Download Unit 1 Brief Review of Algebra and Trigonometry for Calculus

Unit 1
Brief Review of Algebra and
Trigonometry for Calculus
Objectives
When you have completed this unit, you should be able to
1. perform basic algebraic operations and factorizations.
2. define all trigonometric functions.
3. find exact values of trigonometric functions.
Learning calculus requires a knowledge of basic geometry, algebra and
trigonometry. Throughout Mathematics 265, you will be informed about the
background concepts needed in each section. However, to start the course, you must
make sure that you have the minimal basic background to succeed. The time you
spend in reviewing the prerequisite material is an investment. In most cases,
frustrations and disappointments in learning calculus result from deficiencies in
algebra and trigonometry.
In this unit, we help you by briefly reviewing some basic algebra and trigonometry.
We present a series of exercises, and recommend that you submit the first
assignment before starting Unit 2. The textbook presents some necessary material
on geometry, algebra and trigonometry in the “Reference Pages” section (pp. 1-6)
and on pages 336-367.
You may wish to consult one or more of the references listed below. The first three
are available in the Athabasca University Library. See the “Library Services”
section of the Student Manual for instructions for making a loan request.
Drooyan, Irving, and William Wooton. Elementary Algebra with Geometry, 2nd ed.
New York: Wiley, 1984.
Steffensen, A., and M. L. Johnson. Introductory Algebra, 2nd ed. Glenview, IL:
Scott, Foresman, 1981.
Swokowski, E. W. A Precalculus Course in Algebra and Trigonometry. Boston:
Prindle, Weber & Schmidt, 1973.
Blitzer, R. F. Precalculus Essentials, 2nd ed. Upper Saddle River, NJ: Prentice Hall,
2004.
Introduction to Calculus I
7
Connally, E., D. Hughes-Hallett, A. M. Gleason, P. Cheifetz, D. E. Flath,
P. Frazer Lock, K. Rhea, C. Swenson, F. Avenoso, A. Davidian, B. Lahme,
J. Morris, P. Shure, K. Yoshiwara, and E. J. Marks. Functions Modeling
Change: A Preparation for Calculus, 2nd ed. New York: Wiley, 2004.
Slavin, S., and G. Crisonino, Precalculus: A Self-teaching Guide. Mississauga, ON:
Wiley Canada, 2004.
In general, any precalculus textbook will do. If you are not sure about the quality of
a precalculus textbook, consult your tutor.
You may also wish to consult the websites below.
Geometry. Retrieved November 7, 2007, from the website of Math League Media:
http://www.mathleague.com/help/geometry/geometry.htm
College-level Geometry. Retrieved November 7, 2007, from website of The Math
Forum@Drexel: http://mathforum.org/geometry/coll.geometry.html
Kreider, Donald, and Dwight Lahr. Principles of Calculus Modeling: An
Interactive Approach. Retrieved November 7, 2007, from
http://www.math.dartmouth.edu/c̃alcsite/video1.html
This site presents a series of videos with brief explanation of key concepts.
Stewart Calculus Home Page. Retrieved November 7, 2007, from
http://www.stewartcalculus.com/media/1 home.php
This website accompanies your textbook. Bookmark the page, and be
prepared to refer to it often—it contains many useful links and other features.
At this point, pay special attention to the PDF document titled “Review of
Algebra,” which is available through the “Algebra Review” tab on the
right-hand side of the Stewart Calculus home page.
Minimum Requirements
Before you begin this course, you must, at the least, have a firm grasp of the
concepts listed below.
Geometry
• points and lines
• similar triangles
• Pythagoras’ theorem
• perimeters, areas and volumes of regular geometric figures
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Mathematics 265 / Study Guide
Algebra
• the real line
• arithmetic operations: addition, subtraction, multiplication and division
• addition, subtraction, multiplication and division of algebraic expressions
• factoring
• distributive law, factoring polynomials
• completing a perfect square
• quadratic formula
• exponents
• radicals and rationalization
Trigonometry
• angles in radians
• trigonometric functions
• graphs of trigonometric functions
• common values of trigonometric functions
• trigonometric identities
• laws of sine and cosine
The material presented in this unit is only a brief review of algebra and
trigonometry, it is by no means a comprehensive treatment of what is required in
this course; in each unit, we continue working on these prerequisites.
Introduction to Calculus I
9
Algebra
To keep the topic simple, in the following calculations, we neglect the question of
limitations on the values of x for which the statements are valid.
I. Exponents
Pay attention to how the given expressions are simplified using the laws of
exponents.
1.
2x5
3x7
2
x
2x5 + 3x7
=
+
=
+
6
6
15x
15x
15x6
15x 5
2. (a−7 b−2 c3 )−2 = a14 b4 c−6 =
3.
√
√
4.
−1/2
√
3
a3 b2
=√
a14 b4
c6
1
a3/2 b2/3
−1/2
√
3
a3 b2
= a−3/4 b−1/3 =
or
1
a3/4 b1/3
p x5/2
x2
x2 √
3 =
+
x
+
y
y3
y3
y 3/2
II. Operations of Algebraic Expressions
Study the operations and simplification of the algebraic expressions given below.
1.
1
(2x + 3)(5 − 2x) + x + 3x 4 −
x
2
= 10x + 15 − 4x2 − 6x + x2 + 12x − 3
= (−4x2 + x2 ) + (10x − 6x + 12x) + (15 − 3)
= −3x2 + 16x + 12
2.
10
(4x2 + y 2 z)
x
+ 3z 2 − (x + y)(y 2 − xz)
z
4x3
=
+ 12x2 z 2 + xy 2 + 3y 2 z 3 − xy 2 + x2 z − y 3 + xyz
z
4x3
=
+ 12x2 z 2 + 3y 2 z 3 − y 3 + x2 z + xyz
z
Mathematics 265 / Study Guide
3.
8x6 y − x5 yz + 3x4 yz
−x3 y
3
= 48x − 24x2 z + 3xz 2 − 8x3 + x2 z − 3xz
3x(4x − z)2 +
= 40x3 − 23x2 z + 3xz 2 − 3xz
4.
5
1
5(x − 9) + (x + 3)
6x − 42
+
=
= 2
x+3 x−9
(x + 3)(x − 9)
x − 6x − 27
III. Factorization
We will be factoring algebraic expressions continuously in this course, so pay
attention to the following examples.
Similar Terms
x3 − 5x2 − 4x + 20 + 10x
3
2
= (x − 5x + 6x) + 20
2
= x(x − 5x + 6) + 20
do the operations of similar terms
group similar terms
factor the common term with the
lowest exponent
Conjugates
The binomial terms (a + b) and (a − b) are conjugates of each other. Observe that
their product
(a − b)(a + b) = a2 − b2
is a difference of squares. So, for example,
√
4x2 y 4 − 6z 6 = (2xy 2 )2 − ( 6z 3 )2
√
√
= (2xy 2 + 6z 3 )(2xy 2 − 6z 3 ).
Quadratics
A quadratic is an expression of the form ax2 + bx + c with b and c any real
numbers and a a nonzero real number.
√
• A quadratic is a perfect square if b = ±2 ac. If so, then
ax2 + bx + c =
√
ax ±
√ 2
c .
The quadratic 4x2 − 20x + 25 is a perfect square because
p
√
a = 4, c = 25 and b = −2 4(25) = −2 100 = −20.
Introduction to Calculus I
11
Then,
4x2 − 20x + 25 = (2x − 5)2 .
Similarly, 4t2 − 12t + 9 is a perfect square, and
4t2 − 12t + 9 = (2t − 3)2 .
• If we have a quadratic expression of the form x2 + bx + c, and we can find two
numbers k and m such that c = km and b = k + m, then
x2 + bx + c = (x + k)(x + m).
In the quadratic equation x2 − 2x − 35, we see that (−7)(5) = −35 and
−7 + 5 = −2; hence, x2 − 2x − 35 = (x − 7)(x + 5).
• The solutions of the quadratic equation ax2 + bx + c = 0 is given by the
“quadratic formula”:
√
−b ± b2 − 4ac
.
x=
2a
Using the quadratic formula, we find that the solutions of the equation
32y 2 + 4y − 6 = 0 are
y=
−4 ±
√
−4 ± 28
16 + 768
=
;
64
64
1
3
that is, y = − and y = . Moreover,
2
8
1
3
y
3
y+
y−
= y2 + −
2
8
8 16
and we see that
1
3
32 y +
y−
= 32y 2 + 4y − 6.
2
8
Verify the following factorization:
1
3
2
8 y−
= 2(2y + 1)(8y − 3).
32y + 4y − 6 = 2 2 y +
2
8
Let us try another example: the solutions of −8x2 + 10x + 3 = 0 are
√
−10 ± 100 + 96
−10 ± 14
x=
=
;
−16
−16
hence, x =
12
3
1
and x = − .
2
4
Mathematics 265 / Study Guide
Since
3
1
5x 3
x−
x+
= x2 −
− ,
2
4
4
8
we have
3
1
−8x2 + 10x + 3 = −8 x −
x+
2
4
3
1
=− 2 x−
4 x+
2
4
= −(2x − 3)(4x + 1).
Other Factorizations
See the sections titled “Factoring Special Polynomials” and “Binomial Theorem”
on page 1 of the “Reference Pages” at the beginning of the textbook. To apply these
factorizations, identify the corresponding terms of x and y. For example, consider
the following difference of cubes.
The expression (8a3 − 27z 6 ) is a difference of cubes where x = 2a and y = 3z 2 ;
hence,
(8a3 − 27z 6 ) = (2a − 3z 2 )(4a2 + 6az 2 + 9z 4 ).
Rational Expressions
To simplify rational expressions, we factor and cancel equal terms.
1.
5(x + 7)
x+7
5x + 35
=
=
20x
5(4x)
4x
2.
x2 (x + 1)
x3 + x2
=
= x2
x+1
x+1
3.
x2 + 6x + 5
(x + 5)(x + 1)
x+1
=
=
x2 − 25
(x + 5)(x − 5)
x−5
4.
x3 + 5x2 + 6x
x(x2 + 5x + 6)
x(x + 2)(x + 3)
x(x + 2)
=
=
=
2
x − x − 12
(x − 4)(x + 3)
(x − 4)(x + 3)
x−4
We can also simplify rational expressions by long division.
For example, to simplify
Introduction to Calculus I
7x − 9 − 4x2 + 4x3
, we divide:
2x − 1
13
2x − 1 4x3 − 4x2 + 7x − 9
arrange terms in the dividend and
divisor in decreasing order of their
exponents
2x2
2x − 1 4x3 − 4x2 + 7x − 9
divide
4x3
= 2x2
2x
2x2
2x − 1 4x3 − 4x2 + 7x − 9
4x3 − 2x2
multiply
2x2 (2x − 1) = 4x3 − 2x2
2x2
2x − 1
4x3 − 4x2 + 7x − 9
−4x3 + 2x2
−2x2 + 7x − 9
subtract 4x3 − 4x2 − (4x3 − 2x2 )
= 4x3 − 4x2 − 4x3 + 2x2
= −2x2 ;
bring down 7x − 9;
the new dividend is −2x2 + 7x − 9
2x2 − x
2x − 1
4x3 − 4x2 + 7x − 9
−4x3 + 2x2
−2x2 + 7x − 9
we repeat the process:
−2x2
divide
= −x,
2x
then multiply
−2x2 + x
2x2 − x
2x − 1
4x3 − 4x2 + 7x − 9
−4x3 + 2x2
subtract: note that
−(−2x2 + x) = 2x2 − x;
the new dividend is 6x − 9
−2x2 + 7x − 9
2x2 − x
6x − 9
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Mathematics 265 / Study Guide
2x2 − x + 3
2x − 1
4x3 − 4x2 + 7x − 9
repeating the process
−4x3 + 2x2
−2x2 + 7x − 9
2x2 − x
6x − 9
−6x + 3
−6
The remainder is −6; hence,
7x − 9 − 4x2 + 4x3
6
= 2x2 − x + 3 −
.
2x − 1
2x − 1
IV Rationalization
To change quotient expressions with square roots, we rationalize; that is, we
multiply by the conjugate to change the expression into a different, more
manageable one.
For example, to simplify
x2 − 81
√
,
x−3
√
√
it√is helpful√to realize that the conjugate of x − 3 is x + 3, and
( x − 3)( x + 3) = x − 9. So,
√
√
x2 − 81
(x − 9)(x + 9)( x + 3)
√
= (x + 9)( x + 3)
=
x−9
x−3
Similarly,
√
√
√
h+1−1
( h + 1 − 1)( h + 1 + 1)
√
=
h
h( h + 1 + 1)
1+h−1
1
= √
=√
h( 1 + h + 1)
1+h+1
Introduction to Calculus I
15
Trigonometry
Prerequisites
For this review you must be able to
1. locate points on the Cartesian plane (read “Coordinate Geometry and Lines,”
pages 344-345 of the textbook).
2. identify similar triangles.
I. Definitions of Trigonometric Functions
Consider the unit circle (centred at the origin 0 and with radius 1) on the Cartesian
plane. Each point P on this circle has coordinates (xθ , yθ ), where the θ refers to the
angle from the point (1, 0) to P . The angle θ is a positive angle if it is measured in
the counterclockwise direction; it is negative if it is measured in the clockwise
direction, as shown in Figure 1.1, below.
P
0>0
(1, 0)
P
0< 0
Figure 1.1: Direction of movement for angle θ
For each angle θ, consider the corresponding point P = (xθ , yθ ) on the unit circle,
and define:
sin θ = yθ
yθ
xθ
1
sec θ =
xθ
tan θ =
second coordinate
of the point P
cos θ = xθ
first coordinate
of the point P
xθ
yθ
1
csc θ =
yθ
cot θ =
These trigonometric functions are defined for angles, not real numbers. In calculus,
however, we work with these functions defined on real numbers; that is, in radians.
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Mathematics 265 / Study Guide
To do so, we define a “winding function”—a correspondence between real numbers
and angles—as follows.
Consider again the unit circle with centre 0, and let P0 be the point (1, 0). For each
number a, we associate a point Pa on the unit circle using the rules given below.
• If a > 0, then we start at P0 and move around the circle in the counterclockwise
direction, until we have traced out a path whose length is a. The point where our
path ends is Pa . See Figure 1.2, below.
Pa
a
P0
Figure 1.2: Pa , a > 0
• If a < 0, then we start at P0 and move around the circle in the clockwise
direction, until we have traced out a path whose length is |a|. The point where our
path ends is Pa . See Figure 1.3, below.
P0
Pa
| a|
Figure 1.3: Pa , a < 0
In either case, we associate the angle ∠P0 0Pa to this number a.
Remarks 1.1
a. The circumference of the unit circle is 2π, so, to this number, we associate the
angle 360◦ . That is, the angle that corresponds to π (radians) is 180◦ .
b. The numbers a and a + 2π correspond to the same angles, since Pa and Pa+2π
coincide on the unit circle. The same is true for a + 2kπ, for any integer k.
c. When we define a trigonometric function on a real number a, we consider this
association; that is, sin a = sin ∠P0 0Pa . For example, sin π = sin 180◦ .
d. We always work with numbers (radians) in this course.
Introduction to Calculus I
17
As you can see from this discussion, the coordinates of the point P are
(cos θ, sin θ). Furthermore, since θ = θ + 2kπ for any integer k, to show a
trigonometric identity or property for any number θ, it is enough to show it for the
corresponding number between 0 and 2π.
Therefore,
• If θ = 0, the coordinates of P are (1, 0); hence,
cos 0 = 1
and
sin 0 = 0.
π
• If θ = , the coordinates of P are (0, 1); hence,
2
π
π
cos = 0
and
sin = 1.
2
2
• If θ = π, the coordinates of P are (−1, 0); hence,
cos π = −1
sin π = 0.
and
3
π, the coordinates of P are (0, −1); hence,
2
3
3
cos π = 0
and
sin π = −1.
2
2
• If θ =
Consider the right-angle (rectangular) triangle T in Figure 1.4, below, with vertices
ABC, legs a and b, and hypotenuse c > 1. Then, θ = BAC is the angle at the
vertex A.
B
c>
A
1
b
a
C
Figure 1.4: Right-angle triangle, T , with a hypotenuse c > 1
Next, consider the rectangular triangle R with sides x, y, z, hypotenuse z = 1 and
the angle at the vertex Y equal to θ = XY Z, as shown in Figure 1.5, below. The
coordinates of the vertex Y are (cos θ, sin θ); that is, x = cos θ and y = sin θ.
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Mathematics 265 / Study Guide
Figure 1.5: Right-angle triangle, R, with a hypotenuse z = 1
The triangles T (Figure 1.4) and R (Figure 1.5) are similar. [Why?]
So,
a
y
sin θ
= =
c
z
1
b
x
cos θ
= =
.
c
z
1
and
We conclude that
sin θ =
a
c
and
b
cos θ = .
c
Therefore,
tan θ =
a
;
b
cot θ =
b
;
a
c
sec θ = ;
b
csc θ =
c
.
a
You should convince yourself that the same argument can be use to explain why the
values of the trigonometric functions are the same, even if the triangle T has
hypotenuse smaller than 1; that is, when the triangle is as shown in Figure 1.6,
below.
X
B
c<
Y=A
0
b
1
a
C
Z
Figure 1.6: Right-angle triangle, T , with a hypotenuse c < 1
Introduction to Calculus I
19
II. Exact Values of the Trigonometric Functions
See the table of exact values for sine and cosine on page 358 of the textbook.
The exact values for θ = 45◦ can be deduced from the right-angle triangle shown in
Figure 1.7, below. The acute angles are 45◦ each; hence,
√
√
π
1
2
π
1
2
◦
sin 45 = sin = √ =
and cos 45 = cos = √ =
.
4
2
4
2
2
2
◦
Figure 1.7: Right-angle triangle, with acute angles of 45◦
The exact values for 60◦ and 30◦ are deduced from the right-angle triangle made up
of half of an equilateral triangle of side 2 (see Figure 1.8), below.
6
2
3
3
1
Figure 1.8: Right-angle triangle, half of an equilateral triangle of side 2
√
The height of this triangle is 3 and the acute angles are 30◦ = π/6 and
60◦ = π/3; hence,
√
π
π
1
3
◦
sin 60 = sin =
cos 60◦ = cos =
3
2
3
2√
3
π
1
π
sin 30◦ = sin =
cos 30◦ = cos =
6
2
6
2
Study Figure 1.9, below.
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Mathematics 265 / Study Guide
P
angle 0
angle
Figure 1.9: Point P at (cos θ, sin θ)
If the angle θ =
2
1
π, then α = π.
3
3
The coordinates of P are (cos θ, sin θ). On the other hand, from the rectangular
triangle, we see that the legs of P are cos α (horizontal) and sin α (vertical). The
first coordinate of P is negative and the second coordinate is positive. Therefore,
cos θ = − cos α, and sin θ = sin α, and
√
1
2
3
2
.
cos π = − , and sin π =
3
2
3
2
Exercises
1. Read the section titled “Inequalities,” pages 338-340 of the textbook.
2. Do the odd-numbered Exercises from 13 to 23 on page 343.
3. Read the section titled “Trigonometry,” pages 358-365 of the textbook.
4. Memorize Identities 7, 8, 10, 12, 15 and 17, on pages 362-363.
5. Do Exercises 9, 11, 19, 21, 29, 31, 59, 61, 65, 67 and 73, on pages 366-367.
Introduction to Calculus I
21
Finishing This Unit
1. Review the objectives of this unit and make sure you are able to meet all of them.
2. If there is a concept, definition, example or exercise that is not yet clear to you,
go back and reread it, then contact your tutor for help.
3. Definitions are important, for each definition you should be able to give an
example (something that satisfies the definition) and a counterexample
(something does not satisfy the definition).
4. Do the exercises in “Learning from Mistakes” section for this unit.
Assignment for Credit
Complete Tutor-marked Exercise 1, from the Student Manual, and submit your
work to your tutor for grading. Whether you send your work through the regular
post or submit it electronically, remember to attach a “Tutor-marked Exercise”
form, and to retain a copy for your records.
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Mathematics 265 / Study Guide
Learning from Mistakes
There are mistakes in each of the following solutions. Identify the errors and give
the correct answer.
1. If x is any number in the interval [4, ∞), in which interval is 1 −
x
?
2
Erroneous Solution
If x is in [4, ∞), then 4 ≤ x, and −4 ≤ −x, thus −2 ≤
Therefore, −1 ≤ 1 −
−x
.
2
x
x
, and we conclude that 1 − is in the interval [−1, ∞).
2
2
2. Simplify the following expressions.
a. (a + 3b)2 ab−1
b. (x3 y −3 )2
c. (ab2 − bc3 )abc
Erroneous Solutions
a. (a + 3b)2 ab−1 = (a2 + 9b2 )ab−1 = a3 b−1 + 9ab =
b. (x3 y −3 )2 = x5 y −1 =
a3
+ 9ab
b
x5
y
c. (ab2 − bc3 )abc = ab2 − ab2 c3
3. Factor the following expressions.
a. 2x2 − 7x − 4
b. 3x5 + 24x2
c. 81x4 y − y 5
Erroneous Solutions
a. By the quadratic formula,
p
7 ± 49 − 4(2)(−4)
7±9
x=
=
;
4
4
hence, the solutions are
x = 4 and x =
Introduction to Calculus I
−1
,
2
23
and the factors are
1
2x2 − 7x − 4 = (x − 4) x +
.
2
b. 3x5 + 24x2 = 3x2 (x3 + 8)
= 3x2 (x + 2)(x2 + xy + 4)
√
√
c. 81x4 y − y 5 = (9x2 y − y 5/2 )(9x2 y + y 5/2 )
√
= y(9x2 + y 2 )
√
= y(3x − y)(3x + y)(9x2 + y 2 )
4. a. Simplify the expression
x2 + 9x + 20
.
x2 + 5x
√
b. Rationalize the expression
4+h−2
.
h
Erroneous Solutions
a.
9x + 20
x2 + 9x + 20
=
2
x + 5x
5x
√
b.
√
√
4+h−2
2+ h−2
h
1
=
=
+√
h
h
h
h
5. Give the exact values of the following trigonometric functions
7π
a. sec
6
b. sin
7π
12
Erroneous Solutions
a. sec
b. sin
24
7π
6
7π
12
=
1
1
=
π = 2
7π
cos
cos
6
6
π π
π
π
= sin
+
= sin
+ sin
3
4
3
4
√
√
√
3
2
5
=
+
=
2
2
2
Mathematics 265 / Study Guide