 15-2
In each of the problems from 1 to 9, determine if the given sequence is an arithmetic progression. Find the next two terms of the extension of each arithmetic
progression.
-
1. 5, 8, 11, 14.
2.
4.4,12,19,27.
5-
-H'
8.
a
7. a, 6,
-
26
-
36
a,
2a.
1, 7, 13,
12. 22, 18, 14,
14. 2, 4, 6,
to 7 terms.
-
-
to 50 terms.
to 10 terms.
16. a, 2a, 3a,
9.
~-^
,
^
3, 4, 11.
a,
^^
13. 1, 2, 3,
to 10 terms.
15. 1, 3, 5,
to 75 terms.
17. 0.2, 0.5, 0.8,
to 35 terms.
18. 1, 8, 15,
-
10,
^
and S n for the arithmetic progression.
11. 3, 6, 9,
to 26 terms.
19, find l n
to 12 terms.
-
-
-|'fi'-i'-Ig-
'*'
+ 6, a - 6, a - 26, a - 36.
In each of the problems from 10 to
10. 2, 8, 14,
1
3.
19.
19.
-
to 20 terms.
to 100 terms.
1, 2, 3,
In each of the problems from 20 to 27, three quantities relating to an arithmetic
progression are given. Find the other two quantities.
20. a
22.
=
S2
i
5,
=
h =
653,
= 4.
= 21, a = 6.
n
36,
n
21. a
23.
n
= 10, n = 10, d = 10.
= 45, d = L 845 = 63.
i
24.
26.
Z 2l
a
= 8,
= -
n
1
i
28. Find the
29.
= 21,
,
d
=
sum
d
=\
3
= i--
,
27. a
|
= 7, & = 52,
d
=
= 45,
d
=
=
1,
S.
^
1.
of the first 100 even integers.
Find the sum of the
first
31. Insert ten arithmetic
32. Insert six arithmetic
n odd
integers.
means between 20 and
means between 100 and
means between
mean
33. Find the arithmetic
of 10
30.
40.
3 and
and 56 and that
2.
of 4
and
28.
Find the arithmetic mean of 28 and 65 and that of 33 and 78.
35. Insert k arithmetic
36.
Z
12*5
30. Insert five arithmetic
34.
25.
o
means between - and
a,
where a
^
0.
A display of cans in a grocery store is in the form of a pyramid whose base is an
equilateral triangle. If each side of the base contains 20 cans and the number of
cans decreases by one for each successive row, how many cans are in the display?
37.
A
numbered consecutively from 1 to 100. Customers
and pay according to the number of the ticket, except that
numbered above 50 cost just 50 cents each. How much money isi
lottery contains tickets
draw
tickets
tickets
collected
if all
tickets are sold?
x
3x
be an arithmetic progression.
38.
Determine x so that
39.
The sum of the first and fourth terms
sum of the third and twelfth terms is
40.
Find the sum of
all
x,
2,
will
of
an arithmetic progression is 20. The
Find the sum of the first 15 terms.
36.
multiples of 5 from 100 to 1,000, inclusive.
264
Sec.
Progressions
1
5-6
HARMONIC PROGRESSIONS
15-6.
A harmonic progression is a sequence of non-zero numbers whose
reciprocals
form an arithmetic progression. Thus,
harmonic progression
a
if
>
r
o
>
are in
b, c
a,
- form an arithmetic progression.
c
The terms of a harmonic progression between any two given
terms are harmonic means between the given terms.
To insert a desired number of harmonic means between two
numbers, we insert the same number of arithmetic means between
the reciprocals of the two given numbers and then invert the resulting terms.
The harmonic mean
ing manner. If
a, x, b
two numbers
of
found
is
in the follow-
form a harmonic progression, then - - T
a x u
>
form an arithmetic progression, and x
and r Hence,
mean
the arithmetic
is
i
>
of a
o
mean
Solution of this equation for x yields the harmonic
X
Clearly the harmonic
mean
~
__
_2o6_
+
a
b'
a
exists only if
+
=
b
Example 15-6. Insert three harmonic means between
Solution:
Is
=-
>
The corresponding
and n
=
5.
Hence,
-
arithmetic progression
= -
-
-f 4d,
-
It follows that the arithmetic progression is
Therefore, the three harmonic
means
are
=
and d
-
>
0.
3
is
and
2.
Here
>
o
&
a=
$
o
>
'
^1
~
o
'
77;
'
1Z
O
O
TTT
'
>
o
Z
ZT:
24
8, 12, -=-
EXERCISE 15-3
In each of the problems from
progression. Find the next
,111
3'7'U'
L
*5'i'g-
1
to
6,
two terms
9
2
determine
if
the given sequence
of the extension of each
1
1
f
8
'5'l6'l6'
.-2
8,f.
^
-
7.
Find the tenth term of the harmonic progression
8.
Find the seventh term of the harmonic progression
Insert four harmonic means between 1 and 2.
9.
a harmonic
.111
1
-4'8'l6*
5.16
is
harmonic progression.
Z
^
>
=
7
>
U
6, 3, 2,
>
.
.
f
2,|.
Sec.
15-8
265
Progressions
harmonic means between ^ and 7
4
3
11. Insert four harmonic means between 6 and 24.
12. Find the harmonic mean of 6 and 9.
13. Find the harmonic mean of 24 and 72.
10. Insert three
14. Insert nine
2
3
harmonic means between - and ~
15. If a 2 6 2 r 2
form an arithmetic progression, show that 6-f-c,
o
,
,
2t
a harmonic progression.
16. If a, 6, c form an arithmetic progression and
show that ad = be.
17. If
the harmonic
is
a?
18. If a, 6,
19. If a,
a
=
and
b,
show that
x
=
5
r
b
x
_
=
a
h
T
*
b
,
=
form a harmonic progression, show that C
harmonic mean of a and
and conversely.
1
a
j-
~~"
C
6 is equal to their arithmetic
mean, show that
GEOMETRIC PROGRESSION
15-7.
A
by
of a
a+& form
d form a harmonic progression,
6, c,
d form a harmonic progression, show that
c,
6, c
20. If the
mean
a+c,
geometric progression is a sequence of numbers in which each
first is obtained from the preceding one by multi-
term after the
plying it by a
fixed
number; the multiplier
ratio
may be found by
is called
the
common
ratio.
The common
immediately preceding
it.
Thus,
1,
- >
Zi
in
which the common ratio
is
.=
1,
V2
i
2
is
is
a geometric progression
= --f-- =
4
^
-~-l=i
a geometric progression in which the
=
u 4=
V2 4=
2
>
4
o
\/2,
4 ^
O
--r--
is
dividing any term by the one
1
=
i
Also,
Zi
common
ratio
* V2 = -i
A/2
\/2
that a necessary and sufficient condition that three nonC
= R
zero numbers A, B, and C form a geometric progression is
-^
-j
It follows
1
15-8.
THE GENERAL TERM OF
A GEOMETRIC PROGRESSION
Let a denote the first term of a geometric progression, and let r
denote the common ratio. Then the progression may be written
as follows
:
a, ar, ar
Hence,
(15-6)
if
ln
2
,
ar 3
"1
,
,
ar 71
.
represents the value of the nth term,
ln
=
ar*- 1
266
Sec.
Progressions
SUM OF THE
15-9.
TERMS OF
FIRST n
represent the sum of the
progression. Then
ar
ar 2 H
Sn = a
by r, we have
S nr = ar + ar 2
Multiplying
first
n terms
~2
+
+
5-9
A GEOMETRIC PROGRESSION
Sn
Let
1
n
h ar
of a geometric
+ ar"-
1
.
+ arn~ + ar n
+ ar3 +
l
.
Subtracting the first of these equations from the second, term by
term, we have
Sn =
Snr
ar n
a.
Therefore,
n
a
Sn =
(15-7)
1)
^ j" 1
=
a( *
I
(r
/*
^
1).
n
Submultiply both sides of (15-6) by r, we get rln = ar
in
This
for
S
is
form
we
obtain
useful
another
H
stituting
(15-7),
we
If
.
.
S n = 5L=j
(15-8)
(r
*
1).
Note. If r = 1, these formulas do not apply; in this case, however,
the geometric progression becomes a + a H
haton terms, and
Sn =
na.
Example 15-7.
Determine which of the following sequences are geometric
progressions:
a)
2,6,18;
6)
5,10,30;
c)
*,^~y
y
Solution: a) The ratios found by dividing each of the second and third terms by
the preceding one are 3 and 3. Hence, the sequence 2, 6, 18 is a geometric progression.
6) In this sequence, the ratios of consecutive terms are 2 and 3. Therefore, the
sequence
c)
is
30
5, 10,
is
not a geometric progression.
is a geometric progression in which the
The given sequence
common
ratio
x/y.
Example 15-8. Find h and S&
The common ratio
Solution:
is
in the geometric progression 6, 2/3, 2/27,
(2/3) * 6
=
1/9. Since a
=
6 and
n
=
5,
we have
\
o(l
6
1
-
is
-
96.
Solution:
Find the
By
-
(1/9)*)
_
1-1/9
r
Example 15-9. The
term
6(1
r*)
fifth
V
59,049/
1-1/9
term of a geometric progression
ratio and the first term.
_
14,762
2187
is 3,
and the tenth
common
the formula (15-6) for the nth term of a geometric progression,
we have
ar 4
=3
and
ar 9
= -
96.
15-10
Sec.
267
Progressions
Dividing each side of the second equation by the corresponding member of the
5
32. Hence, r = - 2. Therefore, a( - 2) 4 = 3, or
equation, we obtain r =
first
Example 15-10. Find each value
x
+
1
x
of
which the three numbers
for
form a geometric progression.
C = B
-r we
we apply the condition
Solution: If
>
A
-5MJ
we obtain z =4/5. Hence, the geometric
have
is
progression
-
-
=
4/5,
-
x
4- 1
X
~~~
2,
2
=x
Solving,
X
&
x
x,
6/5, 9/5.
GEOMETRIC MEANS
15-10.
Terms
of a geometric progression between any two given numbers are called geometric means between the given numbers. Let a
and ln be given numbers. Then k means may be inserted between
them by using the formula
ln
=
with n
ar"- 1
=
Example 15-11. Insert three geometric means between
Solution:
Let a
=
the three means are 2
'
Zw
4
,
2
=
1
'
and n
2,
2
,
2
=
Then
5.
/5
= ar
mean
of a and
6.
2.
1
and
2.
4
and
r
= 2 1/4
.
b
Since #
X
=
a geo-
called
is
x
-
=
a
db
we note that a geometric mean of two numbers
mean proportional between the two numbers.
same
the
is
Hence,
as a
Hence,
.
3 '4
form a geometric progression, then x
If a, #, 6
metric
1,
1
+
k
EXERCISE 15-4
In each of the problems from 1 to 9, determine whether the given sequence is a
geometric progression. Find the next two terms of the extension of each geometric
progression.
-
1. 2, 8, 32.
2 .\
4.2,4,6.
5.27,18,12.
>
3. 4, 16, 64.
1, 2.
&
*
V3 V6 V3
7
7'
"T
'
'
~6~
8>
In each of the problems from 10 to
&
a2
.
T'
a
'
'
2"
.
Q
9.
3"
15, find
ln
and S n
12. 3, 9, 27,
-
-
;
n = 10.,
n = 45.
13. 100,
;
14. log 2, log 4, log 16,
16.
11. 3, 2, 4/3,
Find the sum of the
;
n
first
=
10.
n terms
-
-L
f
J
1
1,
-
1
1, 1.
1
in the given geometric
progression.
10. 4, 2, 1,
,
-
;
n
10, 1,
15. log 9, log 3, log
=
;
15.
n
=
V5,
of the geometric progression
1, =r
101.
;
n
j
= 6.
.
268
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
15-10
For what values of x do x - 2, x
6, 2x + 3 form a geometric progression?
For what values of x do 3x + 4, x
2, 5x -f 1 form a geometric progression?
1?
For what values of x is x -f 1 a geometric mean of 2x + 1 and a;
Find a geometric mean of 4 and 16. Also, find their arithmetic mean.
Find a geometric mean and the arithmetic mean of 3 and 12.
Insert four geometric means between 1 and 32.
Insert five geometric means between 1 and 1,000,000.
Insert ten geometric means between 1 and 2.
If A, G, and H denote the arithmetic mean, a geometric mean, and the harmonic
2
mean, respectively, of two numbers a and 6, prove that G = AH.
If the arithmetic mean of a and 6, when a, b 7* 0, is A, a geometric mean is G,
H.
and the harmonic mean is H, find A
G, G
H, and A
2 4
Show that the sum of the first n terms of the geometric progression 1> o
>
'
*
*
'
Q
is3(l
28.
Sec.
Progressions
(j
1
Discuss
how
this
sum
varies as
n
increases.
Show that the sum of the first n terms of the geometric
(sli*
16(1
Discuss
how
this
sum
varies as
sum of the first n terms of the sequence
xS n .)
be the sum. Then compute S n
29. Find the
Let
Sn
n
increases.
1,
2x,
15-11.
sum
INFINITE
of the terms of the finite sequence 2,
=
3x 2 4x 3
,
Q
f\
30. Find the
is
progression 8, 4, 2,
>
=^
.
,
11
>
=j
(Hint:
Q<i
>
>
i
O
=^
GEOMETRIC PROGRESSION
A geometric progression in which the number of terms is infinite
an infinite geometric progression.
In Section 15-9, we found an expression for the nth partial sum
Sn of a geometric progression. Hence, Sn is the nth term of the
geometric series based on the given progression. Thus, we have
is called
Si
=
a;
82
=
a
+ ar; 83 = a + ar + ar 2
;
;
Sn = a
+ ar +
+
Furthermore,
Let us consider what happens to the sum of n terms of a geometric progression when the number of terms increases indefinitely.
If a = 0, evidently Sn =
whether r ^ 1 or r = 1. In this case, the
number meets the requirement of a limit of Sn , so that the sum S
of the infinite geometric series is equal to 0.
Now
suppose a
Case
1.
0. For this condition, we consider four cases.
Assume that \r\ < 1. If r ^ 0, the numerical value of r
decreases as w increases. Moreover, by making the number of terms
*
Sec.
15-12
we can make \rn as small
we can make Sn differ from
sufficiently large,
that
if |r
please
<
that
;
269
Progressions
1,
S n approaches
is,
as
\
we
please. It follows
by as
T-^_
little
as
as a limit. This condition
^
we
may
be stated symbolically in the following manner:
S =
(15-9)
where S
lim S n
n-oo
= r^
1
-
>
r
sum of the infinite geometric series. Equation (15-9)
is true also if r = 0, since in this case Sn has the constant value a.
Case 2. If r = 1, then S n
na. Since a
0, |S n increases indefinitely as n increases. Here S n diverges.
Case S. If |r| > 1, then jr increases indefinitely as n increases.
~^Again {Sn diverges.
Hence, so does S n = 7-^
is
the
=
|
{
}
71
]
|
\
|
r
1
Case
(
4-
l)^ 1
a.
If r
then the progression becomes,
1,
n
If
is
\
|
r
1
even,
Sn =
0.
If
n
is
odd,
Sn =
a, a,
a,
a,
In this case,
a.
,
we
say that S n oscillates between and a. Here also the series diverges.
The sum of an infinite geometric progression can therefore be
found by (15-9), but only when \r\ < 1. When r = 1, r
1, or
>
sum.
the
series
no
and
has
that
the
series
we say
1,
diverges,
r\
For a further discussion of this topic, the student may refer to a
treatise on the theory of limits.
Example 15-12. The owner
of a fleet of trucks finds that
refined for re-use, 20 per cent of the oil
gallons of refined
the total
amount
Solution:
We
oil
and
is lost
re-refines this oil
in the process.
each time
it
if
If
used motor oil is
he starts with 100
becomes
of oil he has used before the entire 100 gallons
begin with 100 gallons. After the
first
dirty,
determine
is lost.
reclaiming operation,
we
have 80 gallons of good oil. When this becomes used and is re-refined, we have 64
gallons; and so on. Theoretically, we would never use up the entire amount of oil.
However, the limit of the sum of the amounts of oil reclaimed is approximately
reached after a large number of operations. Hence, we have an infinite geometric
progression in which a
re-refining the oil as
of
=
it is
100 and
r
=
used, the fleet
4/5.
Then S
100
_
=
,
500.
Thus, by
owner has had the equivalent of 500 gallons
oil.
15-12.
REPEATING DECIMALS
JL
a decimal contains a fixed sequence of digits which are
repeated indefinitely, we call it a repeating decimal. Thus,
0.135135
is a repeating decimal. This decimal is written 0.135,
the dots indicating the first and last digits of the sequence which is
to be repeated. Also, 0.34516 = 0.34516516516
repeating
decimal is wholly or partly an infinite geometric series. For
If
.
A
270
Sec.
Progressions
1
5-1 2
0.34516 = 0.34 + .00516 + 0.00000516 +
it is
,
composed of the decimal 0.34 and an infinite geometric series in
which a = 0.00516 and r = 0.001. Hence, since |r| < 1,
since
example,
'
00516
a00516
-
~~
'
34
172
100
0.999
'
33300
33300
Note that if we divide 172 by 33300, we obtain the repeating
mal 0.00516.
Example 15-13. Express the repeating decimal
0.26*3 as
deci-
an equivalent numerical
fraction.
We
Solution:
can write this decimal in the form
0.2 + 0.063 + 0.00063 +
.
Hence, the required number consists of the decimal 0.2 plus an infinite geometric
series in which a = 0.063 and r = 0.01. The sum of the series, since \r\ < 1, is
7
0.063
0.063
-
1
"
110
0.99
90
7
1
~"
~~
0.01
0.263=1+^=^.
Therefore,
EXERCISE 15-5
In each of the problems from
on the given
1.
-
,
1
1
,
.,24
7. 1,
,
g
,
13.
8.
.
5
o
1
,
~
>
.
sum of the
1
>
>
^
3
2
g
,
convergent series based
3. 3,
.
...
6. 8,
V3,
-
-
-
1,
4, 2,
-
.
-...
Qi
* *' - 39 "
>
'
'
5
.".
'
'
25
11. 0.18, 0.0018, 0.000018, ....
+ 0.012 + 0.00012 + 0.0000012 +
Find the sum of the
8,
1
fl
,
g
,
5. 8, 4, 2,
.
10. 0.5, 0.05, 0.005,
12. 0.3
to 12, find the
2. 1
.
|
1
,
1
geometric progression.
-
,
1,
|
4.
infinite
based on the
series
Also, find the
.
sum
infinite
geometric progression 24,
of the first 20 terms of this progression
and com-
S instead of $20What would be the error if S were used instead of S n for the sum of the geometric
pute the error introduced by using
14.
progression 48,
15.
-
36, 27,
-
to 10 terms?
A ball is dropped from a height of 3
feet.
On
each rebound
it
bounces back to
three-fourths the height from which it last fell. Assuming that this bouncing
continues indefinitely, find the distance it travels in coming to rest. How far
has
16.
it
traveled after bouncing ten times?
A swinging pendulum
will gradually come to rest as a result of friction. If, on
each upswing, the pendulum swings through 98 per cent of the arc through
which it fell, and if the initial arc for one complete swing was 20 inches, find the
distance traveled before the
pendulum comes
to rest.
Sec.
15-13
271
Progressions
Convert each of the following repeating decimals to fractional form.
17. 0.1.
18. O.i5.
19. 0.90.
20. 0,243.
21. 0.16.
22. 0.142857.
23. 2.9.
24. 1.1234.
25. 0.11542.
26. 2.123.
THE BINOMIAL SERIES
15-13.
We shall now
number. If we
becomes
/ie IA\
/t
(15-10)
(1
consider the binomial expansion when n is any real
let a = 1 and b = x in (4-13), the binomial formula
1 + nx +
+ x) n = 11
\
i
+
i
...+
n(n
-
L-
n(n
^
1)-
2T
-
1) (n
The right member of (15-10)
We saw in Section 4-6 that,
o
x2
-
+
(n
2)
,
~
n(n
-
1) (n
+
r
1)
2)
x3
+....
x
called a binomial series.
is
n
a positive integer, the series
on the right in (15-10) terminates with x n and (15-10) is true.
Otherwise, the terms of the series continue indefinitely, giving rise
if
is
,
to
an
infinite series.
The question which now
n
is
not a positive integer
arises is
;
that
is,
whether (15-10) is valid when
whether the series on the right
n
It
converges, and, if so, whether its sum is equal to (1 + x)
proved in the study of series that (15-10) is indeed valid if \x\ <
is
.
It follows that, if \x\
accurately as
we
<
1,
we may
obtain the value of (1
please by taking sufficiently
+
many terms
n
1.
as
x)
of the
series in (15-10).
The expansion can
a positive integer. In this case, the
follows
:
(15-11)
Here x
+ 6) n when n is not
expression may be written as
readily be extended to (a
(a
=
a
>
+ &) =
fo(l
L.
\
and the expansion
=
+ -Yr
a/ j
of (a
a" fl
L.
+ -Td-j
+ 6) n is valid
if
\
Example 15-14. Find the first five terms of the expansion of
Solution:
By
a
(1
<
+ z)~ 3 it\x\ <
(15-10),
(-8)(-4)(-6)
3!
ja
(-8) (-4) (-5) (4!
1.
\
1.
272
Sec.
Progressions
15-13
Example 15-15. Find ^L04.
j/TM =
Solution:
pansion
(1
+ 0.04) "3.
1/3 and x
= 0.04,
and the ex-
5
+ 0.04)3 = 1+| (0.04) +
(1
=
Hence, n
is
The approximate value
1
We have
of the
sum
+ 0.013333 -
+
(0.04)2
2!
of this series
3 ^
3
^
(0.04)3+ ....
is
+ 0.000004 =
0.000177
3^
1.013160.
here a case of an alternating
series, that is, a series in which the terms
and negative. It is proved in the study of series that, if in
each term is numerically less than the preceding term and
are alternately positive
an alternating
lim a n :=
0,
series
by using S n
the error introduced
as the
sum
of
the series
is
-
numerically less than the value of the first term omitted; that is, \S n
S\ < \a n +i\.
If in the present example we take for S the value S 3 = 1 + 0.013333 - 0.000177
=
1.013156, the error in so doing
15-16. Find the
Example
F
Solution:
First
we apply
is less
first
than the value of the fourth term 0.000004.
four terms of the expansion of
.,
-
x*
(15-11) to convert the given expression to a suitable
form, as follows:
1
-^
(8
r2\\-i/3
/
/
=
-.)-. = (8(1-!))
1
^T) =K
(/v2\-l/3
Hcnco, by (15-10),
if
x*
<
_i
1
-!-)
8,
a2 \
~
2\-l/3
/
,
(-1/3) (-4/8) /
1-2
V
"*"
s/""
\
i
1-2-3
\
x*\*
(-1/3) (-4/3) (-7/3)7
"*"
x*
8/" ""/
1
V
""
24
288
20,736
EXERCISE 15-6
In each of the problems from 1 to
expansion of the given expression.
i.
6.
^
1
(1
~T"
X
-
z)-i/'.
2 - T1
7.
X
-4+y
x
38.
10, find
vrr^.
4.
'
x
(x
_?
+
v
the
first
vr^.
9. 7
2
(x
i/)
four terms of a binomial
3
+^-r--
5.
(
l/aj
10. (1
.
-r /
v
+ir)
2/)
Find the approximate value of each of the following numbers by means of a
binomial expansion, using four terms of the expression.
11. (1.02)io.
16.
49 4
.
12. (1.01) 13 .
13. (1.04) 8
17. (0.99)'.
18 . 51 3.
.
14. (l.l) 10 .
15. (0.98) 8
19 .
20. (0.97)~ 2 .
(1.03)1/2.
.
16
16-1.
Mathematical Induction
METHOD OF MATHEMATICAL INDUCTION
When
a certain type of formula or proposition has been verified
known to be true in general, the method
of mathematical induction is often found extremely valuable in
in specific cases but is not
determining its validity.
Suppose that a statement involving a positive integer n is to be
proved true for all values of n greater than or equal to a particular
initial value. We begin by showing the result to be true for the first
value of n. We then assume that k is some particular integral value
of n for which the statement holds. With this assumption as a
basis, we establish the validity of the statement in the next suck 4- 1. In other words, we
ceeding case, namely, that in which n
prove that if the statement is true for any specific integral value
of n, say n = k, then it is also true for the next larger value of n,
namely, n = k 4- 1. Suppose for example, that 1 is the initial value
of n. Then the second step establishes that if the statement is
true for n = l, it is also true for n = 1 + 1, or 2 if it is true for
;
n=
2, it is also
of this,
we
true for
ft
=
2
+
1,
or 3
;
and so
conclude that the statement
greater than or equal to the
is
on.
As a consequence
true for
here
all
values of n
A
proof by
mathematical induction, therefore, consists of two parts and a
initial value,
1.
conclusion.
Part 1. Verification that the statement is true for some initial,
value of n, generally n = 1. (This initial value is the smallest value
of n for which the statement is to be proved true.)
Part
2.
Proof that whenever the statement is true for some parn = k, then it is true for the next larger
ticular value of n, say for
value of n, that
is,
for n
=k+
1.
273
274
Sec. 16-1
Mdfhemafi'ca/ Induction
Conclusion. If both parts of the proof have been given, then the
is true for all positive integral values of n greater than
statement
was made
or equal to the one for which the verification
in Part
1.
The reasoning process involved
here, which consists in taking
and then repeatedly taking successors, can be
exemplified in terms of climbing a ladder. Part 1 puts us on the
bottom rung of a ladder. Part 2 shows us how to get from any rung
we have reached to the next higher rung. The conclusion states
that if we know how to get on the bottom rung of the ladder, and if
we know how to get from any rung to the next higher one, then we
can reach all rungs, and hence can climb the ladder.
an
initial integer
The following examples
Example 16-L
If
n
is
will illustrate the
positive integer, prove that
any
1+1+1 +.
1-22-33-4
(16-1)
method.
^
+
n(n
+
n
1)
1
Solution:
Part 1: The formula
is
=
when n
true
1
1-2
=
since
1,
1
1=1.
r
1+1*
2
2*
Part 2: Let k represent any particular value of n for which (16-1)
-
1
tI
I
l-2" "2-3" "3-4"
"*"
tI
I
+
fc(fc
is
Then
true.
-
~~
"
fc
1)
+
1
We now
wish to prove that (16-1) is true also for the next larger value of n,
=
k + 1. The sum on the left in (16-1) when n = k + 1 can be obtained
n
namely,
by adding
we have
its last
term, which
r-m
iwy
+ 1)
(* + 2)
is 7,
.
(*
/LL+JL+JL
+
VI -2 "^2-3 ^3-4 T
.
k(k
+
!)/
>
to both sides of (16-2). Hence,
^
(k
+
k
k
But, since
fc
we obtain
,
lft
ox
(16-3)
+
!
j
+D
1
+2
(t
)
!
+
+ 2oTo3 +
^
^ 3oTT4 ^
.
TTo
1 2
(t
.
^
.
1
(*
+
1) (k
_fc +2fc+l_fc+l_
~
~
~
+ i) + 2) t + 2
2
1
+
+
+2)
l)(k
k
.
(fc
(fc
1
^
(A?
+
1) (k
+ 2)
_
"
(*
k
+
+
1)
(*
+
2)
k
+
l
+
1)
+
1
I
l
+
1
=
k + 1. Hence,
The members of (16-3) are the same as those of (16-1) when n
we have shown that (16-3) is true if (16-2) is true, in other words, (16-1) is true
for n = k + 1 if it is true for n = k.
We
have shown by
Therefore, since (16-1) is true for n
Conclusion:
for every positive integer n.
verification that (16-1) is true
=
1, it
when n
follows from Part 2 that (16-1)
is
=
1.
true
16-2
Sec.
275
Mathematical Induction
Example 16-2. Prove that
-
(x
a factor of (xn
is
y)
y
n
n
if
)
is
any
positive
integer.
Solution:
Part l:Itn
=
Part 2: Let
A:
We
shall
-
(a*M
now
y*+i).
_
x *+i
1,
then x n
We have
*+i _
yk+i = x
By assumption, (x
factor of y k (x - y).
Therefore,
if
-
By
by
n
=x
?/,
which
Hence,
Part
33,* 4. X2/
-
*
y
1
(x
-
is
seen to have
n
=
of the proof, (#
-
true for
virtue of Part
2,
=x
(
7/
,
-
y) as
?/*), it is also
xk
y*). Also,
it is
y) as a factor.
(a;
has (x
)
_
y *)
by
y) is a factor of the left
n
is
*+i
-
y) is a factor of (x*
the conclusion
Conclusion:
Therefore,
y
n
specific value of n for which (x
k
if
is
factor
of
that
a
(x
(x
y)
prove
be a
member
(x
=k +
-
y
n
)
true for
y)
-
k+l
n
n
y) is a factor of (x
is
y).
inspection, (x
also true for
the desired conclusion
_
4. <,*(
a factor.
a factor of
is
y**
a
1
).
1.
when n
any
=
1.
positive
integer n.
16-2.
PROOF OF THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS
We shall now prove that the binomial formula,
n n
n
n
n~ b
T ^ an"
+
(16-4)
(a + b) =a + na
^
l
2fc2
...
- 1)1
n(n-l)...(n-r+l) an_r6r
(r
r!
is
+
.
.
.
+^
true for every positive integral value of n.
Proof. Part
1.
Hence, (16-4)
is
Par #. Let
Then we have
(a
+
fc
6)*
When n = 1, each
true for n = 1.
be any specific value of
=
a*
+
ka k
~l
b
+
k ^k
becomes a
side of (16-4)
~
n
X)
for which (16-4)
is
+
6.
true.
a*~ 2 6 2
1>
Multiplying each
(a
+
6)*+i
=
member
of this equation by (a
'
a** 1
H----
+
ab k
+ab
k
-\
'
+
"" r
(k
&),
+
1
we
obtain
276
Mathematical Induction
Hence, by combining terms,
+
(o
6)*
+1
=
a**
In this result the
follows
-
k(k
+
1
sum
(Jfc
we
+
See.
16-2
get
l)a6
+
.
.
.
of the coefficients of a k
~r+l
br
obtained as
is
:
-
(k
1)
r
+
2) (k
-
+
r
1)
+
r!
fc(Jfc
1)
(fc
-
r
+
2)
(r-1)!
r fc-r+1
~L
_
(fc
r
+
1)
(fc) (fc
-
1)
(k
-
r
+ 2)
r!
We
+
note that the value of (a
1
obtained as the product of
(a + b) and (a + 6), is exactly the same as the expansion which
would be obtained from (16-4) with n = fc + l. Hence, we have
6)**
,
k
shown that
n = fc + l.
if
the binomial formula holds for
Conclusion.
n=
The binomial formula was seen
Therefore, by virtue of Part
every positive integer
2,
we may
k, it
must hold for
to be true for
conclude that
it is
n=
1.
true for
n.
EXERCISE 16-1
Prove by mathematical induction that each of the statements in Problems
20
is
true for all positive integral values of n.
.
2. 1
3.
2
+ 3 4- 5
+4 +6
H---H----
+ (2n - 1) = n
+ 2n = n(n + 1).
2
.
5.3+6
+
12 H----
6.
4
7.
1+4+7 +
9.
10.
8
4-
..
.
+ 4n =
2n(n
+(3W _2)
-f 1).
1
to
16-2
Sec.
12. I 8
+ 2s + 38 +
13. I 3
+3 +5 +
+ 48 + 6 +
+ 22 + 2 +
+2 =
+ 32 + 33 +
+
14. 2
15.
3
2
16. 3
17.
4
18. I 2
277
Mathematical Induction
+
+
3
3
8
3
+ 42 + 43 +
-
(2n
(2n)
8
3 =
2
2 I)
(2r*
= 2n 2 (n + I) 2
rc
2(2"
-
&
+ 4 = | (4
-
1).
= | (3 -
3
+ 3 2 + 5* + 7* +
1).
.
1).
-,,1).
+frt
= n(2n
-%
-
^
19
^'
""
"
1
3
3
1
2ft
5
^2
7
5
2
3
(2n
1
21. Prove that x 2n
1)
+
(2n
"*"
4
~l
2/
"^
i
3
4
5
2n
+
i
"*"
2n is
divisible
+ y*n-i
""
1)
1
i
3
-
1
i
1
"
n(n
by x
+
1) (n
+ 2)
1
n(n+3)
=
4(n
Prove that x 2n
23.
By using mathematical induction, prove the formula for the
metic progression.
24.
By using mathematical induction,
progression.
1} (n
^
+ 2)
+ y for every positive integer n.
22.
{$ divisible
+
by x
-f y, for
every positive integer n.
prove the formula for the
sum
of
an
arith-
sum of a geometric
17
17-1.
We
Permutations, Combinations,
and
FUNDAMENTAL
Probability
PRINCIPLE
begin the study of permutations and combinations by con-
sidering the following principle, which
entire subject.
is
fundamental for the
Fundamental
Principle. If one thing can be done in a ways, and
for
such
each
if,
way, a second thing can be done in b ways, then
the two together can be done in a b ways.
To understand why the principle is true, note that for each of the
a ways of doing the first thing, there are b ways of doing the
second hence, both the first and the second things taken together
can be done in a b ways.
The following examples will illustrate the reasoning upon which
the principle is based, as well as an obvious extension of the principle to the case when more than two things are to be done.
;
Example 17-1. In how many ways can two
officers,
a chairman and a secretary,
be selected from a committee of five men?
By the fundamental principle,
the problem is equivalent to determining
which
the
two
ways
things can be done together. The first
position can be filled in 5 ways; that is, there are a = 5 ways of selecting a chairman.
For each of these possible selections, there are 6 = 4 ways of filling the position of
secretary from the remaining men. Hence, the number of ways of selecting a
chairman and secretary is a 6 = 5 4 = 20.
Solution:
the
number
of
in
How many
Example 17-2.
digits 0, 1, 2,
,
9, if
three-digit
numbers can be formed from the ten
a) repetitions of digits are not permitted; 6) repetitions
are permitted?
Solution:
place can be
a)
Here we have three things to do or places to fill. The hundreds
9 ways, since
must be excluded from this place. The tens
filled in
278
Sec.
17-2
279
Permufaf/ons, Comb/naf/ons, oncf Probability
place can then be filled in 9 ways from any of the remaining 9 digits. Finally, the
units place can be filled from any of the remaining 8 digits. There are, therefore,
648 three-digit numbers in which no two digits are alike.
9.9.3
=
b) If repetitions are permitted, there are 9
ways to
fill
the hundreds place and
10 10
900
each of the tens and units places. Hence, there are 9
three-digit numbers.
10 ways to
fill
=
EXERCISE 17-1
1.
2.
Nine persons apply for each of two vacant apartments. In how many possible
ways can both apartments be rented?
A large room has eight doors. In how many ways can a person enter the room
by one door and leave by a different door?
thrown, in how many ways can they fall?
There are eight men and six women in a club. In how many ways can two
officers be selected so that one is a man and one is a woman?
How many possible four-digit numbers are there in a telephone exchange which
3. If three dice are
4.
5.
uses only four-digit numbers and excludes 0000?
6.
7.
8.
How many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9?
How many of these are larger than 700,000?
How many numbers of six different digits can be formed from the digits 2, 3, 4,
5, 6, 7, 8, 9? How many of these are larger than 700,000?
A woman has seven guests at a party. If she chooses her seat first, in how many
ways can she seat her guests?
17-2.
An
PERMUTATIONS
ordered arrangement of
all
or any part of a set of things
is
called a permutation. Specifically, suppose we have n distinct things
and wish to select r of these to be arranged in a definite order.
Each such ordered arrangement is called a permutation of n things
r at a time. The number of all such permutations is denoted by
n Pr
Thus, 5 P2 is read "the number of permutations of 5 different
things 2 at a time."
In Example 17-1, the possible number of ways of selecting a
chairman and secretary from a committee of five men was found
by means of the fundamental principle to be 5 4 = 20. This is pre.
cisely equal to 5 P2 since it is the number of ways in which two men
can be chosen from among the given five men and arranged in the
two offices. The number of permutations of n things r at a time is
given by the formula
,
(17-1)
n Pr
=
n(n
The truth of (17-1)
is
-
1)
(n
-
2)
(n
-
r
+
1).
readily shown as follows. The first of the
n ways. Then the second can be filled in
r places can be filled in
(n
1) ways, the third in (n
2) ways, and so on. In general, the
280
and
Permutations, Combinations,
number
of
already
filled.
17-2
Sec.
Probability
of filling each place is n minus the number of places
when the rth object is chosen, (r 1)
ways
Therefore,
places have already been filled, and the rth place can then be
r + 1, ways.
in n
1) , or n
(r
In particular,
then have
(17-2)
if
n
r
= n,
Pn =
the last factor becomes
-
n(n
1) (n
-
=
1
2)
n
n+
1
=
filled
1.
We
nl
This formula gives the number of permutations of n different
things taken all at a time.
If both the numerator and the (understood unit) denominator
of (17-1) are multiplied by (n
r) !, we obtain the following alternate formula
:
H7
* P _
(17-6) n r r Here
r
it is
-
n(n
1) (n
-
(n
2)
-r+
(n
1)
-
r)l
agreed that by definition
n\
_
~
( n -r)!
(n
!
=
'
-
r)
!
Hence, (17-3) holds for
1.
~n.
For example, by (17-2),
8
P6 =
n(n
-
1) (n
-
(n
2)
-
r
+
1)
=
8
7
6
5
4
=
6720.
Using (17-3), we have also
aft
17-3.
_
~
_8!_8-7-6*5-4-3-2.1
"
nl
(^=7)1
PERMUTATIONS OF
T2^
8!
n
672
'
THINGS NOT ALL DIFFERENT
required to find the number of indistinguishable
n things all at a time, if n\ things are
regarded as indistinguishable, n^ other things are regarded as indistinguishable, and so on. Let us consider, for example, the number
of permutations of the letters a, a, a, b taken four at a time. For
convenience, indistinguishable objects are given the same notation.
Denote by P the desired number of distinct permutations. Evidently, P is less than ^P^ which is the number of permutations that
could be effected if all the letters were distinguishable. For in any
one of the P permutations, say (a b a a) any rearrangement of the
a's among themselves would not change the permutation. If, how-
Suppose
it
is
distinct permutations of
,
ever, we assigned subscripts to a in this permutation,
(eh 6 02 da), we could permute these three distinct letters
as
in
among
ways. This can be done for each of the P permutations of the letters a, a, a, 6. We would then obtain P 3 permutations of the four distinct letters al9 o^, a 3 b taken four at a
time. There are therefore 4 permutations altogether. Hence,
themselves in 3
!
!
,
!
P-
3!
=4!,
or
P=
Jj
3!
=
4!
~3!1!
Sec.
all
17-4
Permutations, Combinations,
and
281
Probability
In general, the number of distinct permutations of n things taken
at a time, if HI things are alike, n2 other things are alike, n 3 other
things are also alike, and so on, equals
n\
P=
(17-4)
n\\
COMBINATIONS
.
A combination is a set of all or any part of a collection of objects
without regard to the order of the objects in the set. We use the
symbol n Cr to represent the total number of all combinations of n
different things taken r at a time.
The different sets of the four letters a, 6, c, d taken three at a
time, without reference to the order in which the letters are
arranged, are (abc) (abd), (acd), (bed). From each of these four
combinations, we can form 3 !, or 6, different permutations of the
four letters taken three at a time. For example, from the combination (abc) we can form the distinct permutations (abc), (acb),
(bac), (bca), (cab), (cba). Hence, each of the four combinations
9
P3 = 3
permutations to the total number of permutations. Thus, there are 4 C 3 combinations of the four letters if we
disregard order, and there are 3! ordered arrangements or permutations of each combination. Hence, we have J3 = 3 4^3, or
contributes
3
!
!
= 4>3 >2
C3 = 5
o
number we
4
2
1
!
l
'
=
This
4 combinations.
is
the
same as the
o
obtained at the beginning of the paragraph.
In general, if we divide the total number of permutations of n
things r at a time, or n Pr by the number of permutations, or r!,
,
contributed by each combination, we obtain the total
combinations. Symbolically, we have
n4
or
we
=
?
*
of
rbt>
p
C =
7T'
(17-5)
If
r
number
'
note that r
esting relationship
= rPr
I
,
then
we can
:
..
Cr ~
n* r
~~p~
r
r
r
Replacing
n
Pr
by
its
equivalent expression from (17-1) or (17-3)
we have
__
write the following inter-
n(n
-
1)
*
(n
~
r
+
1)
_
'
n\
282
Permutations, Combinations,
If r is replaced
in (17-6)
r)
by (n -r)
n C n_ r
and
Probability
we
obtain
,
Sec.
17-4
n\
=
(n-
r)!r!
Hence,
nC r
(17-7)
Afote.
Having agreed that
!
= nCn-r- 1, we can
to permutations or combinations of
supply (17-1) or (17-3)
at a time. Con-
n things zero
we have
sidering (17-5),
=
n Co
(17-8)
0!(n
l
= L
=
o!
0)I
This result agrees with the intuitive conclusion that there
one such "empty" combination. Similarly, n P = 1.
is
only
BINOMIAL COEFFICIENTS
17-5.
referring to the development of the binomial formula in Sec-
By
tion 4-6, we note that the expansion of (a + b) n involves the product (a + 6) (a + 6) (a + 6)
taken without regard to order. This
fact suggests the use of combinations in the coefficients of the
expansion.
We
a n-r b r
term involving
said in Section 4-7 that the coefficient of the
"
'
-
"
.
is
This
r!
n things r
the combination of
j
g precisely the f or mula for
at a time, or
Cr
obtained in Section
17-4. The binomial formula may therefore be written as follows
n- 2 2
n
b +'
-+ n C n b\
(17-9) (a+b)* = n Coa+nCia -ib+nC 2 a
n
,
:
.+#'"&+
For example, the expansion of (a +
6)
4
takes the following form
:
This reduces to
a4
If
we
let
a
=
6
=
+
4a 3 6
1 in
+ 6a 26 2
+
the expansion for (a
b)
n
in (17-9),
we
obtain
+ 1) = n C + nCl + n C 2 +
+ nCn = 2 Ci + C 2 +
-
(1
Since
n
C =
1,
+
nC n
.
1.
Thus, the total number o;f combinations of n things taken succesn
ft at a time is 2
1.
sively 1, 2,
,
EXERCISE 17-2
1.
2.
Evaluate 5^2, 7^3, 12^6, 21^4,
Evaluate 10^4, nCa, looC's, 21^5,
100^95*
Sec.
3.
17-6
Form
a.
and
Permutations, Combinations,
all
283
Probability
word
possible distinct permutations of the letters of the
How many are
there? b.
How many
begin and end with a vowel?
theory.
c.
How
begin or end with a vowel?
4.
many
How many
distinct permutations are there of the five letters a,
three at a time? Write
5.
them
6, c, d, e
taken
out.
how many different ways can a dime, a quarter, and a half-dollar be disamong five boys?
How many different sums can be formed with a penny, a nickel, a dime, and
In
tributed
6.
a quarter?
7.
8.
8
using combination symbols.
the six digits 1, 2, 3, 4, 5, 6, form all permutations taken five at a time.
a. How many are formed? b. How many begin with 4? c. In how many does
Expand
(a -f 6)
,
From
the digit 3 not appear?
9.
How many
distinct permutations can be
made from
the letters of the
word
probability taken all at a time?
10.
How many
different combinations are there of 4 identical nickels ; 5 identical
dimes, and 6 identical quarters?
11.
How many
of
12.
In
different straight lines are determined
which are in a straight
how many
on a
different
by twelve
points,
no three
line?
ways can a student
select seven questions
out of ten
test?
13.
How many different weights can be
14.
16, and 32 pounds, respectively?
In how many different ways can signals be made with seven different
where a signal is a set of one or more flags arranged in a specific order?
15.
In
how many
different
formed with
ways can the 52 cards
six objects
weighing
1, 2, 4, 8,
of a bridge deck be dealt
flags,
among
four players?
MATHEMATICAL PROBABILITY
17-6.
If,
can
in a given trial, an event can happen in
happen in / ways, and if all the h
fail to
likely,
h
different
+ / ways
ways and
are equally
then the probability p that the event will happen in this
trial is
p
(17-10)
=
The probability q that the event
happen
is
g-
(17-11)
Note that
will fail to
O^p^ 1, O^g^l, and p + q = 1. Two illustrations of
probability follow.
If a bag contains 3 green marbles and 4 yellow marbles, all
exactly alike except for color, the probability of drawing a single
green marble
is
3
3
284
Permufaf/ons, Comfamaffons,
and
Sec.
Probability
17-6
A
die can fall in six ways. The probability of getting 5 or more
2
1
are two possible ways,
th
with
ith one throw of a die is - or - since there
o
o
either a 5 or a 6, for the event to happen.
>
>
MOST PROBABLE NUMBER AND MATHEMATICAL EXPECTATION
17-7.
Let p be the probability of occurrence of some event. Furthermore, suppose that n trials of the event are made, of which h are
Then -
successful.
U
is called
the trials which occurred. It
the relative frequency of success for
h
not to be expected that -
is
=
p.
It is
TV
shown in more advanced treatments of probability, however, that if
n is large, it is very likely that the relative frequency is approximately equal to p. Also, the larger we take n, the more likely it is
that - approximates p closely. Moreover, it can be shown that the
id
most probable or expected number of occurrences of the event for n
trials is up.
For example, when a coin
head
is tossed, the probability of getting a
In 1,000 trials the expected number of heads is thereThis does not mean, however, that if the first 100 trials
is 1/2.
fore 500.
result in 75 heads
tails in
and 25
the next 100 trials.
we
should expect 25 heads and 75
Actually, since one toss of the coin does
tails,
not affect the next one, we should expect about 50 heads and 50 tails
in the next 100 trials. Moreover, we may expect about 450 heads
and 450 tails in the next 900 trials.
the probability of winning a certain amount of money in
case a certain event occurs and
is the amount of money to be won,
If
p
is
m
defined to be pm. For instance, if a
person can win $12 provided he throws an ace with a die, his expectation is - ($12) = $2. Hence, $2 is the fair amount he should be
the mathematical expectation
willing to
17-8.
pay
to
STATISTICAL,
make
OR
the
is
trial.
EMPIRICAL, PROBABILITY
frequently impossible to have sufficient knowledge beforeall the conditions that might cause an event to happen or
fail to happen. In such a case, however, it may be possible to determine the relative frequency of the occurrence of the event from a
It is
hand of
number of trials. Thus, if an event has been observed to
happen h times in n trials, and n is a large number, then until addi-
large
Sec.
17-$
tional
and
Permufaf/ons, Combinations,
knowledge
is
we
available,
Probability
define the statistical probability,
or empirical probability, to be
P
(17-10
where -
l>
the relative frequency.
is
n
=
Example 17-3. A molding machine turns out 12 parts per minute. Inspection
experience has shown that there are 20 defective parts per hour. What is the
probability that a single part, picked at random, will be defective? In a run of
10,000 parts, how many defectives should be expected?
Solution:
defective.
defective
The
parts are produced at the rate of 720 per hour, and 20 of them are
Hence, the probability that a single part selected at random will be
is
=7^
(ZO
=
ob
In a run of 10,000 parts, we should expect
^
oO
(10,000), or
approximately 278, defective parts.
17-9.
MUTUALLY EXCLUSIVE EVENTS
Two
of
more events are mutually exclusive if not more than one
them can happen in a given trial. The following theorem may
or
be stated.
Theorem. The probability that some one of a set of mutually
exclusive events will happen in a given trial is the sum of the individual-event probabilities.
Proof. Consider, for simplicity, a set of two mutually exclusive
events. Suppose that the first can happen in hi ways and the second
can happen in h2 ways, and
n be the
let
which the two events can happen or
and p2 =
fail to
number
of
ways
Then
happen.
p\ =
total
in
are the corresponding probabilities of the two events.
it
Since the n events are mutually exclusive, the hi ways are different from the h2 ways, and the number of ways that either the first
or the second event can happen is therefore hi + h%. Hence, the
probability p that either the one event or the other will happen
.,.&.,..
p.idt*.
(17-12)
is
For example, suppose that a bag contains 2 green marbles, 3 yellow marbles and 5 brown marbles. If a marble is drawn at random,
the probability that
2
it is
green
is
o
yellow
is
Hence
if
either green or yellow
a marble
is
and the probability that
it is
drawn, the probability that
it is
^
is
>
23
51
+^
^
>
1U
lu
or
>
lu
or ^
^
286
Permutations, Combinations,
and
Sec.
Probability
1710
DEPENDENT AND INDEPENDENT EVENTS
17-10.
In case two or more events are not mutually exclusive, they are
dependent if the occurrence of any one affects the occurrence of the
others, and they are independent if the occurrence of one does not
affect the occurrence of the others.
if a card is drawn from a deck of 52 cards and the
not replaced before a second is drawn, then the second
drawing is dependent on the outcome of the first. If, however, the
first card is replaced, then the second drawing is independent of
the first. In the latter case the two drawings are equivalent to
simultaneous drawings from two decks.
We shall now state and prove the following theorem relating to
For example,
card
is
dependent and independent events.
Theorem. The probability that two dependent or independent
events will occur (successively if dependent; successively or simultaneously if independent) is the product of their individual
probabilities.
Suppose that the first event can happen in hi out of a
ways, and that the second event can happen
in ft out of n^ different ways. Then it follows, by the fundamental
principle in Section 17-1, that the two events can happen together
in hih^ ways out of a total of n^n2 different ways. Therefore, the
probability that both events will happen is
Proof.
total of ni different
P
*
(17-13)'
H2
ni
Example 17-4. Two cards are drawn from a deck containing 52 cards. Find the
when the first card is not replaced before the
probability that both cards are aces
second
is
drawn.
Solution:
1)
The
We
shall begin
with a
listing of the following useful probabilities:
probability of drawing an ace from a deck of 52 cards
2) If the first card is
another ace
is
an ace and
it is
is
oZ
not replaced, the probability of drawing
-^
01
3) If the first card is
not an ace and
second being an ace
is
not replaced, the probability of the
is
51
4) If the first card is replaced, the probability of the second being
This drawing
~4
an ace
4
is
-r^
is entirely independent of the first drawing.
the
Consider, now,
given problem. The probability that the first card is an ace is
41
If the first card drawn is an ace, then the probability that the
Pi
16
O/
=
=
Sec.
1
7-1
second card drawn
will
be aces
287
Permufat/ons, Comb/naf/ons, cmcf Probability
1
pl p a
is
is
an ace
is
p2
=-._ =
3
= ^r
= T1
oi
Hence, the probability that both
ii
.
REPEATED TRIALS
17-11.
Theorem. If p is the probability that an event will happen in any
= 1 - p is the probability that it will fail, then the probtrial, and q
ability that it will happen exactly r times out of n trials is
n
r
nC,tr<r*
(17-14)
=
W-
i
-
Proof. The happening of the event in exactly r trials and its failure in the remaining n
r times are independent events. Hence,
~
the probability, by the theorem in Section 17-10, is p rq n r But these
r trials can be chosen from the n trials in n Cr ways. Since these
~r
r
Note
ways are mutually exclusive, the total probability is n Crp q n
that this expression is the (r + l)th term of the binomial formula
.
*
+
for (q
(q
+
p)
M
pY =
,
q
since
n
+ n Ciqn~ p + n C qn~ 2p 2 +
l
+ nC q - p +
n
2
r
r
r
+ pn
.
The successive terms of this expansion give the probabilities that
n times in n trials.
the event will happen exactly 0, 1, 2,
r,
An event will happen at least r times in a given number of n
r + 1, or r times. Since these events
trials if it happens n, n
1,
are mutually exclusive, the probability that an event will happen at
le&st r times is given by the sum
,
,
Example 17-5. What
is
the probability of tossing an ace exactly three times in
four trials with one die?
Solution: Since the probability of tossing
PJ
ability of failure is
formula.
The
^
*
we may
an ace
in
trial is
^
and the prob-
substitute in the term n V<r~ rP r of the binomial
result is
* (f )'(!)'= '(I) (s>
Example 17-6. What
trials
one
is
=
sir
the probability of tossing an ace at least twice in four
with one die?
Solution:
The event
Hence, the probability
will
is
happen at
given by the
least twice
following
-
if it
sum
happens
:
'(!)'=
4, 3,
or 2 times.
288
and
Permufaf/ons, Comfa/naf/ons,
Sec. 17-1
Probability
1
EXERCISE 17-3
1.
A
certain event can
happen
ways and can
in four
to
fail
happen in
six
ways.
the probability that it will happen? If $60 can be won on the event,
what is the mathematical expectation?
2. A box contains 5 white balls, 4 red balls, and 13 black balls, a. If one ball is
What
drawn
that
3. a. If
is
what
out,
it is
the probability that
is
it is
red? b.
What
is
the probability
white or red?
one die
is
thrown, what
is
the probability that a "1" or a "2" will turn up?
What is the probability that a "3" or larger number will turn up?
When a coin was tossed 100 times, 80 heads and 20 tails turned up.
b.
4.
If the
tossing were continued until 200 tosses had been made, what would be the
most probable number of tails in the second 100 tosses?
5.
A
bag contains
drawn, what
6.
7.
8.
five
$1
bills,
ten $5
bills,
and twenty $10
one
If
bills.
bill is
the mathematical expectation?
the probability of throwing a "7" or an "11" on one throw of two dice?
is
What is
An automobile owner
carries $1,000 theft insurance on his car. If, during the
past year, 237 out of 97,864 automobiles registered in his area were stolen,
what is the mathematical value of the policy?
In a city of 77,000 families, a careful sample of 800 families showed that 120
sample families owned their own homes, a. What is the probability
of the
that a family selected at
9.
random
23,100 voted for his opponent.
random voted for the winner?
10.
in the city
owned
its
home?
b.
What
expected number of families in the city who own their own homes?
In a certain city 28,600 persons voted for one candidate for an
What
is
is
the
and
office,
the probability that a voter chosen at
A
bag contains 5 red balls and 9 black balls. If two balls are drawn in sucand the first is not replaced, find the probability that the first is red
and the second is black.
cession,
11.
In a baseball tournament the probability that team
A
will
win
is
=
>
and the
'
probability that team
two teams will win.
12.
The
B
I
will
team
probability that
win
A
is
will
Find the probability that one of these
^
reach the finals of a tournament
2
is
=
>
'
I
and the probability that it will win the
team A will win the tournament.
finals is
-
Find the probability that
13.
Find the probability of throwing three successive fours on a pair of
14.
The
probability of
A
winning a game when he plays
it is
j
He
is
dice.
scheduled
to play four times, a. Find the probability that he will win exactly three times.
b. Find the probability that he will win at least three times.
15.
Three dice are tossed,
turn up. b.
What
is
a. Find the probability that exactly two threes will
the probability that at most two threes will turn up?
Solution of the
18
18-1.
General Triangle
CLASSES OF PROBLEMS
There are certain relationships among the lengths of the sides
and the trigonometric functions of the angles of every triangle. If
one side and any two other parts of a triangle are given, the
remaining parts can be determined; that is, the triangle can be
solved. The three given parts may comprise any one of the following four combinations
:
Case I. One side and two angles.
Case II. Two sides and the angle opposite one of them.
Case III. Two sides and the included angle.
Case IV. Three sides.
In this chapter, we shall discuss methods for treating these four
cases.
For convenience, we shall let ABC denote any triangle
whose angles are A, B, and (7; and we shall let a, 6, and c represent
the lengths of the corresponding opposite sides.
18-2.
THE
Law
tional
LAW OF
SINES
ABC be any triangle lettered in the convenmanner. Then the following relationship between the sides
of Sines.
Let
and the sines of the angles may be written
a
(18-1)
This relationship
.
sin
is
A
=
b
.
sin
commonly
B
=
:
c
sin
C
called the law of sines.
B A
FIG. 18-1.
289
290
Solution of the General Triangle
Sec.
1
8-2
Proof. We first note that all angles may be acute, as in Fig.
18-1 (a), or one angle, say B, may be obtuse, as in Fig. 18-1(6).
(The case where B = 90 entails no difficulties and will therefore
be omitted.) In each diagram, let h denote the altitude from
the vertex
sin
B=
C
to the side
AB, Then,
in either case, sin
h
a
Dividing the
sin
sin
equation by the second,
first
A
B
a
a
_
QJ
j
b
sin
-
A
A=
- and
we have
--B
b
sin
In a similar way, by drawing the altitude from the vertex
we get
A
to
the side BC,
b
c
B
sin
~~
sin
C
The equations thus obtained may be combined
law of
to give the
sines
"
sm B
sm A
sin
C
Note. The law of sines is well adapted to the use of logarithms
because it involves only multiplications and divisions.
Since any pair of ratios in the law of sines involves two angles
and the sides opposite, it may be used in the solution of problems in
Cases I and II.
As noted above, the law of sines also applies in the special case
where ABC is a right triangle. In this case, one of the angles is
90, and
18-3.
the sine of that angle
SOLUTION OF CASE
TWO ANGLES
I
is 1.
BY THE
LAW OF
GIVEN ONE SIDE
SINES:
AND
When one side and any two angles of a triangle are known, the
third angle can be found from the relation A + B + C = 180, and
each of the required sides is uniquely determined. These sides may
be found by the law of sines.
Example 18-1. In a
triangle
ABC, A = 3814', B =
6720',
c
=
Solve
329.
the triangle.
Solution:
The values
of the given parts are indicated in Fig. 18-2. In this case,
C =
c
To
180
find a,
-
+
(3814'
we use the
6720')
329
a
sin
3814
Therefore,
FIG. 18-2.
7426'.
relationship
a
c-329
=
a
_
""
/
sin
7426'
329 sin 3814
/
'
sin
7426'
18-4
Sec.
291
Solution of the General Triangle
and
- log sin 7426'.
log sin 3814'
log a
log 329
indicated operations may be performed as follows:
2.5172
log 329
=
The
+
log sin
3S14'
=
=
-
9.7916
12.3088
log sin
7426'
=
9.9838
10
10
10
2.3250
log a =
a =211.
To
determine
6,
we
use the relationship
329
~~
sin
Hence,
6720'
_
""
sin
329 sin 67 20'
sin
and
log b
follows:
The work
7426 /
7426 /
'
= log 329 + log sin 6720' log 329
log sin
6720'
log sin
7426'
=
=
2.5172
9.9651
12.4823
log b
b
As shown
log sin 7426'.
= 9.9838
= 2.4985
= 315.
-
10
10
10
in Fig. 18-1 (a),
c
= b cos A +
cos B.
This relationship
may be used as a check. Thus,
c = 315 cos 3814' + 212 cos 6720'
=
which agrees
18-4.
(315) (0.7855)
+
(212) (0.3854)
satisfactorily with the given value of
=
329.1,
c.
SOLUTION OF CASE II BY THE LAW OF SINES GIVEN
THE ANGLE OPPOSITE ONE OF THEM
TWO
SIDES
AND
Case II is called the ambiguous case, because the data may be
such that two, one, or no triangles are determined. The number of
solutions when a, 6, and A are given is indicated by the accompanying
table.
TABLE OF POSSIBLE SOLUTIONS
Solution of the General Triangle
292
We
Sec.
18-4
shall use Fig. 18-3 to illustrate in turn the different possiand the sides a and b
considered in the table. If the angle
A
bilities
AX
are given, we first construct the angle A with the initial ray
and the terminal ray AR. Next, we lay off the distance AC = b
along the terminal side. Then, with C as the center and the length
of the side a as the radius, we describe an arc. We mark the point
of the
or points in which this arc intersects the initial ray
AX
angle A.
-X
X
A
(<*)
Figure 18-3 (a) corresponds to (18-2) in the table. Since
BC = a = b sin A, this segment is the altitude of the triangle drawn
from the vertex C. Hence, the arc with radius a is tangent to the
initial side at B, and the triangle is a right triangle.
a
Figure 18-3(6) corresponds to (18-3) in the table. Since
b sin A, the side a is too short to intersect AX, and there is no
<
triangle.
In Fig. 18-3 (c), a < 6 and 6 sin
and the arc will intersect
table,
A < a,
as stated in (18-4) in the
AX in two points marked B and B'.
Therefore, two solutions exist. The angle B' in the triangle
the supplement of the angle B in the triangle ABC.
AB'C
is
Figure 18-3 (d) represents the case in which the side a is longer
than the side 6, as stated in (18-5) in the table. Hence, there is
only one point B in which the arc with radius a intersects the initial
ray AX of the angle A. There is only one solution.
Sec.
1
8-4
293
Solution of the General Triangle
In Fig. 18-3(e), the angle A is obtuse. Since a < b, as stated in
(18-6), the radius a is too short to intersect the initial ray AX,
and no triangle
exists.
A is obtuse and a > 6, as stated in
can
Here
arc
the
intersect
the initial ray AX in only one
(18-7).
point. There is, in this case, only one triangle.
The following examples will illustrate some of the possibilities.
Finally, in Fig. 18-3 (/)
Example 18-2. In a
triangle
,
ABC, A
=
=
3615', a
=
9.8, b
12.4.
Solve the
triangle.
Solution:
Draw
Fig. 18-4 approximately to scale,
showing the given parts. After
A
and side 6 have been drawn, an arc is described with C as center and a as
angle
radius. The arc intersects the side AX in two points, B\ and B 2 and we apparently
have two possible triangles, ABiC and AB 2 C.
,
FIG. 18-4.
To
find sin B,
we use
sin
B
__ sin
A
~T~ ~~~T~
Then
,
10
12.4
~
.
.
.
sm 36
9.8
and
=
The
log 12.4
log sin B
follows:
+ log sin
logarithmic work
log 12.4
log sin 3615'
=
=
3615'
=
log 9.8
1.0934
9.7718
10.8652
log 9.8
-
-
10
10
0.9912
B = 9.8740 - 10
B = 4826'.
A < a < b. The left inequality follows from
6
sin
since
log sin
There are two solutions,
the fact that log sin B
= log sm
<
0,
and so
sm
,
<
1.
If
we
let
BI
= 4826',
294
then
Solution of the General Triangle
- 4826' =
B = 4826',
B2 =
180
l
13134' leads to another
solution.
Thus,
=
180
-
(3615'
+ 4826') =
C2 =
180
-
(3615'
+
Ci
18-4
Sec.
9519',
and
# 2 = ISl^',
To
we
find Ci,
=
3615'
sin
16.5.
we have
" 2 ""
1211'
3615'
9.8 sin
sin
As a
=
/
9519
9.8 sin
Similarly,
13134')
use the law of sines again. Thus,
=
3.5.
= b cos A -f a cos B\ may be used. Thus,
3615' + 9.8 cos 4826' = 12.4 (0.8064) + 9.8 (0.6635) = 16.5.
the same as the value previously calculated. The same method may
partial check, the equation Ci
12.4 cos
This result
is
be applied to check
c2
.
Example 18-3. Given
A = 5630
,
=
a
13.0, b
=
ABC.
10.7, solve the triangle
Solution: Here we clearly have only one solution, since a > b. This can be seen
geometrically if we draw Fig. 18-5 approximately to scale and show the given parts.
Since a is greater than b, an arc with C as center intersects
on opposite sides of
AX
A. Obviously there
is
only one triangle,
ABiC, containing the angle A.
To find BI, we use the relationship
sin
a-
13.0
BI
sin
5630'
13
10.7
whence
=
log sin
log 10.7
-f log sin
5630'
-
log 13.
This gives
B =
FIG. 18-5.
=
Then C
To
find
c,
l
(5630' + 4320') = 8010'.
we use the law of sines and obtain
180
13
c
sin
This gives
c
4320'.
-
=
8010'
sin
5630'
15.4.
Example 18-4.
C
Given
A =
6740', a
Solution:
=
16.0,
b
The given parts
we have
=
17.3,
solve the triangle.
shown
are
in Fig. 18-6.
By
the law of sines,
sinff
17.3
a m 16.0
B =
Hence,
B
FIG. 18-6.
/
16
log 17.3 -f log sin
=
c
6740
Therefore,
log sin
6740'
sin
__
~~
The
6740' -
(1.2380) -f (9.9661
-
10)
log 16
-
(1.2041)
=
0.
B = 90.
be completed by applying the theory
of right triangles. Only one solution exists.
solution
may
18-4
Sec.
Example 18-5. Given
From
Solution:
work
To
A = 4723
it
Fig. 18-7,
;
a
,
= 230,
6
= 720,
appears that no triangle
solve the triangle.
is
possible.
The
following
verifies this fact.
find
B
t
use the relationship
sin
btaining
B
__ sin
A
*720sinV23',
.
Sm
The
295
Solution of the General Triangle
230
logarithmic work follows:
log 720
log sin 4723'
=
=
2.8573
9.8668
-
12.7241
log 230
=
10
10
2.3617
B = 10.3624 - 10
= 0.3624.
sin B would have to be
FlG 18~7
*
log sin
Since log sin B > 0,
Therefore, there is no solution.
'
-
greater than
1,
which
is
impossible.
EXERCISE 18-1
In each of the problems from 1 to 12, solve the given triangle by the law of sines.
a = 12.30, A = 3625', B = 4437'.
1.
2. b
3. c
4. b
5.
a
6. c
7.
o
8. 6
9.
a
10.
a
11. 6
12. 6
13.
= 12.18, A = 4733', B = 6751'.
= 461.3, B = 6719', C = 2314'.
= 0.6384, B = 3939', C = 8716'.
= 6.714, A = 3753', C = 13636'.
= 7832, A = Way, B = 4358'.
= 21.23, c = 64.21, C = 6231
= 0.8146, c = 31.63, B = 1119'.
= 987.4, b = 503.6, A = 5413
= 0.003862, c = 0.0008157, A = 2613'.
= 1.386, c = 2.451, 5 = 83 19'.
= 4.395, c = 9.806, C = 3746'.
/
.
/
.
A surveyor wishes to find the distance across a stream from point A
He
to point B.
from A to a point C on the same side of the stream is
and angles BAG and BCA are 4953' and 816', respectively. Find
finds that the distance
687.4 feet,
the distance
14.
A surveyor
edge of the
ran a line
AB.
was running a
swamp
A
line
N 2723' W. How
original line produced?
15.
due west when he reached a swamp. From the
for 2500 feet, and from this point he
S 63
far had he gone on this line when he reached his
far was it across the swamp?
line
he ran a
How
W
building 63.7 feet high stands on the top of a hill. From a point at the foot
of the hill the angles of elevation to the top and bottom of the building are
4216' and 3831', respectively. Find the height of the hill.
296
16.
Solution of the General Triangle
Sec.
18-4
From a certain point on the ground the angle of elevation to the top of a building
46 17'. From a point on the ground 83 feet nearer the building the angle of
is
17.
6823'. Assuming that the ground
elevation
is
building.
One side
and a diagonal
respectively.
is level,
find the height of the
of a parallelogram are 14.63 inches
and 21.4
The angle between the diagonals and opposite the given
inches,
side is
11623'. Find the length of the other diagonal.
measure the distance between two artillery pieces A and B.
from an observation point C to gun A is 2447'. Sound
travels at the rate of 1140 feet per second, and the sounds from guns A and B
reach C in 2.3 and 1.7 seconds, respectively. Find the distance AB, assuming
that points A, B, and C lie in the same plane.
A body is acted on by two forces, Fi = 2643 pounds and F 2 = 2341 pounds.
The resultant F 3 lies on a line making an angle of 4633' with Fi. Find F 3 and
18. It is necessary to
The angle
19.
of depression
the angle between the lines of action of Fi and
forces is their vector sum.)
20.
A buoy,
F2
(The resultant of two
.
located at a point B, is 6 miles from a point A at one end of an island
is
C at the other end of the island. If the angle
BAG
and 10 miles from a point
132
18-5.
16', find
THE
Law
the distance between the points
A
and
C
on the
island.
LAW OF COSINES
ABC be any triangle. Then
a = b 2 + c 2 - 2bc cos A,
b 2 = a 2 + c 2 - 2ac cos
2
2
2
c = a + b - 2ab cos C.
of Cosines* Let
2
(18-8)
(18-9)
,
(18-10)
These relationships constitute the law of cosines.
FIG. 18-8.
Proof.
We shall establish
(18-8) by considering the case when A
and the case when A is obtuse as in
= 90 involves no difficulty, and it will
acute, as in Fig. 18-8 (a)
is
Fig. 18-8 (&). The case A
therefore be omitted.
Let h denote the altitude from
denote the length AD. Hence,
DB
c
+ x in Fig.
18-8(6).
C
to the side
is c
-
AB.
Also, let x
x in Fig. 18-8 (a) and
is
Sec.
1
8-6
Solution of fhe General Tr/ongfe
297
In Fig. 18-8 (a),
(c
- xY +
h2
=
+
=
62
a 2,
and
x2
Subtracting,
we have
c
2
Since x
=
6 cos
=a
= 62 +
2c#
a2
A,
a2
h2
2
6
c
2
-
2
=
62
+ c2 -
(c
+
x)
.
or
,
2cz.
26c cos A.
In Fig. 18-8(6),
+ h2 =
2
a2,
and
+
x2
Subtracting,
we
find that c 2
+
2cx
a
Since x
=
b cos
A
2
in this case,
a2 = 62
=
=
+
=
h2
a2
6
62
.
6 2 or
,
+c +
2
2
c2
2bc cos A.
and proceeding in a
obtain (18-9) and (18-10).
Note. For a simple algebraic proof of the law of cosines, see problem 22 in Exercise 18-4. The law of cosines applies equally well if
ABC is a right triangle. In this case, one of the formulas reduces
By drawing
similar manner,
altitudes to the other sides
we
=
to the pythogorean theorem, since cos 90
18-6.
SOLUTION OF CASE
III
AND CASE
Since the law of cosines
IV
BY THE
0.
LAW OF COSINES
expressed by formulas involving addition and subtraction, it is not well adapted to logarithmic computation and its use is not recommended unless the given sides are
is
easily squared.
Example 18-6. Given
Solution:
By
b
=
9.0, c
=
13.0,
A =
11510', solve the triangle
the law of cosines,
a2
52 _|_ C 2
_ 2bc cos A
=
= 9 + (13) - 2(9) (13) cos 11510'
= 81 + 169 - 234( - cos 6450')
= 250 + 234 (0.4253) = 349.52.
2
2
Hence,
a
We
=
18.7.
employ the law of sines to find angle B. Thus,
sin
_
""
9
Therefore,
C=
sin
11510'
_
~~
sin
18.7
5=25'50'.
/
180
(11510' + 2550 ) = 39.
6450
18.7
/
ABC.
298
Solution of the General Triangle
Example 18-7, Given a
Solution:
From a
2
=
62
=
+c
2
A =
8-6
Find the angles.
7.
-+
52
A =
Similarly,
=
1
we have the formula
C2
_
a2
-T
>
Hence,
We
5, c
26c cos 4,
cos
Therefore
=
3, 6
Sec.
B =
3813' and C
=
21*47'.
120.
can check these by the equation
A
-f
B +C =
180.
EXERCISE 18-2
In each of the problems from
to
1
solve the given triangle
6,
by the law
of cosines.
l.o= 300, 6 = 250, C - 5840'.
2. a = 50, c = 240, 5 = 11050'.
3. 6 = 65, c = 310, A = 6710
4. o = 130, c = 90, B = 10020'.
5. 6 = 50, c = 110, A = 150.
6. a = 1.63, 6 = 3.45, C = 2610'.
=
7. If a = 15, 6
12, c =5 20, find A.
8. If o = 25, b = 30, and c = 35, find B.
9. If a = 100, 6 = 300, c = 500, find C.
10. If a = 15, 6 = 12, and c = 20, find B.
11. If o = 16, 6 = 17, and c = 18, find A, B, C.
12. If o = 260, 6 = 322, c = 481, find A, B, C.
13. The distance between two points A and B cannot be measured directly.
Accordingly, a third point C is selected, and it is found that AC = 3000 feet,
BC = 4500 feet, and angle ACB = 4620'. Find the distance AB.
/
.
14.
15.
Two sides of a parallelogram are
125 feet and 200 feet, and the included angle is
11030'. Find the length of the longer diagonal of the parallelogram and also
the angle between that diagonal and a longer side of the parallelogram.
Two
16.
Two
is
feet
and 200
feet,
and the
6733'. Find the perimeter of the plot.
and 420 feet, and one diagonal is
Find the length of the other diagonal.
In a triangle ABC, a = 25, 6 = 27, and the median from A is 20. Find c, A, B, C.
600
17.
ground are 250
sides of a triangular plot of
included angle
18-7.
sides of a parallelogram are 700 feet
feet.
THE
Law
LAW OF TANGENTS
ABC
of Tangents. Let
relationships exist between
Ma
in
(18-11)
a
a
~
+
,
b
b
be any triangle. Then the following
sides and the angles opposite them
two
=
:
tan i (A
2
-
B)
=
tan
>
(A
+ B)
Sec.
18-7
299
Solution of the General Triangle
tan
(B
-
C)
(B
+
C)
(18-12)
b
+c
C
+
^
2
=
5_IL
(18-13)
tan
a
1
x
tan s /^Y
(C
X
+ A)
\
,
These relationships constitute the law of tangents.
Proof. Let us denote the common ratio of the law of sines by
Thus, a = r sin A, 6 = r sin JS, c = r sin C. Then
a
a
b
+
_
~~
A
A +
r sin
6
r sin
r sin J? __ sin
~"
B
r sin
sin
A
A +
sin
sin
B
B
^
Substituting from (7-27) and (7-28) of Section 7-4,
D
sm g
.
sm AA
sin
2 cos
.
A+
sin
B
2
+ B)
(A
y
sin
2
(A
-
r.
we have
J5)'
~~
_.
+ JS) cos z- (A -
1 ,
2 sin - (A
A
.
,
1
,
,
.
Hence,
tan
A
similar procedure
may
(A
-
be followed to prove
(18-12)
and
(18-13).
We
two
shall
sides
now
use the law of tangents to solve a triangle in which
and the included angle are given. Note that
this
law
is
well adapted to logarithmic computation.
Example 18-8. Given
Solution:
Here c
To
-6=
find
123, c
C -
6
=
=
249, c
A = 5622
372,
;
,
solve the triangle
B, we use the formula
+ b = 621, C + B =
180
-A =
12338',
(C
+ B) =
Therefore,
tan i (C
-
B)
=
g tan 6149',
and
1
log tan
i (C
-
B)
= log
123
+ log tan 6149' -
log 621.
ABC.
300
Solution of
The
logarithmic
work
Genera/ Triangle
f/ia
Sec.
1
8-7
follows:
=
=
log 123
log tan 6149'
2.0899
0.2710
-
10
-
10
12.3609
-
B)
=
=
-
B)
= 2017'.
log 621
log tan
i (C
2.7931
9.5678
i
1 (C
Hence,
1
C=i(C+)+i(C-)=826',
and
B = ~ (C +
)
-
~ (C
-
B)
= 4132'.
Cfodk:
4
To
find a,
we
4.
5 +C=
5622'
+ 826 + 4132 =
/
180.
use the law of sines. Thus,
249 sin 6622'
sin
18-8.
7
4132'
Q1Q
=313
'
THE HALF-ANGLE FORMULAS
The following relationships are very convenient for the logarithmic solution of Case IV, where the three sides a, 6, and c are
known
:
2
In these formulas, 5
Proof.
From
=
r
- (a
2
+
b
(*
+
6)
c)
c).
(7-16) in Section 7-3,
we have
A ~ l-cosA
1 + cos A
2
Also,
from the law of
cosines,
cos
Therefore,
_i
-l
and
&2
A =
62
+ c2 -
2 + c2 - a 2 _
-a
25^
(6
-a2
- c) 2 _
~
2bc
(a
+ b - c) (a - b + c)
2bc
Sec.
18-8
we
If
Solution of the General Triangle
let
a
301
b + c = 2s, then
a
b - c = a
6
a - 6
c = a
6
+
+ + c - 2c = 2(a - c),
+
+ + c - 26 = 2(s - 6),
=
6 + c
a
a + 6 + c - 2a = 2(s - a).
.A _ (a + & - c) (a - 6 + c)
_
2
" (g c)2
(6 + c + a) (b + c
t(
a)
a)
+
Therefore,
(
6)
and
(18-14)
v
'
We
r
2
s(s
a)
can derive (18-15) and (18-16) in a similar manner.
= 379, b = 227, c - 416, find the angles of the triangle.
Here 2s=a+b+c = 1022. Then
Example 18-9. Given: a
Solution:
=511,
8
- a = 132,
- 6 = 284,
- c = 95.
s
s
s
Hence,
A
.
The
calculations
by logarithms
log 284
log
95
log (284) (95)
log (511) (132)
log tan
.
.,
,
Similarly,
/(284) (95)
follow:
= 2.4533
= 1.9777
= 24.4310 = 4.8290
y ==
A
A
-
log 511
log 132
20
19.6020
-
20
9.8010
-
10
6438
4.8290
;
.
.
B _ ^7 (132)
2--r
w"
Hence, we
= 2.7084
= 2.1206
find that
B =
C ""
_
(95)
(511) (284)'
|/(132)(284)
y
2
3246' and C
V
(511) (95)
= 8236
/
.
A + 5 + C = 6438' + 3246' + 8236' =
180.
EXERCISE 18-3
In each of the problems from 1 to 10, solve the given triangle by the law of
if an angle is given, or by the half-angle formulas if three sides are given
/
2. 6 = 17.1, c = 22.3, A = 2116 .
a = 50, b = 60, C = 60.
/
=
=
=
=
=
=
B
6
c
A
4.
a
9510 .
79.3,
113,
230, c
106,
tangents
1.
3.
Solution of the General Triangle
302
5. 6
7.
a
9.
a
11.
= 41.82, c = 75.89, A = 7849'.
= 625, 6 = 725, c = 825.
= 67450, 6 = 84380, c = 98630.
6.
a
8.
a
10.
a
Sec.
1
8-8
= 0.1028, 6 = 0.8726, C = 14S13'.
= 60.65, 6 = 38.64, c = 23.57.
= 0.1146, 6 = 0.3184, c = 0.6379.
diagonals of a parallelogram are 6 inches and 10 inches,
at an angle of 63. Find the sides of the parallelogram.
The
and they
intersect
A and B are separated by an obstacle. In order to find the distance
between them, a third point C is selected and it is found that AC = 126 rods
and BC = 185 rods. The angle subtended at C by AB is 9614'. Find AB.
Two circles whose radii are 14 and 17 inches respectively intersect. The angle
12. Points
13.
14.
15.
between the tangents to the circles at either point of intersection is 3846'.
Find the distance between the centers of the circles.
The sides of a parallelogram are 13.4 inches and 18.5 inches, and one diagonal
is 15.6 inches. Find the angles and the other diagonal of the parallelogram.
Three circles whose radii are 10, 11, and 12 inches, respectively, are tangent to
each other externally. Find the angles of the triangle formed by joining their
centers.
16.
The sides of a triangular field are in the proportion 4:5:6. The area of the field is
18 acres. If there are 160 square rods in an acre, find the length of each side
of the field in rods.
17.
In triangle
ABC, prove
the following:
a
sin
-6
^-*)
n
cos-C
I
a
+b
-s
C
1
sm-C
.
These formulas are called Mollweide's equations. They
may
be used in checking
the solution of a triangle.
18-9.
We
AREA OF A TRIANGLE
can readily see that the area
K
(18-17)
In either triangle, h
=
=
\ch
&
K of the triangle
= JcbsinA.
<u
6 sin A. In like
K
(18-18)
=
in Fig. 18-1 is
manner, we obtain
lac sin B,
Zi
K
(18-19)
c
By
substituting
sm
sm C
=
^ab sin C.
for 6
from the law of
transform (18-17) to obtain
~o(\\
n
(18 20)
*
A sin B
KK -~ c sin
2sinC
sines,
we may
Sec.
1
8-9
303
Solution of the General Triangle
cyclic interchanges of letters, we obtain
= a2 sin B sin C
(18-21)
By
K
K=
(18-22)
A
2 sin
A
b 2 sin
C
sin
2smB
To derive a formula for finding the area of a triangle when its
three sides are given, we first transform (18-17) in the following
manner
'
:
K
=
ybc
sin
A =
sin
^bc
= 1,/ sm A
(2
A
.
y cos
rpc
=
,
A
.
oc
sm
-jrZi
yJ
(2
2
A
cos
-^-
But, from (7-14) and (7-15) in Section 7-3,
=
A
cos
/I
v
A
,
and
2
cos
2"
Using the values from Section 18-8 for
have
.
sm
A 2
A
/(s
^
be
c) (s
A/
V
A =
6)'-
cos
1
^
A
A -
,
and
/I
y
cos -^
2
2
and
1
-f
.
/s(s
--r
\/
V
be
Consequently, the formula for area in terms of the sides
= \/ s
(s
The following examples
a) (s
=
372,
A =
a)-
is
c).
triangle in
Example
18-8, in which b ='249,
5622'.
Solution: Since
two
sides
and the included angle are given, (18-17) may be used*
Thus,
K = ^bc
sin A = ^(249)
Z
2
The
we
illustrate the use of the area formulas.
Example 18-10. Find the area of the
c
t) (s
cos A,
logarithmic work follows
(372) sin
5622 /
:
= 9.6990 log 249 = 2.3962
log 372 = 2.5705
sin
5622' = 9.9205 log
log K = 24.5862
=
K 38,600.
log 0.5
10
10
20
.
304
Solution of the General Triangle
Example 18-11. Find the area
of the triangle in
Sec.
1
8-9
18-1, in which
Example
A
= 3814', B = 6720', c = 329.
Solution: Since
two angles and a side are given, (18-20) may be used. In
sin B _ (329)* sin 3814' sin 6720' _ Q0 inn
"" * 2 100
~
2 sin 7426'
2 sinC
_ c 2 sin A
*^ -
= 227,
c
in
which a
= 379,
= 416.
Solution: In this solution, (18-21)
K=
'
'
Example 18-12. Find the area of the triangle in Example 18-9,
6
this case,
Vs(s -
a) (s
-
6)
(
-
is
c)
We
used.
=
have
V(511)
(132) (284) (95)
= 42,700.
EXERCISE 18-4
1.
3.
5.
In each of the problems from 1 to 8, find the area of the given triangle.
/
a = 12.30, A = 3625', B = 4437'.
2. c = 461.3, B = 6719', C = 2314
=
=
=
=
=
=
3746'.
4. 6
a
5413'.
9.806, C
4.395, c
503.6, A
987.4, b
6 = 65, c = 310, A = 6710'.
6. a = 300, 6 = 250, C = 5840'.
=
=
7.
a
9.
In triangle
15, b
Au
j
=
12, c
ABC,
r
20.
let r
j.1.
x
and, therefore, that r
=
/4
/-($
/I/
~
a)
r
10.
Find the radius of the
feet,
=
a
8.
100, 6
--
be the radius of the inscribed
6) (S
(
-
=
=
300, c
500.
Prove that
circle.
.
K = rs
C)-
circle inscribed in the triangle
whose
sides are 48.92
63.86 feet, and 72.31 feet.
11.
A cylindrical tank is to be built on a triangular lot having sides whose lengths
are 200 feet, 186 feet, and 176 feet. Find the radius of the largest such tank
which can be built on the lot.
12.
In triangle ARC,
let
R
be the radius of the circumscribed
a
b
2B= sin A
13. In triangle
ABC, show
scribed circle and
that
=
sin
R = -r^
>
B
=
circle.
Show that
*
sin
where
R
C
is
the radius of the circum-
K is the area of the triangle.
15.
The sides of a triangle are 23, 29, and 46 feet. Find the areas of the triangle
and the inscribed and circumscribed circles.
The sides of a triangular plot of grass are 42 feet, 65 feet, and 87 feet. Find the
minimum radius of action of an automatic lawn sprinkler which will water all
parts of the plot from the same point.
16.
An
14.
arc of a circle of radius r subtends a central angle
bounded by
this arc
and
its
chord
is jr
r2
(6
-
0.
Show
that the area
sin 8).
t
Find the area of the largest pentagon which can be cut from a circular piece of
metal 4 feet in radius. How much metal is wasted?
18. In triangle ABC, prove that the median from any vertex to the side opposite
17.
divides the angle at that vertex into two parts whose sines are proportional to
the lengths of the parts into which the side opposite is divided by the median.
Sec.
1
8-9
19. In triangle
ABC, prove
A
cos
a
20. In triangle
21. In triangle
2
-f 6
(6
ABC, prove
+ c2 =
a 2 (cos 2
+
B
+
6
cos
C ""
_ a2
-f
fr
2
-f c
2
1
2a6c
c
+ c)
cos
A +
(c
+ a)
cos
B +
(a
+ b)
cos C.
2
2
that
C +
ABC, show
22. In triangle
that
cos
that
ABC, prove
+b +c =
a
a2
305
Solution of the General Triangle
sin 2
B)
+
6 2 (cos 2
4
2
-f sin
C)
+c
(cos
B + sin 2 A).
that
a
6
c
= b cos (7 -f c cos 5,
= c cos A -f a cos C,
= a cos B + 6 cos A.
Multiply the first equation by a, the second by 6, and the third by c, to give a
second proof of the law of cosines; that is, prove (18-8) by showing that
62
+c 2
23. Consider
a
>
=
a2
any
2bc cos A. Similarly prove (18-9) and (18-10).
triangle ABC. If a > b, prove that A > B. If A > B, prove that
b.
24. In triangles
ABC
and A'B'C',
let
A
and
A',
B
and
B',
C and
C" be pairs of
corresponding vertices, and let the corresponding sides be a and
;
c and c'. If a = a
6 = 6', and C > C', prove that c > c'. If a
,
and
c
>
c',
prove that
C >
C'.
a',
b
= a',
and
6
b',
= &',
Appendix
A
Tables
Appendix
TABLE
A
I
FOUR-PLACE VALUES OF FUNCTIONS OF NUMBERS
30*
310
Appendix
TABLE
A
I (continued)
Appendix
TABLE
A
I (continued)
311
Appendix A
TABLE
I (continued)
Appendix A
TABLE
II
FOUR-PLACE VALUES OP FUNCTIONS
313
314
Appendix
TABLE
A
II (continued)
Appendix
TABLE
A
II (continued)
315
316
Appendix
TABLE
A
II (continued)
Appendix
TABLE
A
II (continued)
317
318
Appendix A
TABLE
II (continued)
Appencf/x
TABLE
A
II (continued)
319
320
Appendix A
TABLE
II (continued)
Appendix A
TABLE
III
FOUR-PLACE LOGARITHMS OF NUMBERS
32V
322
Appendix A
TABLE
III (continued)
Appendix A
TABLE
III (continued)
323
324
Appendix A
TABLE
IV
FOUR-PLACE LOGARITHMS OF FUNCTIONS
Appendix
TABLE
IV
A
(continued)
325
326
Appendix
TABLE
IV
A
(continued)
Appendix A
TABLE IV
(continued)
327
328
Appendix A
TABLE IV
(continued)
Appendix
TABLE IV
A
(continued)
329
330
Appendix
TABLE IV
A
(continued)
Appendix
TABLE
IV
A
(continued)
331
332
Appendix A
TABLE V
SQUARES AND SQUARE ROOTS
Appendix
TABLE V
A
(continued)
333
334
Appendix
TABLE V
A
(continued)
Appendix A
TABLE V
(continued)
335
336
Appendix
TABLE V
A
(continued)
Answers
fo
Odd-Numbered
Problems
PAGE
EXERCISE 1-1.
1.
(a)
Commutative law
11
of addition.
(c) Associative law of multiplication.
(e) Commutative law of multiplication.
2.
(a) 2.
(c) 1.
-35.
3. (a)
-6.
(c)
4. (a)
-5.
(c) 0.
5.
(a)l.
(g) -5.
1.
(e)
-10.
(e)
(c)5/17.
(g) 2.
(e) 1/1.02.
2a.
1.
(a) -3, -2, 0, 4, 5.
>
3
PAGE
1/3.
3. (a) False,
(c) 0.
4. (a) 0.
5. (a)
6. (a)
7.
1.414.
(e) False.
(e) 2.
(g)
^6.
7
-1.
3.
x
a.
<
7.
5.1,000,000.
57.
59.
69.
IT.
-2 < x <
(e)
< V3 <
(e) 1
1.
4.
2.
PAGE 20
-216.
9.2401.
21. a 7 .
~4
23. a 4 6 4
11.
13.
II.
19.
a".
27.
33.
35.
a
4
2
.
2
2
2
2
2
2
).
),
2
2
3
3
2
2
2
).
14
7
a:
3
3. x*y*.
+
-2a
3^
2
+ 26
+
35. 2x 2
43. 2z?/.
.?
3
63. x 3
-
a-?/
5.
3.r
-
+
3?y
45. -1.
66 2
PAGE 24
13. -4.r//.
21.
z2
2.
3
2
?/
4.T
47. 2.
+ z y + xy* +
+ 3x 2 + 3a: +
3
2
2/
1
;
2a;?/
29. 24a
.
37.
-
.
x2
4
-
-4a.
7.
.
2.
27. -lOx-
-
-
5a 2
2c.
2
2
),
2xy.
2
),
?/
25. -18x?/.
55.
25. a 4 6.
.
37. 5 - 6.
41. 3a - 46.
39. 46.
^a
-2a - 36. 45. (a - 2c)x* - (a + 2b)xy + (6 - I)?/
47. 4a - 56 - 2c.
a - (6 - c
a + (6 + c), a - (-6 - c). 51. a + (c - 6
z
z - (y + z ).
2a + (6
55. x + ((3c
6).
3c), 2a
-a 6 + (a& + 6
-a 6
6
(- ab
61. 12.
63. 20.
65. 17. 67. 1.
2.r + (- 3y - 4z), 2x - (3y + 4s).
-6. 71. 8. 73. 10. 75. 13. 77. 13. 79. |- 81. ~31. a 4 .
EXERCISE 1-4.
I.
x/3, 3.
lu
19. 125a.
17. 32.
29. a 2 6 2 *.
53.
1,
10.
15.
49.
0.1.
-2, -3/2,
(e)
01
43.
-
(k) r
(i) 0.
EXERCISE 1-3.
1.32.
x.
15
10.
>
22/7
(e)
True,
-1 < x < 1. (c) -a
4 < 5 < 6. (c) -1 <
+
a
V2 >
(c)
(c)
-
(k) 3
x+y.
-4, -2, -1, -1/3,
(c)
-a6.
(i)
(g)-||-
EXERCISE 1-2.
2. (a)
29.
(i)
-
(g) 36
-2/3.
(e)
(k) -7.
-2.
(i)
5x
15.
/>
Sx*y
2
4.r
49. -z?/ -h 2x
57. x 3
-
+ 2x +
.r?/
1.
2
//
+ 2xy.
2
^x
2
?/
+y
65.
339
59.
,
2.r
4
^
-
-
+ 26 2
x 2 + ~ o^ +
y y.
.
4jy.
-
33.
2xy* -f
.r
9a:
2
3
i/
-
41.
-2*.
+ 5. 53. + y.
+ xy + y 2 61. 2y - xy
+ 6. 67. Odd values.
51. x
3y.
2
-
3
39. x*
.
-
3a 2
17.
.
31. -Go;*
c.
-
2
23.
-f y*.
4
z2
9.
.-c
.
340
Appendix B
PAGE 30
EXERCISE 1-5.
1.
+ 2y). 3. 2(2x +
- 2?/ + 3z). 11.
3(x
9. a(x
21.
a2
2
?/
-
15. (2x
z
2
(z
-
25. (7xy
33. (5p
3
2
2
*/
2x
-
2
(*
)
2/
w)
4
+
.r
4
4- * ).
2
4- 3z) 4- (2y 4- 3z)
-
(z
Z/)
w)
.r
2/
].
-
-f (z
.
)
?^)
2
].
.
(.r?/
.
3.
-
x
5.
1.
-
19- x 2
SOxtyV.
3x 3 ?/.
7.
y.
f.r
-
(a) 12.
(c)
(a)
.
4a.
(C)
347'
x2 -
(y
(e) l&r//
3a:
.
3)
1.
(e)
2
,
_
(g)
i^T5*
,
^
94
3
24"
x2
/.
(x
+
+x +4
-I
\
/
.
1) (x
-
\
I)
o
2
(*+y)(-y)
>-
J.W.
*"'
/
.
+
__
(x
21
2(a
/
2)
(a;
- ab
2
+
rtx
+
o\
3) (x
2
4- 6 )
a
n
-5.a
x
(x
-
1) (x
7) (*
+ 2)
-
2)
'
1& "
(9x
2
3>c
_
x
6
7a
+
64
M.
1
2
-
(2.t
2
'
^2
*
- xy
x2
4-
-y
2
41
2(x
-
5)
'
4
2
,
2j?4-?/4-3
'
"'
+
5
-
I
4)
-
3x
..
'
+ (to + 4)
+
?
1+*
^\
i
PAGE
3
x2
,,
^?y
n
9x
4-2
W 2*+8*
*
27
(g
x).
,,
x3-l
q
**
-6 2
EXERCISE 1-9.
54
/_.
"
2
4).
^r+,-2^
+5
12f
.
_
3a+b,
(s)^ 6
9-
'x^T-
+
x) (b
PAGE 38
_ x' - 3.r 2 -
-2
_
7
--20'
3* 2
t
- x(x + 5y)
iq
57
_
5
--2l'
-
(.r
- 7
x4-2*
W
_
,
(q)i.
EXERCISE 1-8.
,17
L
15. 24,r?/z.
2
8)
3>c
...
^i'
_
.
_
(0)^-3,.
WT^&rr^r-
-
(a
(i)
1
().r?y
,
-
3) (x*
PAGE 35
-
3
?/
13. 24.
11. 1.
-
49) (x
.r
(g)
.
2
~IT'
-
3
1).
PAGE 33
9. 2xy*z.
21. (x 2
4.
.
)
.
101
.
a;
).
EXERCISE 1-7.
***
- 2 ).
- ay* 2
(6
)
- 9) 2
- y) (36a: 2 4- fay 4- 2 ). 43. 3(6 - 2 2 45.
2
2
_
2
_
_
_
x
r
X
49. (4.
51.
-f 1).
3) (2
2?/)
(X
3).
5) ( X
3 2
2
2
57. (1 4- 15.r
55.
4- 1) (* 4- 2).
4) (X + 3).
(x
63. (2x - 1) (x + 3\
61.
+
1)
6).
2).
(x
(2x
5x(x 4- 4) (x
2
69. (a 4- 3)
67. (x
4- 2y).
0.6)
6).
(6.r
1) (x
2 - 3). 73. (x 4- 3) (2a + y - z). 75. (x 2 - 3) (x +
(4a;
5) (2x
1. 2.
2'
a) (3y 4- a).
?/ ).
EXERCISE 1-6.
1.
2
(a-
59.
17.
-
8
4- (x 4-
2/)
4) (x
19. (3y
11).
(
53.
65.
4-
+ x\
+ 4).
2c
6) (0.1 4- 6).
r 10
3x 2 (2y
2
41. (te
71.
ry
4-
+
r6
3
*V +
[9.r
[(x
-
-
(.r
+
11) (7*
27. j(0
2
5p
44
(2y 4- 3z)]
4-
-
fl
13.
-
-x(a
31. (jy 4- z 2 ) (* 2 ?/ 2
4- 4).
(25p g
)
39. (x 4- y
47.
-
7.
2).
a).
23. (0.1
3a).
4
r6
(x
)
-
37. 3[3s 2
-
(x*
4-
<?
35. (3 4-
(5
I2ab) (7xy 4- 12a6).
+ 2)
29. 2(x
+
3a) (x
+
a(3x
+ 3?/ -
17. (7x
3) (2* 4- 3).
-
5.
7).
5)
(4x
(fa
+
+
1) (x
1) (*
-
6)
'
5)
y
2
341
Appendix B
-
+ 5)
3(4s
+-3)
2L
(2s
28
V
5)
23
70
37.2i.
.i|.
35.1^.
x + 5
5z zy
33
2
tK
15.
x
=
23. y
17.
~
=
y
'
,'
4
39.
^jp
5.
tft
19-
39.2.
PAGE 47
=
x
y
-2.
41. 4.
(e)
V2.
(c)
=x 1
~y~~
27.7.
V34.
The area
of a circle
3.
The area
of a trapezoid
5.
The volume
is
(g) 4.
II.
~
oZ
S =
-2, 2
3,
288, Cy
,
23. 5/7.
29.
2
49. 40, 58.
35.
~
4
51. 6, 7.
20, 40, 120.
57.
PAGE 52
(i)
V^T^.
(k) 2.
PAGE 56
a function of
its
a function of
is
altitude
its
A = Trr 2
bases, A =-(bi
circle,
and
.
height and the radius of
life
insurance policy
is
+62).
its
base,
a function of the applicant's age
of the type of policy, of the
27.
25. 0.
^/3(57r
F2
.
43. all x.
53. 3
-2.
5* 0,
V2
-
-3, -3, 3/4, 2y
+ 7.
41. all x.
A =
HT*,
45. s
*
C =
2xr,
0, 1.
57.
a:
3,2.3,
|
x
|
.
3. 0,0.5.
.5.
3,
17.
|
p
2
A =
33. all x.
31. all x.
55. all x.
15. 14.
13. 0.
company's rate
policy, etc.
-1.
1,0,1,0,1.
3,^3'
+ q* + r 3
,
C =
3.
49.
\
59.
.
19. 0.
x
-2 ^ x g
PAGE 57
21. 30.
2
37. all x.
35. all x.
47. \x
EXERCISE 2-3.
1.
33.-^J
formula can be written.
-3, -5,
-
is
of a
and physical condition,
9.
31.3.
irr*h.
The annual premium
No
-3
60;
2
a function of the radius of the
of a cylinder
13.10.
11.7.
(g) \/29.
(e) 7.
I.
7.
=
47. --
45. 2.
43.
-11.
9.
01
21. y
'
29.1/4.
EXERCISE 2-2.
V =
-
15
55. 170 adults, 330 children.
ft.
(c) 4.
3. (a) 5.
^2
7.
EXERCISE 2-1.
2. (a) 5.
41.
4-^8
-5.
25.
~
53. 921,600 sq
-3/7.
^
f 17
2g
007
37.
=
-= 4 -23s
-.
-2.
x
3.
-5/2.
32/
'5T
1
EXERCISE 1-10.
1.
25
'59*
= 0.
2.
39. all a.
51. all s.
342
Appendix B
PAGE 59
EXERCISE 2-4.
1.
=
y
y
3. 567.
x.
15. 1000/1, 100/1.
7.
5. 5.
327C.
17.
-2.
9.
10/3.
1.
(a) (1, 0).
3. (a)
PAGE 67
sin
7.
(e) (1, 0).
(c) (0, 1).
V5/2.
(c) (-0.99,0.14).
(c) 1.
-1/2.
(e)
cos
t
(e) (-0.65, -0.76).
-2.
(g)
tan
t
V3/2
(a)
I/
(i)
cot
t
V3
- V5/2.
sec
t
V%
esc
t
2/V3
t.
2.
(c)
5/13
12/13
5/12
12/5
13/5.
(e)
A/3/2
-
V
V V3
2/ V3.
(g)
1/V5
1/2
(i)
2/ \/5
1/2
4/5
3/4
V5/2
(c) 0.58.
1. (a) 1.
2. (a) 0.9927.
PAGE 74
(e) 0.86.
(e) -1.500.
(c) 24.52.
(g) 0.2571.
1.
3. 7T/6-
7T/3.
45.
27.
7.
5. 27T/3.
270.
29.
15.
31.
33.
35.
0.5925.
5. 0.7412.
3. 1.092.
15. 9.010.
25. 0.2930.
27. -9.462.
45. 5520', 23520'.
47. 427', 2227'.
61. 0.8016.
63. 3.079.
73. 0.220, 6.060.
81.
55. 730', 18730'.
1.143, 1.997.
75.
65.
-100.00.
1.120, 4.260.
43. 8310', 26310'.
7.81.
9.
;
51.
3135 32825'.
57. 9736', 26224'.
59.
9311' 273
67. 0.3459.
77.
~
0.755, 3.895.
7
13.1.
!!'.
71. 1.214.
79. 1.158, 5.122.
PAGE 96
11. ay".
,
69. -0.7073.
1
5.100.
23. 0.5154.
49. 189', 34151'.
EXERCISE 4-1.
3.^.
11. 0.2504.
83. 0.574, 5.706.
64
1. 3y.
-0.2462.
21. 0.9831.
37. 3520', 14440'.
41. 26450', 27510'.
.
39. 4343'.
31. 17310', 35310'.
29. -1.059.
39. 5630', 23630'.
r
9.
19. 0.8437.
35. 310', 18310'.
2474
37. 47023'54".
25. 1.6323.
PAGE 85
-1.453.
7.
17. 0.6817.
33. 4230', 22230'.
53. 674',
27.
13. 437T/36.
23. 3.2107.
(c) 0.04 radian.
(b) 16/9 radians,
EXERCISE 3-5.
13. 1.181.
11. 27T/5.
21. 1.4358.
630.
(k) 1.011.
5.798.
59. 14.74 in.
57. 147T/3, 1.25 radians.
61. (a) 4 radians,
9. 47T/3.
7T/15.
19. 0.8090.
17. 77T/20.
(i)
PAGE 80
EXERCISE 3-4.
15. 1077T/60.
V5.
-5/3.
5/4
4/3
EXERCISE 3-3.
1.
21. 0.0324 in.
19. 99.5 Ib, 95.2 Ib.
EXERCISE 3-1.
2. (a) (0.54,0.84).
13. 9/4, 27/8.
343
Appendix 8
21. 4(5"*).
33. (5292) ".
-
2
49. 2(2
-
x2) 3
'
5. 6,652,800.
15.
-
-
+
-f
-4+
7*8
^--4-+6
27. x 12
29. x
31.
2
z8
X4
?/2
-
+y + z
2
-f
4x
7.
15.
8=3.
Iog 256 2 =
10x6 -f 16z
-14x 2
80x~ 1/2
+
5
2-
27. 3.
=~
I
80ar 3
39. log* (u
41. (a) log*
-
+
16x 3
35
5
\/u
+
TT
-
- 0.5.
10
31. 5/2.
2
1.
3. 4.
19. 7.314.
5.
1.5441.
a
No
33.
-
1
=
9.
=
y
M-
1.
-8x7.
41. 924x 3 i/ 3 .
3.
Iog 10 x.
/
2
=
82
13.
21. 4 3
10,000.
=
64.
23. 6.
8.
35. Iog6
25. 2.
37.
T
fe
7 -f Iog6 g
-
2
log*
TT.
PAGE 107
13.
11. 5.
-1.
-1.
15.
17. 7314.
PAGE 110
7.
4.3636.
17. 7.8452-10.
15. 0.4536.
45. 0.06114.
47. 4.554.
57. 3.728.
9. 9.5490-10.
19. 9.8908-10.
39. 69.2.
31. 46.4.
41. 292.3.
49. 0.00001072.
59. 2.5023.
11. 8.8215-10.
21. 0.4972.
29. 9.1306-10.
27. 9.9476-10.
37. 0.0000000000276.
55. 1.585.
-
(b) 2 log
g.
-1.
7.7931-10.
5.
25. 9.9824-10.
53. 0.4266.
.
23. 0.007314.
13. 9.9279-10.
43. 5,454,000,000.
+ 4* +
13x 33 ?/ 8
solution.
1/2 logb
-3.
21. 7314000.
35. 504.
10x 2
4-
).
7.
3. 2.0212.
-f-
1000
11.
19. 10 4
23. 8.9439-10.
33. 0.262.
-
PAGE 104
EXERCISE 5-3.
1.
7x
2
1/2 Iog6
-1.
-f-
2
/
7
EXERCISE 5-2.
1.
l)n.
-
3
2x0.
5. logio
= 343.
17. 73
29. 1/10.
+
11. (n
-f
19x 4
= 4.
9. Iog
100
-
1).
^8
39. 2 11
.
81
3. logs
Iog 2
-
n(n
35a;
EXERCISE 5-1.
1.
)
276*.
2?/2:
37,
9.
_
2/
2
+
35. 7920a 8 6 4 .
X
7
+ 15z 8 4 - 2
+ 2xy +
+
Qx l y 2
7
2
PAGE 100
7. 5/24.
19.
x 612
i/4
a;
t
~T~
21. 8a<>
25.
+
.
71
23.
_^
x 2 (l
2
3. 3/2.
13.
s
35.
-...- m
x
EXERCISE 4-2.
1. 54.
23.
61. 1.6297.
51. 0.6021.
344
Appendix B
PAGE 112
EXERCISE 5-4.
3. 0.04292.
1. 8.540.
5. 3.183.
15. 11,670.
25.1.708.
27. -1.021.
33. 127,900,000 sq
7.
17. 0.1795.
13. 48.91.
29.1.249.
35. 12.62
ft.
ft.
0.0008416.
0.1104.
9.
19. 0.02950.
11. 54.61.
23. 538,100.
21. 20.56.
31.0.4343,0.2171,9.5657-10,23.1,22.46.
37.
6,070,000 sq
39. 16.5
ft.
amp.
41. $1,074.00.
PAGE 114
EXERCISE 5-5.
1. 2.3026.
13. 1.4307.
7. 6.0001.
5. 0.8735.
3. 1.4429.
11. 2.0794.
9. 1.5373.
15. 0.8228.
PAGE 119
EXERCISE 6-1.
B = 57, 6 = 18, c = 22. 3. A = 1S50', a = 21, c = 66.
7. A = 2244', 6 = 10.30, c = 11.17.
5. A = 27!', B = 6259', c = 7.012.
11. B = 4643', a = 73.66, c = 107.5.
9. B = 5339', 6 = 13.40, c = 16.64.
=
=
=
B
A
c
13.
15. A = 493', a = 2.663, c = 3.528.
793.0.
858',
812',
= 33 ft. 19. 113 ft. 21. 227 ft. 23. 1334'. 25. 4 ft.
17. h = 29 ft,
1.
Z
PAGE 124
EXERCISE 6-2.
3.
6326', 11,000ft.
80 ft, 173 ft.
1.
9.
60.
5.
N1124'W, 74
PAGE
EXERCISE 6-3.
1. (a) 5.
2. (a)
3. (a) 5,
5. 12,
2 x/5.
5 [3/5, 4/5].
(g) 4
4. (a)
(c)
2
538'.
(c) 5, 1438'.
9.8733-10.
18, 198.
3. 8.8059-10.
5. 0.7391.
13. 0.0030.
15. 0.1004.
23.
/
5416 23416'.
29. 5340', 23340'.
,
25.
5.
9.
020'.
11. 671 Ib,
15. 1815 Ib, 1962 Ib.
117.
a
=
20040'.
3.
B =
5.94.
13.
17. 49.19 in.
7.
PAGE
Ib,
25045
r
,
.
9. 9.9427-10.
17. 120', 18120'.
64 2
/
133
7. 9.3661-10.
31. 2724', 15236'.
B = 59, 6 = 60, c =
A = 7115', b = 2.01,
49
11.
Ib.
EXERCISE 6-5.
1.
131
V
VS, 6326'. (c) V73, 15927'.
0. 7. 22 knots, 20 knots. 9. 60
II. 9.5906-10.
21.
400ft.
[0, 1].
EXERCISE 6-4.
I.
7.
(g) 4.
(e) y/2.
(e )
nautical miles.
29558'.
19.
830', 17130'.
27. 3011',
33. 127', 181 27'.
210
35. 6547', 29413'.
PAGE 134
6440', 6
A =
=
133.9, c
6251', b
607 mph,
=
N3642'W.
19. 16,900 Ib.
=
11'.
148.2.
18.53, c
= 40.61.
345
Appendix 8
PAGE 138
EXERCISE 7-1.
-.
-3.
11.
13.
12-VS-
.
(V5 -2 V2),
J
g
+ 2 V).
(1
-
2
7.
15. 4/5,
-3/5.
17. 33/56, 63/16.
PAGE 142
EXERCISE 7-2.
26
169'
239
11. (a)
-3/5,
-^
(b)
25. cot
?
27. tan*
-
17. sin 2 20.
tan 60.
15.
^
(g),
(0-4/5,
,
13. cos 30.
^
2t
7.
70
-
sin
C oS
48
-
[sin
_
1
^
[
PAGE 144
cos 8].
+
100
-
9.
\
2i
sin 20.
[cos 60
5.
-
^ cos ^
2*
0.
3,
8 radians,
15.
-
2 sin 50 sin 30.
3. 27r, 1/2,
1,
0.
0.
17. 1 radian,
0.
11. 27T/5, 3,
radians.
,
19.
0.
x
7. 5?r/2,
13. ir/6,
2 radians,
0.
,
=
3.
x
=
-=-
2
13. 7T/3.
T/2.
23. 12/5.
37. -7T/2.
53.
1
,
0.
15. Tr/4,
,
0.
7
radians.
CD,
21. 2ir, 1,
0.
7T
PAGE
5.
m
x
=
-12T/
161
-
22.
7. Tr/6.
2ir
11.
|.
2i
^ radian.
3,
EXERCISE 8-2.
1.
^& cos
2 sin 3230' cos 730'.
17.
7T
TT,
2 sin
PAGE 154
5. Sir/3, 1/3,
4
23.
11.
t
EXERCISE 8-1.
9.
cos 40].
43 cos 3.
19. 2 sin
1. 2?r,
[cos 60 -f cos 20].
^
0/3
Q/a
13. 2 sin
23. cos 9.
-
3. sin
0].
OOy21. tan 20.
19. cos 20.
EXERCISE 7-3.
1.
(d) 4/3,
(c) 3/4,
f/50+5x/I6,
-
25. 5/13.
39.
-w.
17. -7T/3.
15. 7T/6.
27. 3/4.
41. u.
29. 0.3919.
43. w.
45.
19.
y
31.
5\/6
21.
9. ir/6.
-2427'.
33. 3/4.
47. u.
35.
49. u.
-3/5.
51. 1/u.
346
Appendix 8
PAGE 168
EXERCISE 9-1.
1.x- 10/7,
=
y
= -==-,
13.
23. x
27. z
_
j.
2L
*
=
10 / 7
= -21/11, z = -9/2. 25. x
= 4, z = 20/3. 29. x = 0, y = 5/7.
10/7, y
35. Consistent
33. Inconsistent.
and independent.
=
41. Inconsistent unless c
9. (5/3, 0), (0,
-5).
11.
x
=
=
5/8.
17.
x
= -
x
15.
1,
=
y
3.0.
6/5, y
j
= 25/13,
x
15.
19.
7.7.
5.0.
y
=
17.
-5/13.
(-3, -1), (2,4),
5.
-4).
x
5.
z
= 2,
= 2,
= 3,
= 3,
y
y
z
z
=
=
3. 0.
5.
7.
3
II.
19.
+ 4t, - 4i.
\/2 + 3 V2i,
-2.
3.
2.
7.
(y,
=
=
x
-
3 \/2
\/15 +8|a|Va6i,
f.
21. 1.
31. x
=
39.
+
|
1,
y
^
23. 0.
=
t.
-4.
41.
o),
1,
y
=
80,
7.
=
x
13.
(-4/3,
y
=
0), (0, 4).
22/7.
4.1,
= 0.3.
y
-17).
(0,
PAGE 185
+ 3f.
3i,
-3
\/2i.
=
33. x
+ K.
2/3.
No nontrivial solution.
9.
3z.
PAGE 187
PAGE 193
-
5.
9.
1
+
4
V&,
13.
2,
y
=
5/2.
43. 25
1
-i.
=
35.
+ Oi.
+
6|a|t,
-8|a|Va6*'25. 0.
27. & = 3/2, y
=
=
2
-2,
y
VT5
-1
1.
9. 2184.
7. 308.
3.
=
if c
= 23/7,
x
13.
11.22.
EXERCISE 11-1.
I.
1.
-3).
(6,
-110.
o
PAGE 180
EXERCISE 10-3.
1. 0.
~
=
31. Inconsistent.
(0, 0).
= 4/5.
EXERCISE 10-2.
1.
2;
|,y=|.
-11.
9.
7/5,
PAGE T72
EXERCISE 10-1.
1.5.
*/er
39. Inconsistent.
3. (4, 0), (0,
(1/4, 0), (0, -1/3).
=
y
and dependent.
EXERCISE 9-2.
1.
,
Consistent and dependent
1.
1.
= 25/7, * = -1/7.
= 9/5, y = -1/5, z =
*/
>
16/11, y
37. Consistent
-
=
o
^=
=
-2, y
-10
IT
17. x
T
x
15. Inconsistent.
=
x
5.
15/7.
O
=J
y
=
1
-ir
J?H
2/
75
=
=
=
-9/7, y
-47
~
to
=
x
3.
-6/7.
-1
18
a?
45.
4
15.
6|fl|i.
Vf.
-t.
29.
z
7/3, y
=
4/3.
-1
+
f.
-i.
17.
=
1/3.
=
-
47.
6,
1
y
=
-5.
37. 7
+ Oi.
+
3i.
347
Appendix B
-3 +
49.
-
57. 11
-2 -
75 '
~
1+
,
,
+\
25
-
2i.
23. 5
+
(2
-
29. ~^(('os
5444'
+ t sin 0).
33. 3 (cos
2
V2(cos
15
+ t sin
-8 +
V
+
2
71. 33
t.
55.
-
22t.
4t.
+ 5 <>/)
73. ~ +
^
(2
PAGE 197
-
35.
+ i sin 45).
(cos
bl
32012'
+
+ 3f.
21.
4f.
+ t sin 270).
27. 3(cos 270
6723'
31. 13(cos
5444').
-
19. 6
3f.
+ f sin
f sin
6723').
32012
/
).
PAGE 202
EXERCISE 11-3.
a
5,
2i.
,
^
A
+
7. 1
-1.
9.
i.
11.
-1/2 -
QSO
-2o.
17.
19.
+
^
i.
0.7951
A +A
13.
21.
t).
i.
ot
31.
- -
fc
/<*
amperes.
,
=-=^
PAGE 206
EXERCISE 12-1.
-7.
9. sin
3.
= 0,
5. 3,
-1/2, -3/4.
1;
13. sec
0=2,
15. cot
=
17. cot
0=7,
19. sin
= 0,
0, 90, 180.
-2.
11. sin
7. 3, 3,
=
1,
in
10,
9.
tan
II. sec
o
2.
-3.
-2; 90.
-8; 00, 300, 9711', 26249'.
-3, 2; 16134', 34r34', 2(>34', 20634
2,
|
Q in
3.
10,
=1
= 1,
/
.
-17, 88', 1888', 17G38', 35638'.
;
0, 180.
PAGE 209
EXERCISE 12-2.
i
I.
1.
A-
--
1. 0,
1+ 1
-1.
fc
29.
15.
+ 90) + f sin (9 + k 90)], k = 0, 1, 2, 3.
= 0, 1, 2, 3, 4.
2[cos (36 + k 72) + i (36 + k 72)],
= 0, 1, 2.
cos(SO + k 120) + f sin (80 +
120),
14 +
18t
828
154i
+
00
33.
ohms.
23. 2[cos (9
25.
(0.6065
|
i.
'
+
17.
9f.
25.
V5)*.
-
1019.
+ 3233 *
1233
-1 -
15.
lit.
63. 5
IGt.
EXERCISE 11-2.
13. 3
+
53. 13
g-~
69.
i.
625
77 '
*'
25
+ 24t.
67. ~
2(K.
+ Of.
61. 28 +
51. 6
\/3)i.
27
59.
3f.
65.
+
(\/2
\/2;
Q
3.
K
5.
--1
= 6730',
=fc
V7i
^
7.
-1
24730', 15730',
-3; 0, 10928', 25032'.
-
=fc
TT
\/21
27.
1, t,
-1, -t.
348
Appendix B
Q
= ==
36 23', 14337', 23730', 30230
4
= 0.4142, -2.414, 6532', 29428'. 17. (x - 3) 2 + 4(y + 2) = 4.
cos
- 5) 2 + 4(y - 5)2 = 16. 21. 4(s - 2) 2 + 9(i/ - I) = 36.
(x
2
2
4(s + 4) - Q(y - 2) = -36. 25. 3(s + 10/3) - (y + I) = 64/3.
/
13. esc
15.
19.
23.
27.
.
;
2
2
V2[(z-4) 2 +9/2].
33. [9((o;
+
4/3)
2
+
2
*
29.
V (3
I)]' 1 / 3
3)
31. [2((x
2
II.
-
-1,
-1
3.
13.
11/8.
-1, -1.
5.
>/6.
=
-r
o
,
no values of
;
0.
;
,
23. esc
3.
-4.
5.
7. 5.
9.
PAGE 215
EXERCISE 12-5.
2i.
A/3,
=fc
\/15, 3
3.
i,
it
A/6.
5.
2, db3.
Conjugate imaginary.
3. Real,
5. Real, unequal, irrational.
9. Real, equal, rational.
11.
13. Real, unequal, irrational.
17.
-2, -1.
2 J7
s2
+
1
5S
PAGE 216
unequal, irrational.
7. Real,
unequal, rational.
Conjugate imaginary.
15.
Conjugate imaginary.
PAGE 218
EXERCISE 12-7.
1.
7.
2 A/2.
EXERCISE 12-6.
1.
.
~
13
4 x/5.
8
/
PAGE 213
EXERCISE 12-4.
-4
-1/2.
;
= 0.414, -2.414; 11428 24532'.
= -0.6972, -4.3028; 19326', 34634
21. sec
9.
32)]~" 2
-1, -15/7.
19. cos
1. d=
9. 5/2,
^
17. sin 6
1. 18.
-
3
7.
= 1, -5/2, 45, 225, lllW, 29148'.
= 3.7913, -0.7913; 23218 30742'.
15. tan
2
PAGE 212
"""
-7.
7)
.
EXERCISE 12-3.
I. 1,
+
16
3.0,2.
5.3/2,6/5.
7. 6/5,
-1/5.
_
= 0.
19.
z2
- (\^ -
\/5)x
=0.
21. x*
2
-
2 A/5 z
_
+ 8 = 0.
.
349
Appendix B
PAGE 224
EXERCISE 12-8.
3. Intersecting lines.
1. Circle.
5.
13. Intersecting lines.
11. Hyperbola.
15. Intersecting lines.
17. Intersecting lines.
(-3/2, 3/4).
7. (3, 2),
(-1, -6).
No
13.
(4,
5. (2, 4),
(-4, -3).
(2 V,
11.
9. (2, 3).
15.
solution.
3. (3, 4),
7. (4, 0),
13.
(-5,
9. (4, 6),
3).
"^
(
3. (3, 4),
(-3/2, 3/4).
(3, 3),
19. (4, 1),
(-4, -1),
21.
(-4, -2), (V6,
(4, 2),
(14,
+ i V2,
2
, 5.
.
5.
(3,
11.
(2)
4)
-3), (-2, 6).
(4,
(3 VI6i,
1)
.
.
(1> 2)> (2> 1} .
-2
(- V6,
V6,),
2 VG).
23. (5, 5),
i
-3 -
\/7,
-
i \/7), (2
i
V2, 2
27>
7. 1.682.
5. 2.292.
9. 1.
Q
Q
II. Q
15. Q
:
:
-
x
7,
R
:
Q x -
3.
0.
:
+ 3z + 4, 72
+ 2* - 15, R
a;*- + x n ~*y +
2
2*3
x2
:
:
:
0.
13.
-f
R
:
2.
PAGE 239
:
:
x*
2x*
-
2x
-
3.
3.
Not a
r* #
1
:
,
+ 2xV + 4a*V
7.
cc'
9.
Not a
15. 12a
3
factor.
-
22*
2
-f 8j:
11. 5x*
-
34o:
factor.
4
2/
+
+ 2a6(5 -
+ 60.
17.
x*
2/
24s
3x 2
+ 6x - 24, R
:
:
17. 52, 2.
0.
-f
2z
-
32x^i<>
-f
-
19. 24,
-36.
PAGE 241
3c* 2 )x
3
:
x2
5.
16x 3
:
5x + 8, R 11.
+ 3s - 6, R 0.
EXERCISE 13-2.
1.
Q
5.
+
x2
Q
Q
0.
1
:
1,
9.
3h06M4>
^-
EXERCISE 13-1.
I.
15. 8.547.
25. -0.44.
^
t
log c
13. 2.718.
11. 1.836.
23. 0.7S74.
21. 10, 0.1.
19. 49.3.
log(l-ac)-log6-logc
7.
*
PAGE 233
EXERCISE 12-11.
3. 4/3.
+ i v^),
'
'
17. 2.944.
(-5, -5).
(-4, -3).
- i V), (-3 +
(-3-tV7, -3+iV7).
- 3 V93 31+3 Vm\ /31 + 3 x/93 31-3 \/93\
^' /31
V
31
31
)'\
31
31
/
1. 6.
-10).
-4), (-14,4).
25. (3, 4) (4, 3), (-3, -4),
27. (2
9).
PAGE 231
(-4, -3).
(-3, -1).
3^85)
i}
(-3,
1).
2).
EXERCISE 12-10.
1.
19. Parallel lines.
PAGE 227
EXERCISE 12-9.
1. (3, 3),
Parabola.
7.
Hyperbola.
9. Intersecting lines.
-
9(te
5.
+ 64.ri/ +
12a 3 6 4
2
.
13.
128y".
Not a
+ 39z + 45.
factor.
78.
Appendix B
350
PAGE 244
EXERCISE 13-3.
1
1.
-3.
\/3i,
-
7. x*
3. 1
+ 9x -
4x*
5.
3.
=fc i,
2
-\/2
1, 1, 2.
*',
= 0.
10
PAGE 247
EXERCISE 13-4.
"1
1.
15. 1/2,
-1.
9.
-1/3, 1/2, 5/3.
17. 3/5, 1
t.
-2, -2.
11. 2, 2,
PAGE 253
EXERCISE 14-1.
< 3.
>
-5.
II.
-i<z<i.
&
&
13.
19.
-4
25.
No
x
I.
81.
a;
z
21.
4.
values of x.
|
|
3.
2
33. x
5.
5.
x
< 4.
7.
<
a;
-|<*<i.
O
o
-1 < x <
1/3.
4fl
~
23.
x
,
&
=
15. /7s
=
21. /io
31.
~
o
d
=
S 75 =
S 10 =
100,
1,
Not an
5.
36
ll.
&
149,
n
= -^5
=
.
11
5625.
J 26
=
/2o
23. a
13 or a
=
33. 33j i 6 .
35.
4
=
-
n
60
3. 1/20, 1/25.
IT" '
T
120
'
,
II. I,.
Ifcfc
= |J
a;
<
>
x
2,
3.
=
ho
13.
61.
7.
19.
46
30, 5b
= 55.
10, Sio
i 100
=
40.
Sioo=5050.
100,
o
27.
16.
^~
~T~
5. 4, 16/5.
/
= 9,
n
= 9.
29.
n2
.
1
.
37. $37.75.
39. 282.
ly
PAGE 264
7. 1/47.
9.
^, |
,
y
,
|.
,
"IT '
3. 256, 1024.
,
Si,
= 9[1 -
5. 8, 16/3.
(2/3)'*].
=!,&
'"' ={.
16
<
=~
Z 45
EXERCISE 15-4.
I. 128, 512.
2.
fi9
EXERCISE 15-3.
120
1053.
S 2Q =
5.9,
= -^
-1/2, n
>
16
13.
PAGE 263
40
550.
-13 < x <
17.
PAGE 260
= 78, S 26 =
17.
1
arithmetic progression.
tt\K
1. 1/15, 1/19.
-5/3, s
-1.
7. 2, 5, 11, 23, 47; 2, 7, 18, 41, 88.
3.1,2,3,5,11,35.
.
9.
25.
5a
<
< -3, -2 < x < -1. 29.
< -2, s > 0. 35. x < -1/2.
27. x
3. 18, 25.
26
<
x
9.
-i<*<5.
&
A
EXERCISE 15-2.
1. 17, 20.
-1/2.
15.
EXERCISE 15-1.
1.1,2,4,7,11.
'
V-
19. 1/2,
t.
t&
13. 2/3,
7.
PAGE 267
,
^jJL&
13. ljol
17. 3,
-14.
=
^j"
JL&
10-",
9.
-1,1.
S101 =
19.i^.
21.
6,
15/2.
351
Appendix B
23. 10; 100; 1000; 10,000; 100,000.
As n
27.
sum approaches
increases, the
3 as a limit.
+
9
3.
13. 36,
5 ' 16
2
36
-
[1
19. 10/11.
(1/3)
20
36(l/3)
],
23.3.
21. 1/6.
7' 3
'
20 .
9 - 5/8.
'
15. 12
1.1*
,o
-+-.
2
*
7 .i_4+*!_y!.
X2
X*
X3
X
9
.
3
3. 216.
5. 9999.
C
a8
4-
+ 8 C2a
sCitfb
C6 8
Ca/>7
X
12
[1
-
(3/4)
10
]
17. 1/9.
feet.
PAGE 772
^
-
8- 1
a?
3z 2
5z 3
-2 + -T-lG'
,
11.1.2190.
13.1.3684.
PAGE 279
7. 20,160; 7560.
6
62
+
PAGE 282
3. (a) 720, (b) 48, (c) 480.
20; 210; 95,040; 143,640; 970,200.
7. 8
*
19. 1.0149.
EXERCISE 17-2.
1.
ft,
6
&
EXERCISE 17-1.
1. 72.
JL
11.2/11.
+ 6_i?f.
.J__f
X
X*
X
X
17. 0.9415.
15. 0.8508.
3
#
s
A
i.n.--_+_.
*
X)
25.
EXERCISE 15-6.
-.
~~
PAGE 270
EXERCISE 15-5.
1.64/65.
29.
sC 3 a 5 6 3
5. 125.
+
.
f
11.66.
9.9,979,200.
13.63.
15TJ
jy
EXERCISE 17-3.
1.
2/5, $24.00.
3. 1/3, 2/3.
A
15
15.
5. $7.29.
7. $2.42.
5.
9.
9. 0.553.
11. 16/63,
215.
,
EXERCISE 18-1.
1.
PAGE 288
PAGE 295
= 14.55, c = 20.46. 3. A = S927', a = 1169, b = 1079.
B = 531', 6 = 1.051, c = 7.513. 7. A = 173', B = 10026', 6 = 71.18.
B = 2427 C = 10120 c = 1193. 11. No solution. 13. 615.3 ft.
C =
15.
9858', b
;
;
,
449
ft.
,
17. 12.6 in.
19.
3158
Ib,
EXERCISE 18-2.
PAGE 298
= 270, B = 5130', A = OO^O 3. a = 290, B = 1150', C =
9. No solution.
5. a = 100, B = 10, C = 20.
7. 4820'.
13. 3257 ft. 15. 700 ft,
II. A = 5420 B = 5940', C = 66.
17. c = 20, A = 63, 5 = 73, C = 44.
7
I. c
.
/
,
101.
352
Appendix 8
EXERCISE 18-3.
1.
5.
PAGE
301
A = 50, B = 70, c = 56. 3. A = 61, C = 2350', 6 = 262.
B = 3111', C = 70, a = 79.25. 7. A = 47, J5 = 58, C = 75.
A = 4220', B = 5730 C = 8010'. 11. 4 in., 5 in. 13. 11 in.
/
9.
,
15. 55 30', 5950', 6440'.
EXERCISE 18-4.
1. 88.41.
17.
38 sq
3. 243,900.
ft,
12 sq
ft.
5. 9285.
7. 90.
PAGE 304
11. 54
ft.
15.
45
ft.
Index
Abscissa, 50
Absolute value, 14, 57
Absolute value function, 57
Addition, 2
of algebraic expressions, 21
of fractions, 36
fundamental laws for, 2
of real numbers, 2
Addition formulas, 135
Algebraic expressions, 17
Binomial theorem, 97
combinations applied to, 282
general term of, 99
proof of, for positive integral
exponents, 275
Braces, 18
Brackets, 18
Cartesian coordinates, 50
Characteristic of a logarithm, 106
Characteristic of the general
quadratic equation, 222
Circular system, 77
Classification of functions, 61
Coefficient, 17, 18
Combinations, 278, 281
and the binomial coefficients, 282
Common difference, 260
Common logarithms, 105
Common ratio, 265
Commutative law, 2
Completing the square, 206
Complex fractions, 40
Complex numbers, 189
absolute value of, 196
addition and subtraction of, 192
amplitude of, 196
argument of, 196
congugate of, 191
definition of, 189
De Moivre's Theorem, 199
division of, 193, 198
graphical representation of, 195
modulus of, 196
multiplication of, 192, 198
roots of, 200
trigonometric representation
of, 196
Composite number, 25
addition and subtraction of, 21
division of, 23
multiplication of, 22
symbols of grouping of, 18
transposing terms of, 43
Amplitude of a trigonometric
function, 149, 152
Angle
coterminal, 76
definition of, 75
degree measure of, 77
of depression, 120
of elevation, 120
initial side of, 75
measurement of, 76
negative of, 75
positive of, 75
radian measure of, 77
standard position of, 75
terminal side of, 75
trigonometric functions
vertex of, 75
Antilogarithms, 109
Arc of a circle, 78
of, 81,
82
Area
of a sector of a circle, 78
of a triangle, 302
Arithmetic means, 262
Associative law
for addition, 2
for multiplication, 2
Axes, coordinate, 49
Axis of symmetry of a parabola, 219
Composite (reducible) polynomial, 26
Conditional equation, 18
Constant, 53
Constant function, 53
Convergence of series, 259
Coordinate systems
Cartesian, 50
one-dimensional, 49
rectangular, 49
Coordinates, 49, 50
Cosecant, definition of, 65
Base
change of logarithmic, 113
exponent and, 16, 86
of logarithms, 101, 113
in navigation and
surveying, 121
Binomial, 17
Binomial series, 271
Bearing
353
354
Index
Cosine, definition of, 64
Cotangent, definition of, 65
Coterminal angles, 76
Counting numbers,
1
Defective equations, 45
Degree
of a polynomial, 17
relation to radian, 77
of a term, 17
term of highest, 17
unit of angle measure, 77
De Moivre's Theorem, 199
Dependent events, 286
Descartes, Rene, 50
Determinants, 173
expansion of, 176
minors and cof actors of, 175
principal and secondary diagonal
of, 174
properties of, 177
of the second order, 173
solution of linear equations by
means of, 181, 183, 184
sum and product of, 185
of the third order, 175
Directed line segment, 122
Discriminant, 215
Distance between points, 50, 51
Distributive law, 3
Division, 5, 16, 23, 25, 39
of a function, 53
Domain
Double-angle formulas, 139
Empirical probability, 284
Equality symbol, 12
Equations
conditional, 18
consistent, 163, 164, 165
defective, 45
definition of, 18
dependent, 165
equivalent, 43
extraneous roots of, 45, 213
inconsistent, 163, 165
independent, 164
linear, 42, 43, 163, 166, 167
in quadratic form, 204, 214
solution of, 42, 163, 171, 181, 183,
184, 212, 224, 227
Exponential equations, 231
Exponents, 16, 86, 88, 90, 92
Factor theorem, 240
Factorial symbol, 97
Factoring, 25, 27
Fractions, 9
addition and substraction of, 36
Fractions (Cont.)
complex, 40
fundamental operations on,
9, 10
multiplication and division of, 39
reduction of, 33
signs associated with, 34
Functions
absolute value, 57
algebraic, 61
classification of, 61
constant, 53
definition of, 53
domain of, 53
exponential, 233
graphs
of, 146, 147, 148, 169, 218,
233, 244
greatest integer, 57
identity, 53
inverse, 155
irrational, 62
linear, 54, 169
logarithmic, 233
multiple-valued, 53
notation for, 55
periodic, 149
point, 63
polynomial, 61
of, 53
rational, 61
rule of correspondence of, 53
single-valued, 53
transcendental, 61
trigonometric, 63
Fundamental assumptions, 1
Fundamental operations, 1, 9, 10, 20
Fundamental theorem of algebra, 241
range
General term in the binomial
expansion, 99
Graphs
of exponential functions, 233
of inverse trigonometric functions,
158, 159
of linear functions, 169
of logarithmic functions, 233
of polynomials for large values
of x, 244
of trigonometric functions, 146,
147, 148, 149, 151, 152
Greater-than symbol, 12
Greatest
common
divisor, 30
Greatest-integer function, 57
Half-angle formulas, 140, 300
Highest common factor, 31
Highest degree term, 17
Horizontal line, 50, 169
Identity, 18
Independent events, 286
355
Index
Inequalities, 248
absolute, 248, 254
conditional, 248, 249
involving absolute values, 14
properties of, 248
solution of conditional, 249
Inequality symbol, 12
75
Intercepts, 170
Most probable number, 284
Multinomial, 17
Multiplication
of algebraic expressions, 22
of fractions, 39
fundamental laws for, 2
Mutually exclusive events, 285
Initial side,
Interpolation, 83, 108
Inverse functions, 155
Irrational functions, 62
Irrational number, 1
Irreducible polynomial, 26
Law of cosines, 296
Law of sines, 289
Law of tangents, 298
Law(s) of exponents, 86
Least common multiple, 32
Less-than symbol, 12
Like terms, 21
Limit of sequence, 258, 259
Line
horizontal, 50, 169
vertical, 51, 169
Linear equation, 42
graphs of, 163, 169, 171
in one unknown, 43
Literal parts, 17
Logarithmic computation, 110
Logarithmic equations, 231
Logarithms, 101
base of, 101
change of base, 113
characteristic of, 106
common, 105
computation by, 110
definition of, 101
laws of, 102
mantissa of, 106, 108
Napierian (natural) ,105
tables of, 108
of trigonometric functions, 131
Mantissa of a logarithm, 106
Mathematical expectation, 284
Mathematical induction, 273
Mean
arithmetic, 262
geometric, 267
harmonic, 264
Meaning of a m/n 94
Meaning of a, $8
Measurement of angles, 76
Monic polynomial, 18
,
Natural (Napierian) logarithms, 105
Negative exponents, 90
Negative numbers, 7, 8
Number
algebraic, 62
complex, 189
irrational, 1
negative, 1, 7, 8
positive, 1, 7, 8
prime, 25
real, 1
transcendental, 62
Number scale, 6, 49
One-to-one correspondence, 49
Operations
on fractions, 9, 10
with zero, 5
Order of fundamental operations, 20
Order relations for real numbers, 12
Ordered number pairs, 50
Ordinate, 50
Origin, 6, 49
Parentheses, 18
Period, 150
Periodicity, 149, 150
Permutations, 278, 279
Phase, 149, 152
Point function P (t), 63
Polynomial, 17, 18, 26, 61
Positive integer, 1
Positive integral exponents, 16, 8
Positive numbers, 7, 8
Power, 16, 17, 86, 87, 88
Prime, 25, 26
Principal branch, 158
Principal value, 158, 159, 160
Probability, 278
empirical, 284
mathematical, 283
Product, definition of, 2
Product formulas, 143
Progressions, 256
arithmetic, 260
geometric, 265
harmonic, 264
infinite geometric, 268
356
Index
Projections, 122
Tables
Quadrant, 49
Quadratic equations
methods for solving, 204, 209
in one unknown, 204
in two unknowns, 221, 224, 227
Quotient, 5
Radian, 77
Radius vector, 52
Range of a function, 53
Rational exponents, 92
Rational function, 61
Rational number, 1
Real number, 1
Reciprocal, 4, 5
of logarithms, 108
of trigonometric functions, 71,
of,
Terms of algebraic
expressions, 2, 17
Theory of equations, 235
Transcendental functions, 62
Triangles, 115
solution of general, 289
solution of right, 117, 133
Trigonometric functions, 63
of angles, 81, 116
definitions of, 64, 65
of, 146, 147, 148, 149,
151, 152
of important special numbers, 65
inverse, 146, 156, 158, 159, 160
graphs
logarithms
Rectangular coordinates, 49, 50
Redundant equation, 45
Remainder theorem, 240
Repeated trials, 287
Repeating decimals, 269
Root of an equation, 42, 45, 213, 217
Scalar quantities, 125
Scientific notation, 92
Secant, definition of,
2
64
Tangent, definition
Terminal side, 75
of,
131
of sums and differences, 135
variation of, 146
Trigonometric identities, 68
Variable, 53
dependent, 53
independent, 53
Variation, 57, 58
Vector, 125
65
Second degree equation, 204
Segment, line, 122
Sequences, 256, 258, 259
Series, 256, 257, 259
Sexagesimal system, 77
Simultaneous equations, 163, 171,
181, 224, 227
definition
of, 64
Sine,
Special products, 22
Standard position of an angle, 75
components of, 125
magnitude of, 126
multiplication of, by a
normalization
of,
scalar, 126
127
projection of, 125
representation of, 125, 128
sums and differences of, 127, 129
Vertex of an angle, 75
Vertical lines, 51, 169
#-axis, 49
as-coordinate, 50
y-axis, 49
^/-coordinate, 50
Sum, definition of 2
Symbols of grouping, 18
Zero,
Synthetic division, 235
Zero polynomial, 61
,
3,
5
