Download An alyti cal Trigonome try

An alyti cal Trigonome try
R_elationships between the trigonometric functions, known as trigonometric
identitieso play a vital role in ca-lculus. Also, an alternative to the rectangular
coordinate system, called the polar coordinate system, often is better suited for
certain curves. Both of these concepts occur in the following example.*
Find the points of intersection of r _ cos 20 and r
-
cos g,
for0<0 a 2n.
Solution
Set cos 20
-
: 2 cosz 0 - l, we get
2cosz0_ l:cosg
2cosz e - cos 0 - | - 0
(cos0-l)(2cos0+1):0
cos 0. Using the identity cos 20
Thus,
l
cos0_
and
cose_
L
-2
Hence, 0 0, 2n13, 4n13, so that the points
of intersection are
(- U2,2nl3),
(1, 0),
To check the pole, we set r
For r - cos 20 with r 0,
-
cos 20
Forr-cosgwithrcos 0
:0,
-
and
(-
U2,
4nl3).
0 in each equation:
so 0
:1
4
Pole is a point on
the graph
0
-
0, so 0
-L2
Pole is a point on
the graph
The identity cos 20 _ 2 cosz 0 _ I is one of many
we will encounter in this
chapter. The polar graph of r : cos 20 is calle d a
teaf rose, and the polar
four
graph of r_ cos g is a circle. These are illustrated
in the figure. Note also that
the solution of a trigonometric equation is involved
in the example. All of these
are things we will study.
* Abe Mizrahi and Michaer
Sulrivan , carcurus and
Geometry (Belmont, Calif.: Wadsworth, Publishing co., @ L9g2 wadsworth, Inc.), p. -Analytic
657. Reprinted by permission.
309
310
1
B/ANALYTTcAL rRrGoNoMErRY
Trigonometric Functions of Real llumbers
So far our definitions of the trigonometric functions have had as domain the
set of all admissible angles. The values depend only on the angle, and not on
whether the angle is measured in degrees or radians. The next stage of this
course is an important one, although it is somewhat subtle. We wish to gire
meaning to such expressions as sin x, cos x, and tan x, where x stands for a reai
number.
Toward this end, let us recall that in the definition of the radian measure of
an angle, we used an arbitrary circle with its center at the vertex of the angle.
We find it useful in the present case to specify a circle of radius I , with center at
the origin, and we refer to this as the unit circle (Figure 1). If we are given an
rtt"
angle of 0 radians, then the arc length subtended on a circle ofradius r is s
1, the arc length s equals the angle 0. In other words, the linear measure
so if r
(using the measurement units of r) of the arc length equals the angular measure
:
:
of the central angle in radians. This relationship provides us with a way of
defining the trigonometric functions of real numbers that is consistent with our
previous definition.
Figure
1
a A,
1
\
x
+
Figure 2
0
Given any real number 0 (not at present to be thought of as the measure of
an angle, but just a number), we mark offa distance of 0 units on the unit circle.
starting from (1, 0) and moving counterclockwise when 0 is positive and clockwise when 0 is negative. Let P(0) be the point arrived at in this manner with
coordinates (x, y), as shown in Figure 2.* With this established, we can define
the following:
* P(0) is thus a function of the real variable 0, whose range consists of ordered pairs (x,.rr
representing points on the circle.
m
fFl
IiTRIGONOMETRIC FUNCTTO\S OF RE{L
fu
EiM
D
EFII\ITION
sin0
1
1
-y
fiEffi.r0M
mff'ffihmrl
cOS0:x
[$ gmwr
CSC0-v
(r'*
sec0-1
x
@*01
0t
l' &,,ttiumu
tan o
ilUWG
-Lx
(x
+0)
cot
o:! (l*o;
v
'iilfiir'
ryurwlitu,
m@"wl
b,@,M
frffisMmrH
The relationship between the functions defined in terms of numbers and those
defined in terms of angles should now be apparent. we simply consider the
angle with initial side the positive x axis and terminal side connecting the origin
and P(0), as shown in Figure 3. Then the radian measure of this angle is precisily
0, the same as the number of units of length of arc. Now by the definitions of
the trigonometric functions of the angle 0 given in Section 4 of chapter 7, we
Fffifrrmtlmrfifltr
m'y
Note that for the tangent and secant, 0 cannot be any number for which
and for the cotangent and cosecant, 0 cannot be any number for which
:0;
:0,
!
x
r:,idit
utrtffi:
Ufoi,gnruU,"
have:
sin 0
cos 0
tan
0
yy :y
r1
xx
r1 -x
v
x
csc0:T
v
sec 0
cot
ofi'
wmfu
kdfi*lruP
r1
I for the functions of the number 0, we see
that they are precisely the same.
Suppose, then, that we are given the expression o'sin 2." How shall we interpret it? Does it mean the sine of the number 2 or the sine of the angle whose
radian measure is 2? The answer is that it does not matter, because the values
are the same. conceptually these are quite different, but sin 2 where 2 is a
number has the same value as sin 2 where 2 is the radian measure of an angle.
More generally, any trigonometric function of a real number has the same value
as the same trigonometric function of the angle of g radians. Thus, insofar as
M,&#nmch-
v
0:!v
comparing these with Definition
[rc
1
Figure 3
units
0 radians
312
8/ANALYTICAL rRrcoNoMErRY
values of the trigonometric functions are concerned, it makes no differene
whether we think in terms of an angle measured in radians or in terms of thr
real number that equals the number of radians in the angle.
EXAMPLE
1
x coordinate of the point P(0) on the unit circle is
and P(0) is in the
second quadrant. Find the y coordinate of P(0) and write all six trigonometric
-!,
The
functions of0.
Solution
By the Pythagorean theorem we have
(-+)'+v2:1
16
,,2
1
J12525
-
9
y -*
By Definition
7,,
sin 0
'5
at
EXAMPLE
2
csc 0
4
x-
-t
:L :
3
cos0_
tan 0
3
v
1
5
v
3
1
sec 0
x
5
-4
4
-x
- v-_ -;J
cot 0
4
Determine the value of e, where 0 < 0 <2n.
P(0) (-,fr12, -+)
b. P(0) (-
a.
Solution
b.
-
-
rlo,uo)
We observe that the reference angle of the angle determined by the radius
drawn to the given point is 30", or 7116 radians. So the angle itself is 7n 6
radians. Therefore 0 :7n|6.
The reference angle in this case is 45", or nf4 radians, and the angle is
therefore 3nl4 radians. So 0 :3n14.
Figure 4
Figure 5
I/TRIGONOMETRIC FUNCTIO\S OF REAL
qE@
\L\TBERS 313
If you are using a calculator, you can find trigonometric functions ol real
numbers by considering the number as the radian measure of an angle- and this
can be done regardless of the size of the number. For example. using a cale-utator
we can redd
trrfu
sin 2
-
cos 10
0.9093
:
-0.8391
tan(-3.5;
-
-0.3-46
aqd so on. However, to read these values from a table such as Table \- ar the
end of this book requires so{ne preliminary steps. This is because the table onllgives functions of angles between 0' and 90o, o1 equivaleptly, between 0 and
nf2 radians. For larger angles, we make use of reference angles. Since z 2 =
1.5708, we cannot find'functions of numbers grepter than this directly from the
table. To get sin 2, for example, we proceed as follows: As ap angle in radians
j hes in the second quadrant (between nl2 x 1.5708 and n x 3.1416). We shorv
this in Figure 6. The refergnce angle is thereforg n - 2 x 1.1416. Since the sine
is positive in the second quadrant, we have that sin 2 is approxirnately equal to
sin 1.1416. Reading from Table V (and intelpolating between values found
there), we see that this is approximately 0.9093. We could also obtain cqs 2 by
observing that the cosine is negative in quadrant II.'Thqn, again usipg the
reference angle, we find cos 2 x -0.4l6L Thus, the coordinates of the point
P(2) on the unit circle are approximately (-0.4161, 0.9093).
Figure 6
P(2)
r ( - 0.4161, 0.9093)
:
:
tan2 sin 2
0.9093
coS 2
-0.4161
-2.185
To find the functions of 10 using Table V, we again must find the reference
angle. Since 10 exceeds 2n,the first thing to do is to subtract 2nftom 10, in order
to obtain an angle in the first revolutipn that is coterminal with 10. This gives
l0
-
2n nv 10
-
6.2832:3.7168
This is seen to be in the'third quadrant (Figure 7), and so the reference angle is
3.7t68
-
3.1416
:0.5752
From the table we can flow obtain sin 0.5752 : 0.5440 and cos 0.5752: 0.8391.
Since both sine and cosine are negative in the third quadrant, we have finally
that
sin
10
t -0.5440 and
cos 10
=
*0.8391
The advantage of a calcplator for finding functions of real numbers should be
evident.
314
8/ANALYTTcAL TRrGoNoMErRy
Figure 7
x
P(10)
-P(10-2n)
! (- 0.8391,
EXERCISE SET
A
3.7169
_ 0.5440)
1
In Problems 1-10 flnd all six trigonometric functions of
r. p(0): (-+,
3' P(0): G, -P.)
-zrfztzl
0.
z. rp1:1_z1ls, U\fs)
4. P(0)= (-0.6,0.8)
5. The x coordinate of P(0) is f, and p(0) is in the fourth quadrant.
6. The y coordinate of P(0) is f , and rl2 < 0 < n.
7. The abscissa of P(0) is -1, and n < 0 < 2n.Whatis the value of 0?
8. The ordinate of P(0) is uE1z, and the abscissa of p(0) is negative. what is the value
of 0?
9.
10.
The abscissa of P(0) is {, and the ordinate is negative.
The ordinate of p(0) is
and 3n < 0 < 4n.
_*,
ll and 12 give the coordinates of the point p(0) for the given values of 0.
ll. a. P(nl2l b. Plrl
c. P(3n12)
d. P(5n16) e. P(anl3)
12. 1 P(snl4) b. P(8n13) c. p(-7nla)
d. P(5n) e. P(3lnl6)
In Problems
ln Problems 13 and 14 indicate the approximate lgcation of p(0) on the unit circle for tbe
given value of0.
13. a. 0:0.5
14. a. 0:8.2
In Problems 15 and
b. e--2
b. 0:L2.75
16 give the value
15. a.
P(0)
61 312- r)
c. P@) -- (+, ,lZlz)
e. P(0) - (1, 0)
16. a. p@) - ( - ,,_- ,ll tZt
c. P(0) - (- J 312, +)
e. P(0) - (- 1, 0)
In Problems 17 and
18
c. 0:4
c. e--I.54
d. e- -15
d. 0-4.72
of 0 if 0 < 0 <2n.
b. P@)= (1 IrE,rlJ2)
d. P(q -. (0, 1)
b. P(0) _ (0,_- 1)
d. P(0): (J212, _rlZlz)
flnd sin 0, cos 0, and tan 0 using either a calculator or tables.
17, a. 0 : 0.3
b. 0 : 2.718 c.
18. a. 0 : 7 .1432 b. 0 : 0.2967 c.
0
0
- t4.03 d. e - -3.625
_-5.1324 d. 0:10.5146
2/SOME BASIC TRIGONOMETRIC
19. Find all values of 0 such that 0
a, sin0:l
d. sin0-0
g. sin0--l
B
20.
Use the
21.
< 0 < 2n for which:
b. cos0-l
e. cos0-0
h. cos0:_1
c. tanl:l
f. tan0_0
i, tan0__l
unit circle to show that:
n. sin(zu - 0) _ sin 0
c. sin(z + 0) _ - sin g
e. tan(n+q_tan0
b.
d.
IDENTITIES 315
cos(zu
cos(zr
-
g) :
-
+ 0) : -
cos g
cos 0
Show that for all integers n, each of the following
is undeflned for the stated values
of 0:
a. tan0 for 0- 2n*l
b. cot? for 0:/ur
-,-n
c.
sec0
for 0- -2n*1
2 "
22. Show thar lsec 0l >
23, Prove that the range
24. Let
1
.s-
d. csc0 for 0:fin
and ncsc 0i > t for all 0 in the domains of these
functions.
of tan 0 is the set of all real numbers.
{rr^ :
t
T - {0 e R:
U-{keR:
^ (2n * ltn
U+
,il:0, rl, -2,
Z
"
0+nn,n:0, +1, +2, ,JI
lr'l
]
>U
Prove that if f(0): Sec 0 and g(0): csc 0,
then:
a. "f is a function from S onto'U
b. g is a function from T onto U
Is either f or g a l-1 function? Explain.
7
Ll
Some Basic Trigonometric Identities
You may have already noticed the following relationships
among the trigonometric functions:
tano-sino
cos 0
cotU-
cos 0
sin 0
SeCU--
1
cos 0
These are seen to be true from Definit ion 2 of
chapter 7 or Definition I of this
chapter . They are examples
trigonometric identities-so-called because they
-of
are true for all admissible values
of a. There are many other identities, and sorrie
316
8/ANALYTTcAL TRTGoNoMETRY
are so basic that they should tje committed to mbmory. These
will be indicated
as we proceed.
The Pythagorean theorem is the basis for another important set of identities.
Using the theorem, we have, for (x, y) on thb unit circle (see Figure 8),
x2+Y2:1
Figure 8
But x
-
cos 0 and
y
-
sin 0. So
tn2 0
If we divide both
*
cos2 0
-
1
sides of the flrst equation by x2 (assuming
x
*
0), we obtain
which, according to our definitions, is
1*tan20-sec20
However. if we divide the first equation by
y2 ,
we get
(r'*1 :6)'
or
co( o +1
-
cscz o
We collect these results together:
sin2 0 + cosz 0
t + tan2 0 -
1
sec2 g
1+cot20-csc20
EXAMPTT
These are known as the Pythagorean identities and should be learned. Notice
the similarity between the last two. In applying these, it may be necessary to
write them in various equivalent forms, for example, cos,
I sin2 6 or
sec2 0
| tanz 0, but it is probably besi to concentrate on learning them in
- :
b: -
2/SOME BASIC TRIGO\O\IETRNC
Ibe indicated
IDE\TITNES 317
just one form, such as the one given. you can then mentallr renrire
then: in
various ways.
Starting from any point on the unit circre, if we go 2zr unirs along rhe cir;lc
either clockwise or counterclockwise, we wind up ut th. same poini. srnce rhe
circumference of the unit circle is 22. Thus, fo. any 0, pldl: prd 3:-. .
where n is any integer, positive or negative. Since the vaiues of the trigonomeri:;
functions depend only on the coordinates of p(0), it follows that
cf identities.
rre 8),
:
sin(0
*
cos(0
+ 2nn):
2nn)
si^ g
sss 6
and so on, for the remaining trigonometric functions. These can be expressaJ
briefly by saying that each of the trigonometric functions is periodic, rvith
period 22.*
Figure 9
v
P(0\
\
-(r-v
(1, 0)
t//
h. n-e obtain
P(- t )-(x
x
-v)
consider next the relationships between the functions of 0 and
-0. By
symmetry we see that if the coordinates of p(0) arc (x,y), then those of p( 0;
(x, -y) (Figure 9). The sine of g is by deflnition the coordinate of p( - g)
7re
7
but this is the negative of the y coordinate of g. Thus, sin( _ 0) : _ sin 0. Since
the x coordinates of P(0) and p(-0) are the same, it follows that cos(-0):
cos 0. similar relationships for the other functions can now be obtained:
-0): -sin g
cos(- 0) - cos 0
tan(- 0) : - tan 0
sin(
-0):
sec( - 0)
csc(
-csc
0
sec 0
cot(-0): -eote
:
It
:
is most important to learn the relationships sin( 0)
- sin 0 and cos( _ gt
cos 0, since they occur more frequently than the others and since the others can
be deduced easily from these two.
EXAMPLE
ned. Notise
rcessary to
sin2 g or
img them in
-
3
Make use of the basic identities of this section to find the other fir.e trisonometric functions of 0 if tan 0 :
-2 and cos 0 < 0.
:
+ In gen^eral,
if /lx + k) f(x) forall x, then is said to be periodic with period k
The smallest
positive fr for which this is true is called the rr"i"."nta period,
For the sine, cosine. secant. and
cosecant, the fundamental period is 22, but for the tangeni
and cotangent,
/
;ti, ;--'
318
8/ANALYTTcAL TRTGoNoMETRY
Solution Firstwe observethatcot 0 : lltan0 : -L.Next, since sec2 0 : | + tanz 0 :
I + e42: 5, we have sec e : fJ5. But cos g is negative, so that sec 0.
which is 1/cos 0, is also negative. Thus, sec e: -J5. Therefore, cos 0:
_rtJs.
: - :
: -
Nowweuse sin2 0 | cos2 0 I + f. Sincetan 0 andcos 0 areboth
negative, and tan 0 sin 0f cos 0, it follows that sin 0 is positive. Thus, sin 0
2l'f 5. Finally, csc 0 l/sin e {512.
:
:
:
:
Note.
We could also have determined the signs of each of the functions b1'
observing that for both tan 0, which is y f x, and cos 0, which is x, to be negative.
we must have y positive, so that the point P(0) is in the second quadrant, or
equivalently, the angle of 0 radians terminates in the second quadrant.
EXAMPLE
4
Solution
In each of the following use the basic identities to show that the first expression
can be transformed into the second:
a.
sin 0
coso
tan U'"
a.
Since tan 0
-
cos 0
b. --- ^" csc0- sin0
tan U'
sin 0/cos 0, we have
sin 0 sin 0
cos 0
:
,-sin0._-cos0
slnU
tanA smU
C"S
O
cos0
^ cos0
:
cos '
t^" e ' ,i., o
b.
cos2g
,ir,
o
- 1 - sin2 0 andcsc g : l/sin 0 to get
cos2 g
1 srnz0
: 1-sin20:rirr0s", 0
,irr0
U"0-csc0- sin 0
Now we use the identities cosz 0
By using the basic identities we can proye a multitude of others, but fortunately it is not particularly useful to try to memorize any of the results. B5proving an identity, we mean showing that the given equation is true for all
admissible values of the variable or variables involved. The general procedure
is to work only on one side of the equation and, by use of the basic identities.
transform it so that in the final stage it is identical to the other side. Often the
more complicated side is the better to work with, since it offers more obvious
possibilities for alteration, but there are times in calculus when it is better to
change a simpler expression into a more complicated one. We illustrate the
procedure by means of several examples.
EXAMPLE
5
Solution
Prove the identity tan 0
+ cot 0 -
sec 0 csc 0.
Perhaps it would be better to word the instructions, "Prove that the following
is an identity." For we cannot prove it is true by assuming it is true, and this
is an important point of logic often missed. This precludes, for example, working
E\A}IPI
2/SOME BASIC TRIGO\O\IETRIC
-{l-
t-m] # ,:
thr
re'
on both sides of the equation, unless we verify that each step is rerersible. A
proper approach is to begin with the left-hand side and try to obtain the rigtrthand side:
,rru d&"
$cr$
f-
tano+coto:sino*coso
r # are thrmfu
[m- su:: *
IDE\T[T[ES 319
:
cos
0
sin 0
sin20+cos20
mtiors
cos 0 sin 0
mS
henrymru,e-
uadran:*
1
cos 0 sin 0
$m
ntnL
ll
:"or0'rirr0
I enqwesisnilcmml
and now the given.equation is verified. It is true for all admissible values of
0, and in this case this means all values except 0, nf2,3nl2, and any angles
coterminal with these since, at each of these values, two of the given functions
are undefined. usually we will understand that such exceptions are necessary
without mentioning them explicitly.
EXAMPLE
6
Prove the identity
1
sec, +
r#to
ger
1
:
cotx
csc Jr
-
csc2
x+
1
Solution It would be possible to transform the right-hand
side into the left, and may
even be easier, but we choose to work from the left-hand side because in later
applications the right-hand side will be seen to be the more desirable final
form. We begin with a commonly used trick, that of multiplying numerator
and denominator by the same factor.
; but frorrnnrmuhs" mB
trffi flcr effi
I pooedrrru
r
idemtiuifis"
r. Oftem thc
srg obr-iloum
ir
better ts
hstrate
secx+1
1
secx-1 secx-l
secx+1 secx-1 sec2x- 1
The object here is to bring sec2 x 1 into the picture, because this is, by one
of the Pythagorean identities, equal to tan2 x. So,
1 :...,-secx-1
secx+1 tanzx
cos2 x
cotz x
x sin2 x
cos x
(csc2 x
- 1)
-:Tsln' x -
COSX
and fihis
te m orking
tanz x
cos
:
ncn
1
1
tk
n foflomring
secx
tanz x
sin
x
1
sin x
-cotxcsc x-
csc2
x+
1
320
8/ANALYTICAL TRIGoNoMETRY
EXAMPLE
7
EXE
Prove that
2 sin3 x
1-cosx 2 sin3 x
1-cosx
Solution
EXAMPLE
8
Solution
2 sin x(1
* cos x)
2 sin x(1
- cos2 x)
1-cosx
2 sin x(1 * cos xX1 1-cosx
- 2 sin x(1 * cos x)
cos x)
Pror,e the identity
csc0-cot0 :Cot0
sec0-1
1 cos0
*r0-rir0
csc0-cot0
sec0- |
1
1
cos 0
1-cos0
rir,
0
1-cos0
cos 0
:- 1-cos0
sin
0
cos 0
1-cos0
cos 0
sin 0
:COt0
A word is in order about the utility of proving identities. In calculus, as well
with an unwieldl
expression involving trigonometric functions. Often, by use of the basic identities, these can be transformed into more manageable forms. So in practie
we usually do not have a ready-made identity to verify, but rather an expression that is to be changed to some other, initially unknown, form. It would
be more accurate to describe this procedure as deriving an identity. The value
of proving identities already provided lies in the fact that it gives a goal tc-r
as in more advanced courses, we are frequently confronted
shoot for in transforming an expression. Without this and with little or no
experience in such activity, a student may go in circles or change to a less
desirable form.
R
IDE\TrrrES 321
2/SOME BASIC TRIGONON,{ETRTC
EXERCISE SET
A
In Problems 1-8 find the rbmaining five trigonometric iunctions of 0. usine the basic
identities of this section.
1:
3.
5.
7.
sin0:-!,' tang>0
tan0: -4, nl2<0<inl2
cos?:2lJi, cot0<0
sin0:*, 5nl2<0<7n12
i. r."b:g, cscg<0
4. eotl:#, n<0<2r
6. csc0--+, tan0>0
8. secg:3, 3n<0<4x
In Problems 9-18 show that the first eipression can be transformed into the secon,J.
using basic identities.
9.
tan 0
10.
sec0
SIN H-
-,
sin 0
11. ----- .
sec 0
cot e'
13. cot 0 sec 0,
15.
SCC O
,
-
cos 0
csc 0
11
- -.
17. --- nr
sec'0
cscz
cos 0
12.
sec2
14.
tan 0 csc
18.
I
0'
.
0'
csc
0 stn2
0,
g,
sin 0
16. 1-
tan 0
CSC O'
cot 0
_.
sec2 0
sec 0
sin2 0 sec2
-
- l
cosz0
0'
csc
sec2 0
0,
1
In Problems lg -67 prove that the given equation is an identity by transforming the
leit-hand side into thi right-hand si&.
19. 1+sin0 :cos?+cot?
20.
sinxcotx-sinx
zl.
]
zz.
2i.
24. (tan
tan 0
COSXCSCX
- sec2 o
xtanfi:cosx
(1 + co* o) tan2 o
25. secx-sin
sec0-cos0
--sin0
tan 0
sin2 x-l
,
-:CSC2J-1
cos'x-I
x+cotx)2:sec2 x+ csc2x
tan 0 cos 0
26. 1--:COSZ0
CSC O
-
: L-
27.
(sin
29.
sin0tan0:sec0-cos0
31.
(tan2 0
33.
35.
37.
@
cos
0)'
+ lxcosz
sina x-cos4
0
2 sin
@
- 1) - 1 -
cos
d
sec2 0
ff-2sin2x-1
tan0+l
^--sin0
sec0*csc0
sec2
x
39.
j
I*cot'x
sin 0
32. csc 0
34.
38.
41.
(1
*
-tan2 x
csc 0)(sec 0
-
tan 0)
40.
: cot g
sec2 a
30.
1
.-sinOcos0
-
1 .
cos- d" - | -sec2z
36.
tan0*cot0
1-
28.
42.
*
-
cos 0 cot 0
cos 0 cot 0
-
csc 0
sin 0
(1 - tan 0)2
\;z,:-1-2sin0cos0
cos
x cot x
-==---- 1csc x
cos 0(tan e
-
-sin2 x
sec 0)
-sin0
1-csc0
secx*tanx _
coS x(1 + sin -rl
-,
SEC- X
1-cosx
sinx
Stnx +-_?csc.r
1-cosr
322 8/ANALyrrcAL TRrcoNoMErRy
43, tan2x-
sin2
tan2x-l
a45.
:: sln' x
47.
sec4
.
SeC2
ft:tan2
L+tanx
:
49. ;I*cotx
11,
: :2sec2-r
1-sinx -: 1+sinx
46. cot4 A - l: cot2 g csc2 0 - cscz 0
ff:tanz xsin2x
x
-
44.
x
csc2
t^ tanz a 48.
xsec2 x* sec2x
t€Ifl
sec-
r
1
SeC x *tanx
- tanx
sinx L -cosx
53. ;--.--:2cotx
I_COSX
SlNX
oo...?
-:
r1
.
- oww- X
--
51.
55.
sec
x
;---;
1 - csc'x -
56. (cot g B
csc
tan2
0)'
'
.rr
sec2
:1 -
.\
-
sec2 .y
+
d.
sin2 a
-
I
-
sin:
z
sin x
50.
cotx*cscx -l-cosx
52.
tana
s4.
$an2
ft-
sec2
xtanz x-sec2
o-- l) co: o:
sin0-cos0
sec o
+
x*l
csc o
1
cos 0
l*cos0
57. (1 - cos2 0)(2 * tanz 0) : sec2 0 - cosz 0
58. tan g(sin 0 + cos 0)2 + (1 - sec2 g) cot 0 :2 sinz 0
59. 2 cos x - sec x(l - 2 sin2 x): sec x
sin 0
60.
;-;:cSC0+cot0
I-cosU
62.
1-sinx
, . ,tr, -
63.
-1rsin0 *sec20-tangsec0+l
M.
66.
67.
68.
t
2 sec2 x
-
61.
1
,
1-sin0
sec2 0
+
tan g scc g
-:
2 sec x tan x
- !
sin 0
sin: .r tanl
.x
. :toox*s1nx
tanr-sln.r
sin3 0 - sins 0 cos 0
sin 6
-
sec0-cos0+tan0
-sin0
tan0*sec0
65.
sin g cos2 g + cos3 0
-r- cos 6
-
(sin 0
-
cos 0)2
/i -cosd
sin 6
l0<0<n)
,
l+cosU
V1+cost,- _,
By making the substitution t:2 sin g, where -trl2 < 0 < 7rlz, show that
t_
t
------:
"rl4 - t2
69. By making
the substitution t
-
tan 0, where
JTTT:
-
Tl2 < 0 < nl2, show that
csc o
t
70. Iftan0:
tan 0
t, prove that
singcos
O:-J-
71. Bymaking the substitutio nt:*sec0, where'o*.1'r.nl2if
if < o' show that
'
^trF
t
-
9
t>0
and z
<0 <3n,2
3/THE ADDTTTON FORMULAS FOR SrNE AND COSTNE
72. Make the substitution: x : sin aQnp < 0 < nl2) in the
73.
74.
75.
x21{t
*}
and show the result can be written in the form sec 0 - cos 0,
Let x : 2 tan 0(-nl2 < 0 < nl2) in xll[ a], and.show the result is sin 0.
Substitute x : 3 sec 0, where 0 lies either in the first or third quadrants, in (x2 * 913t2 1x,
and show the result can be written in the form 9(sec 0 tan 0 - sin 0).
Substitute I : I tan 0(-n'2 < 0 <nl2) in xl1@p a 9 and show that the result is
1sin
3
expression
323
0.
The Addition Formulas for Sine and Cosine
We will develop in this section formulas for the sum and difference of two
numbers (or of two angles). These are of fundamental importance in deriving
other identities. We designate by z and p two numbers which we will initially
restrict to be between 0 and
2r
and for which e
>
B. These restrictions
will soon
be removed.
In Figure l0 we show typical positions of P(z) and P(p). Now the distance
along the circle from (1, 0) to P(e) is a, and the distance to P(p) is B. So the arc
length from P(B) to P(a) is a - B. If we think of sliding this latter arc along rhe
circle until the point originally at P(fi coincides with (1, 0), then the distance
a - p is in standard position and the end point of the arc is correctly labeled
P(a - h. Since the x and y coordinates of a point P(0) are cos 0 and sin 0,
respectively, we can show the coordinates of the three points P(a), P(p), and
P(a - fi) as in Figure 11. Now, since the arc from (1,0) to P(u - fr) is equal in
length to the arc from P(B) to P(a), it follows that the chords connecting these
respective pairs of points are also equal. We can get these lengths using the
distance formula:
Distance from (1, 0) to P(u
-b-
Distance from P(p) to P(a)
(cos
a-
cos
p)'+
(sin
a-
sinB)2
Since these distances are the same, their squares are the same; so we obtain
the equation
[cos(a
Figure 10
:3m;3
-
P)
-
1]2
+
[sin(a
-
P)f'
:
(cos
o(
-
cos p)2
+ (sin a - sin
B)2
324
8/ANALYTICAL TRIGoNoMETRY
Figure
11
(cos B, sin p)
P(p)
-
(cos(a
'P(a)
p), sin(a
-
p))
(cos e, sin a)
-
On squaring and collecting terms, and making use several times of the identity
sin2 0 + cos2 0 : 1, this gives
cos2(a
-
frl
-2
-
cos(a
P)
+1 + sin2(a -B;
: cos2 - 2 ccis a cos B * cos2 f + sin2 a - 2 sina sin B * sin2 B
2 - 2 cos(a - $ : 2 - 2lcosri cos B * sin a sin B)
o(
or finally,
cos(a-f)=cosacosP+sindsinp
(1)
Now we remove the restrictions on o( and B. Suppose first that B > a, with
u and P again in 10, 2n]. Then, since cos( - g) : coS g, we can apply (1) with a
and p interchanged:
cos(a
-
f)
:
cos(f
-
a)
:
+
cos B cos a
sin B sin a
:coso(cosf+sinasinB
So (1) still holds true. If a:p,it again holds, since the left-hand side is cos0,
which equals 1, and the right-hand side is cos2 o( + sin2 a, which also equals 1.
Finally, let r and fi be any two numbers whatever. Then we can write oc:
a, * 2mn and, p : frr * 2nn, where m and n arc appropriately choben integers,
and a, and f, lie in the interval l0,2rf. Then, by periodicity of the sine and
cosine,
cos(a
-, :
:
:::i[l
1,,
cos(ar
- f r)
r ::: ::::
],,:,,:;:ffo
BY PeriodicitY
i;ilTh"''
lI l'1,"."*,
Thus equation (1) is true for all real numbers a and p.
From equation (1) we are able to get d number of other results. Consider
first cos(a + f). We write a + 0: u - (-0 and apply (1) with p replaced
bv -fr:
cos(a + P) - cos[a- (-p)]: coso(cos (-fr) * sinasin(-fr)
Since cos( -
P):
cos
B
and sin(
cos(a
- fr): -sin B, this gives
+ p) -
cos n cos p
- sin
oc
sin B
(2)
3ITHE ADDITION FORMULAS FOR
SI\E {\D CCSI\E 325
Next, in equation (1) let n : nlz. Then, since cos n 2 - 0 amci sr-n a:-
i.
we get
cos
In this, put p -
Tclz
-
(;-,)
-sinf
l_.tl
e to get
'ln ,)
-')] : sln\z
l; - (;
Simplirying, we
see*;
r) :
""(; -
r ideriuil
Now we can get sin(a
$." srn: j
sin(a
[t)
wryrr
m-ilTh u
fr): cosl;-
:
sin
oc
(-$
Ir
0) as follows:
(a
+r,]
Bv (3)
:cosl(;-.) -p]
:.*(; - o).o, r + sin(; - -)',,,
I
>&
+
*
cos g
cos B
+
cos a sin
p
Bv (1)
By (3) and
()
Notice that in applying equations (3) and (4) we have substituted different
values for B and 0. Thus,
; fu ccqs ftcquals I
vrite n
i
-
imtegr,ru-
fline effic
sin(a
* 0):
sinacos B + cos a sinB
- B : a + (- fr) and using (5), we get
sin(a - B) : sin[a + (-0)] : sin a cos(-B) + cos d sin(-B)
or, since cos(-fl : cos f and sin(-p) : -sind,
sin(a - P): sinacos P - cos a sinB
(5)
Writing u
(6)
Formulas (l), (2), (5), and (6) are known as the addition formulas for the
sine and cosine. They are of fundamental importance and should be memorized. We summarize them below. Notice that for the sine, the sign between
Cmmdm
rcpleffi{i
u
l,l,,r
The Addition Formulas for Sine and Cosine
+ p) -sindcos fr+ cosasinB
sin(a - p) -sinncos p- cosasinB
cos(a + p) :coSflcos p- sinasinB
cos(a - p) :coSflcos fr+ sinasinp
sin(a
326
8/ANALYTICAL TRIGoNoMETRY
the terms on the right agrees with the sign between a and f on the left, whereas
for the cosine these are reversed. So there really are just two basic patterns to
EXAMPL
be learned.
EXA
MpLE
9
Solution
Find the value of each of the following without the use of tables or a calculator
a. sin(5nl12) b. sin(n/l2) c. cos(l3nll2) d.
a.
We can write 5nll2 as2nll2
:
cos(-nll2)
+ 3nll2: nl6 + nl4. So
.infi:,*(;.;)
:.in1"o.1+"o.1ri,1
---6---4 ---6 4
Now we know that we may treat
of angles. Thus,
sinsi=(;)
#:""(#
smr
b.
nf
6 and nl4 as if they are radian measures
H-g) G) ::f
t
?)-sin G
: sin 1.o* Y * cos !
4646
rinT
f-r
tr)
G)0: 2J'
G)
C.
1,3n
COS
n?:cos tr.?):cos
tr.oJ
: cosf .o. ;-
$,)
d.
cos(
-
n 1,2121 -
cos (n rL2),
since cos ( - 0)
sin+sin
0 H€) =-1+.F
")L
o\o
:
cos 0 ; so,
co
",(- ?):cos #:cos tr-T):cos
TE
f
TE
TE
G-t
)
7T
:[l Hd;d) :
.,F+ t
zrfz
EXERCN
3ITHE ADDITION F'ORMULAS FOR
S[\E A\D COSI\E 327
n,yM.mm
ntficrmm
rum,
ekufri&Irrr
EXAMPLE
If - rcl2 < x < nl2and
10
a.
Solution
y)
b.
y <7t, sin x
0<
cos
(x
- y)
-
f,, andcos )
- -;.
find:
We conclude that x is between 0 and nl2, since if it were between 0 and
l.
sin x would be negative. Similarly, we must have y between nl2 and. z in order
-;:
l2I
r
sin(x +
for cos/ to be negative. By the pythagorean identity,
sin2
we get
x + cos2.r: l.
cos2x:l-sin2x
So
cosx-
t@
The positive sign must be chosen, since 0
<x<
7T12.
cosx-
ffittlnlnurr$
Similarly,
(-;
We now have all the necess ary ingredients for obtaining the solutions.
sin(x
-t
/
+y)-sinxcos y+cosxsi
ny: 43 ('_
:)
-15 + \m
'fr
b
24
b.
cos(x
- y):
cos x cos
y +sin x siny
:{f-:)
- 4 \i)
:- -5\
'', ;
%
..lllil
. (+)$)
t
\-"
,r a
'\y' -
)
EXERCISE SET
A
3
In Problems 1-8 evaluate without using a calculator or tables.
l.
2.
sin(a
sin(a
+ P) and cos(a + P),where a - Tll|and fi - ni3
- P) and cos(a - P),where a - nl4and fr /5n z\
3. a. ,in(!- I\
T) b'
\r-
4.
L.',,(?.T)
3
5. a. .o*tn
'"'(?.;)
b. ,",(+
r7n
b. sin,
-+)
nl3
.1(#)
+
I ({
7 + 3\,
24
11
328
8/ANALYTICAL TRIGoNoMETRY
6. a.
/ z\
r*(-r,/
b.
/ 5z\
*r(-ri
b. cos 15'
7, L, sin 75"
8. a. cos 255'
b. sin 195'
9. If sin a : -!, sin B: -*, P(a) is in the third quadrant, and P(p) is in the fourtL
quadrant, flnd the following.
a. sin(a - fl b. cos(a + p)
10. If cos a : {t, cos P : -1, P(a) is the fourth quadrant, and P(fl is in the second
quadrant, flnd:
a.
B) b. cos(a - B)
andsin p: -J, flnd:
b. cos(a - B)
a. sin(a + fl
12. If tanx: -|, cscx > 0,sec y : +,and coty < 0, flnd:
a. sin(x - y) b. cos(x * y)
13. If sin a : -2l,,,B,tand> 0, cos B: -+, and csc B > 0, find:
a. sin(a - fl
b. cos(a + B)
14. lf cot A : -+, sec ,4 > 0, tan -B : t, and sin -B < 0, find
a. sin(l + -B) b. cos(l -,8)
15. If sinx: -J, tanx> 0, secy: -+, and coty < 0, find:
a. csc(x * /) b. sec(x - y)
sin(a
+
11. If sina<0, cos oa:$,cos p<0,
:
Establish the formulas in Problems 16 -20 by using addition formulas.
b. cos(z+0)- -cos0
16. a. sin(z+0)--sin0
17.
a.
18. a.
,,r(;.r-coso
,rr(*- o) :
b.
o b. .", (+ - g) :
\2 ) 19. a. sin(e - 6) - sin 0
zo. a. rrrfi-r)-coso
\2/\2/
cos
."'(;*r) :-sing
sin o
\2 / b. cos(n - A): -cos 0
b. ."r(i-r)-sing
Prove the identities in Problems 21-29.
21.
sin a cos p(cot a
-
tan
$:
cos(a
+ f)
P
_t ""t("-0-1+tarr.ta"P'r'
sin(a*f)+sin(a-f)
cos(a * p) + cos(a - p)
sin(a * B)
26. :.:
cot a + cot p
sln o( srn p
28. sin(x * y) cos y - cos(x + y) sin y :
29. sin x sin(x * y) + cos x cos(x * .y) :
..
"'
cos(o(+ 0l
n tand-tanf
tandtan
tuno+unB-sir4a+p1
-
cos(a-B;-cos1a+p)
*
p)
cos(a
-
B)
sln
sln l,
sin(a
27.
sin
cos
x
l,
o(
-:
B 30. Prove that, in general, sin(a * f) + sin a * sin B.
31.
Find all values
ofx for
sin(a-0)
which 0 < x < 2n and
sin 5x cos 4x
:
cos 5x sin 4x
-
sin(a
1
-
p)
+ cot acot
B
4/DOUBLE-ANGLE, HALF-ANGLE.
32.
Find all values
33.
Derive a formuler for
ofx for
x: 1 -
2 sin 2,x sin -t
a. sin(a+f+y) b. cos(a+f+y)
Hint.
Ure : r*;;1n
REDLCTNO\ FI]R,\[I-'tr..{S 329
which 0 < x < 2n and
2cos2x cos
in
A\D
First use the associative property for addition.
Prove the identities in Problems 34-36.
34. sin(a * B) sih(a - 0) : .irr' a - sin2 B
35. cos(a * B) cos(a - f) : cos' a - sin2 B
36r sin(a * B) cos(a - f) : sin o( cos o( + sin B cos B
im rhe ffi::mt[
: 4
Double-Angle, flalf-Angle, and Reduction Formulas
The importance of the addition formulas in Section 3 lies primarily in the fact
that so many other idehtities can be derived from them. we will carry out the
derivations for some of the most important of these.
The double-angle formulas are obtained from the addition formulas for sine
and cosine of a and p by putting ll : a.If we denote the common value of a
and p by 0, we obtain
sin(, +
0):;::
sin
0
sin 0 sin
0
;::: ;'o'g
So,
sin 20
Also,
cos(O + 0)
-
2 sin 0 cos 0
:
cos 0 cos e
:
cos2 0
So,
cos 20
-
-
sin2 0
= cos2 e -
sin2 0
(7)
Two other useful forms of cos 20 can be obtained by replacing, in turn,
by 1 - sin2 g and sin2 0 by 1 - cos2 0. This gives
cos2 0
ffil*
and
ffih
cos20:(I -sin2 q-
0- 1 -2sin20
cos 20 : cos2 0 - (1 - cos2 0):2 cosz 0
sin2
1
::
lin'$m
|liry]lr*
rffir d
rr
So we have
cos20=1-2sinz0
(8)
|
(e)
and
cos 20
= 2 cos2 e -
whether to use (7), (8), or (9) depends on the objective. we
some examples where a particular form is clearly preferable.
If we solve (8) foi sin2 g and (9) foi cos2 0,we obtain
sin2 0
-
l-cos20
wiil shortry see
330
8/ANALYTICAL TRIGoNoMETRY
and
- = I +cos20
2
cos'0
These forms are employed extensively in calculus.
By replaiing 0 by ul2 in the last two equations and then taking square roots.
we get the half-angle formulas:
,ofr
sm,
rt- [1
d,
cos,=
*ai[+cosd
z
:{
-cosd
2
The ambigirity of sign has to be resolved in each particular instance according
to the quadrant in which al?lies.
From the addition formulas we cari obtain a class of identities sometimes
called reduction formulas, of which (3) and (4) of Section 3 are special caser
Here are some others:
sin(rc
+
0)
tl,;;ot
0
+ cos tr sin 0
.o,
e
_
0
- sin
0
+
=
and
sin(n
_
q:
cos z sin o
:i, ;
Similarly,
cos(z
and
cos
(n
+
0)
:
.:X;'
- r, .:X;'
I
'*(|
and
sin
0
sin z sin 0
these, we have used the facts that sin z
si:n(3n12): - L and cos(3n12): 0, we also have
In
zz
: 0 and cos fi : - 1' Sine
* r) : rinf .o, o + cos !,in
: -cos o
,) : ,i, f
,r"(!t
:
g
"o,
o
- cos ! "i, o
-cos0
Likewise, tL"r" ur" similar formulas for cos[(32/2) t 0].
To generalize, it appears that we should consider two cases:
Case
1.
Functions of (nn
case
2.
Functions
* 0)
j
6l@!!l * 6]
L2
I
[x
4IDOUBLE-ANGLE, HALF-ANGLE. AND REDLCTTO\ FCR\IL
LqS 331
l
where z is an arbitrary integer. Note that in case 2. an odd multiple of r
nor*-be
is involved since (2n + 1) is always odd. The following idsntitiEs can
established.
B
rosilt$,
Reduction Formulas for the Sine and Cosine
Case
L.
Case2.
sffidrre
t g) :
t 0):
*
sin
0
*cos
0
,rrff-{:tcosg
coswt{:*sing
the
To determine the correct sign on the right, it suffices to determine what
mtirrnmr
rI
sin(ruz
cos(nn
cusfrr
sign is when 0 is aqute.
rs
These are called reduction formulas since in elch case- the given expression
right
the
on
function
1
the
in
Case
that
Note
reduced to a simpler expression.
ii1i" ,,o*" as that o1 ih. left, whereas in Case 2 the function on the right is
the cofunction of the one on the left'
gthers as exercises'
We wilt pgove the first of these formulas and leave the
when n is even'
1X
sinpe,
Secondly,
cos
first that iin nn
:0.
We note
nn:'(-
nz is coterminal with 0 so that the cosine is * 1 and, when n is odd, nz is
the
coterminal with z so that the cosine is - 1. Using the addition formula for
sine, we have
sin(run
l-
+
0)
Srfiuffi
:
-
sin nIT cos 0 + cos nn sin 0
(0) cos 0
+
(-
1)' sin
0
-(-1)'sin0
is
This shows that sin(nn + 0) : * sin 0. Furthermore, the sign on the right
g.
when
sign
the
determine
ijil.*ti"t, is independent of the size of ofSo0' if we
0 is acute,
EXAMPLE
11
Solution
it will
be correct for all values
Express each Qf the following in terms of sin 0 or cos
(, .|
: t sin 0
(Case
sin(3ru
- 0)
L.
sin(3ru
-
0)
/n
ror
L.
If 0 is acr1te,3n
b.
0.
c. ,ir(-;*,
1)
- 0 is in *1ff:;:1;jT - 0) is positive' rhus,
332
8/ANALYTICAL TRIGoNoMETRY
b. .",(; * ,) : tsin o
(case 2)
For 0 acute, q2 1- jis in quadrant II, and tt,.
*,(; * r) :
'
c. '* (-1* ,) :
*cos
g
-sin
.o.i,.
is negative there. So
o
(case 2)
For0acute,_nl2*0isinquadrantlV,wherethesineisnegative'So
'*(i*r):
EXERCISE SET
A
-cosg
4
In Problems 1-6 flnd sin 20 and cos 20'
1. sin 0 : ! and 0 terminates in quadrant II
2, cos g : -t *d 0 terminates in quadrant III
4' sec0:^'fandcsc0<0
3. tan|:Sandsin0<0
6' cscg:{"noi=e=3}
s. cotg: -2ald0<0<n
7.Ilsin0:xald_n1230<nlL,findsin20andcos20intermsofx.
8.
If
cos
x and 0 < 0 < z, flnd sin 20 and cos 20 in terms of x'
0:
In Problems g-12frnd sin 0 and cos 0'
g.
cosz
-
+and
ft <20 <2n
812
b.
-
-!^',L7r
<" =+
forrnulas to flnd
13. Use the half-angle
sin 1
srn20
12. tan20--+and3n<20<4n
-7r<20<0
11. secz1-+and
a.
10.
cos
c.
:
sin
75'
d'
cos 67'5'
In Problems 14-17 flnd stn(ul2) and cos(a/2)'
14. cosn
:*,0< a<ft
a.: -*,
/ -I
18. If P(0) : I
16.
tan
2n
a u .,-3n
2\
-.),
15' sinal7'
nnd P(20).
lg. If 0 < 20 < 7r and P(20) - ( - *,
T), flnd P(0)'
sec fr
-#) TtSa=+
- 3, -n 4d < 0
4IDOIJBLE-ANGLE, HALF-ANGLE, AND REDUCTION
FORMULAS 333
In Problems 20-22 express the given function ln terms of sin 0 or cos 0.
a.
20.
frsre-
sin(2n
I
21.
d.
tth€" fu,
- o)
b.
0)
"rr(y\2 /
d. cos( -n r 0) e. ,in (-i - ,)
\2
)
a. sin(g - n)
b. .o, ('l - d )
$u,
22.
cos(g
- 3n)
a. ."'(+ *
e.
;.
(; . ,
,) b. sin(z +
,i,(-;.I
d.
sin
e.
c.
/5n \
sin(r+0)
\-
't
,/
.l--1;
c.
stn
c.
cos(O
I
,)
\-
\l
i
g)
/
-
2n)
,",(, +)
In Problems 23-37 prove the identities.
23.
25.
27.
29.
31.
33.
r cos2x-l :SeC-r
,
r--2 cos"' x
cos20_ sin20
srn 0 cos 0
(sinx-cosx)':1-sin2x
tanz 0
II + tan'z 0:
2
r"tr
35.
-t'
<4m
37.
B
38.
sina 0 cot2
26.
,
I-COSZX^ -2cos2.r
29.
- sina x - cos2x
sin 20
-.--r^
SlN'U_T"*3tan0-tan?
2
30.
1+cos2x:sec2x
--2cot20
tana * cot a:2csc2u
32.
cos20
34.
2
,or'9
21
:-
cosa
isin2 20
2x
stnz
x
cot0*tan0
cot0-tan0
I
9"rr'e:
2
2(l + cos 0)
-:SeC1e
.0
2stn';*cos0
sin2
/.
36.
sin2g 1-cosg
l-tan2x
.---- r
l*tan'x -cos?x
0:
24.
:coSo
ta.o
Prove the reduction formula for each of the following:
a.
sin(nn
-
0)
b.
cos(nn
+
0)
,,"Y#.r] e. '*lq+!r r]
d.
*
c.
cos(nn
f.
,,,YTu.r]
0)
s. ,,,1*+r_r]
39. If tan 0 : x and -nlT < 0 < rl2, find sin 20 and cos 20 in terms of x.
40. If sec0:xand0<0< n,0lnl2,fnd:
sin20 b. cos20 c. sinj0 d.
^. writing 30 :20 + 0, find formulas for
41. By
of 0.
cosj0
sin 30 and cos 30
in terms of functions
334
8/ANALYTICAL TRIGoNoMETRY
42. Derive the formula sin 40:4 sin g cos 0 - 8 sin3 g cos g.
43. Derive the formula cos 40 :8 cosa 0 - 8 cos2 0 +
1,.
Prove the identities in Problems 44-50.
44.
cos4
3 cos 2x cos
=8*
z *
x=
4x
s
4x r-sin2 2x cos 2x
5 cos 2x
*
sinolr162168
1 - cos 2nx
: tan nx
sin 2nx
3 cos
49.
sin3x*sin ff:4sin
xcos2
50.
By calculating sin(zi 12) in two ways, using the half-angle formulas and using the
addition formulas, prove that
45.
46.
-
47.
x
49.
lz-V3:
ikl
sin
srn
30
0
cos
30 .
- /.
cos 0
sin3x+cos3x
slnx*cosx -
1
-I
sin 2x
6-Ol
-
3-
Further Identities
To obtain addition formulas for the tangent, it is necess ary only to use the
fact that tan 0
-
sin 0/cos 0. Thus,
tan(a
sin(a + f) _
+ B) _
cos(a + fi)
sin a cos 0
+ cos a sin f
cos n cos p
-
sin
ry
sin B
This can be improved by dividing numerator and denominator by cos a cos
tan(s
#
fr)
-
cos e cos
*
:
sinecosF*cosasinf
cos r cos p cos fl cos B
cosscos0_ sinasinf
or
tan(a
f
tan
fi')
--
B
cos a cos B
a*tanfi
1-tanatanfi
(10n
And in a similar manner,
tan(e
tan
u-tanp
- fr) = 1 + tan utan fi
(11
r
The double-angle formula for the tangent is obtained from (10) by takin-e
: fr. If we call this common value 0, we have
q,
tan(0 +
0)
tane+t'an0
1-tan0tan0
So,
tan 20 =
2tan0
1-tanz0
5/FURTHER
rDENrrrrES 335
We can derive half-angle formulas for the tangent as follows:
tan Lo-
|*
cos |e
sin
Iz
cos |z
2 sin
2
sin
cos
la
^ cos
2
Ia
lx cos la
2 cosz ta
Since 2 sin(a/2) cos (ut21
-
cos'
2(al2)
s1n
-
sin a and
'(PI:1*cosa
** -
we obtain
o
tan *o
;$i.: r
"'-il
:
sln d
1+cosd
If in this derivation
we had multiplied numerator and denominator by 2sin(al2)
2
instead of cos(2,2), we would have obtained the equivalent formula
1
tm!a.= --cos
c
sln 6
(See Problem 35, Exercise Set 5.)
uSe
.{,,
llt;
L,Lr'} *
We could obtain analogous formulas for the cotangent, secant, and cosecant
in a similar way, but these are so seldom used that we will not clutter up our
already formidable list with them.
We do choose to list one more group of identities known as ths sum and
product formulas for the sine and cosine. These can be obtained from the
addition formulas (see Problems 32 and 33 in Exercise Set 5). Although these
are used less frequently than the other identities we have considered, they are
indispensable at times.
Sum Formulas
sinA*sin B-2sin
i1
I
tt-rl
+
siny' - sin B-2cos
ryrrnry
A +cos B -2
rycos T
cos
,n
1L
cos A-cos
--''t-.*-.1 .t\.,l. .:
rycos
cos
A+B A-B
B- -Zsin2sm2
Product Formulas
sin a cos p
cos a sin p sin a sin B :
cos d cos P -
|[sin(a + p) * sin(a - il)
I[sin(a + p) - sin(a - p))
+[cos(a - p) - cos(a + p)]
f [cos(o( + P) + cos(a - P)1
336
S/ANALYTTcAL rRrGoNoMErRY
EXERCISE SET
A
5
1.
Use (10) and (11) to find:
5nn b.
a. tarlr,
tan
U
2. Find tan 20 if P(0t is the point e+,2OBl.
3. Find tan 20 if sin 0 : -i and cos 0 : -2.
4. Find tan(a + f) if sin o: t, o is in the second quadrant, cos p : -*, and B is ic
the third quadrant.
5. Findtan(a - fl ifseca : 3,csca < 0, csc P : -.t6,andcosB < 0.
6. If seca:f,sina<0,csc P:+,andcosp<0,find tan(a+ fr).
7. For a and p as in Problem 6, find tan 2a, tan 28, tan !a, and tan tB.
8. Find the value of each of the following without using tables or a calculator.
a.. tan
b.
105"
7T
tan
-t2
c. tan
lln
U
d.
e.
tan I95'
9. Find tan 20 and tan \0 if csc 0 : * and sec 0 < 0.
10. For the angles shown in the figures, flnd:
a. tan(e + 13) b. tan 2a c. tan(a - {J) d.
f. tan lB
e. tan tu
tan
,
tan 2B
In Problems 11-14 evaluate by use of the sum and product formulas.
11. a.
13.
14.
15.
16.
17.
t.
a.
a.
b.
a.
b.
sin
)rft
*
-11
sin
i-
1-
1l
ll
,7nTt
D. COS
-1l
,3nTt
D. COS
-
sin 105' - sin 15'
sin 105' sin 15'
Write as a product:
COS
88
-
b.
b.
cos 165' + cos 75'
cos 165' sin 75'
sin 5-t # sin 3r
sin 5-t cos 3r
Write as a sum:
Write as a product: cos 7.r - cos 5x
Write as a sum: sin 7x sin 5x
Derive the formula
cot(e
18.
-12
t2
If
cos 0
:
+fr)-cotucotfr--l
cot d. * cot B
6
x and sin 0 ) 0, f,nd tan 20 and tan t0 in terms of x.
In Problems 19-29 prove the identities.
19.
_-2
tan 2x
cotx
-tanx
20.
o
csc0-tan 1:
cot
0
Tr
6/TRTGONOMETRTC EQUATTOT\S 331
11
21. l-tan?
23.
25.
27.
tan 20 :
22. 1+tan atanl-secn
l+tan?
2
- sin20 ti.n-^ -0
2
2sin0+sin20sinl0 - sin -i0 :lan?
26.
cos 70 * cos 50
cot20-l
cot20:-
24.
2cot0
sin ,4
-
sin B
cosA*cos.B
cos 30
*
_ tan * r-l -Bl
cos 0
sin 30 * sin g
vvr -!ih
ar/..\t
-
'
r
28.
29.
B
30"
,,,(**g),',(;-r)
\4 / '.+
30p
2cos-sin--slnl:2:
.
\
{i
^
rantttn+0)
b. tu"[fZn-lt:
Z
:+tan0
I :-s.1 t
rd_.1
where the sign on the right-hand side can be determined by considenng
Prove the identity
sin 0 sin 2e + sin 30
*
cos 0
32.
1.I
-
Pror-e that
a.
31.
2 sin 0
*
cos 20
+
cos 30
-
H tcr
be acute,
tan 20
Use the addition formulas for the sine and cosine to derive the product formulas as
follows:
a. Add the formulas for sin(d + P) and sin(e - P) to get the formula for sin a cos B.
b. Subtract the formula for sin(a - fr) from that for sin(d + P) to get the formula
c.
d.
for cos a sin
B.
Add the formulas for cos(a + fi) and cos(a - P) to get the formula for cos n cos B.
Subtract the formula for cos(a + fr) from that for cos(e - P) to get the formula
for sin e sin p.
33. In the addition formulas for the sine and cosine make the following substitutions:
a + P - A and u * B - B. Solve these two equations simultaneously for u and fr in
termd of A and B, and obtain the sum formulas.
34. Prove the identity
,h
sin(r
*
lt1
-
sin
r:
sln --1/
taorl[.,
It\
.;)
2
35.
Derive the formula
a
L-cosa
tan-- _-.
2
6
sln d
Trigonometric Equations
The solutions of most trigonometric equations cannot be obtained exactly,
and we have to settle for approximate solutions obtained through some numerical procedure, usually with the aid of a calculator. There are, however,
enough such equations for which elementary techniques will yield exact
answers that soine time devbted to these techniques is justified.
338
8/ANALYTICAL TRIGoNoMETRY
In order to make clear the sorts of solutions we are looking for, let us consider the very simple equation
2sin0
- 1 :0
The problem is to discover all real numbers 0 for which this equation is true.
We write the equation in the equivalent form
sin 0
and then rely on our knowledge
satisfied if, aqd only if,
:;
of special angles to
0
conclude that this is
Tc 5n
- 6,
6
or any other number obtained from these by adding integral multiples of 2r
to each. To see this, remember that the sine of a number 0 is the y coordinate
of the point on the unit circle that is g units along the arc from (1, 0). Thus.
*. u." seeking those numbers 0 for which the y coordinate of P(0) : !.
There are only two points on the unit circle with y coordinate !, and our
knowledge of the 30"-60' right triangle tells us that these points ate P(n 6t
and pl5nl6); see Figure 12. Since P@): P(0 +2nn), we conclude that all
solutions of the equation are given by (nl6) + 2nn ot (5n16) + 2nn.
Figure
12
"(?)
: (+
;)
EXA\[F
it is sufficient to give only
the primary solutions, that is, those
lying between 0 and 2n. We would know then that all other solutions are
obtainable from these by adding multiples of 2zr.
We can usually condense the above reasoning as follows:
Usually,
1.
2.
Determine the appropriate reference angle.
By the sign of the function, locate all primary angles having this reference
3.
Write the answer as the radian measure of these angles.
angle.
In our example, since we know that sin 30" :!,the reference angle is 30'.
The sine is positive in quadrants I and II, so the angles are 30'and 150". In
radians these are nl6 and 5n16.
It might be useful to review at this time the values or the functions of 30=45', and 60'.
6,TRIGO\O\IETRIC
le[
r-
emr*
l.l,5
tkom N
sin 30"
-
cos ,9"
:
.77
Slfl-:
6
7T
cos
"m-{.}c
tan 30o : tan
1
,
tl3
6
2
7T
---
6
45'
-
sin
cos 45"
:
cos
sin
T
tan 45" - tan
lt
7T1
EQrn-
.\]tO\S 339
\-
n
arr
4lz
n1
4lz
It
1
4
-
sim,
-:
cos 60'
:
ccs
-:-
tan 60'
:
tem
''
sin 60'
<!
-!!\
+l
11.
"
,lt
<ii
<tl
-\
i
a
Also,
t'ha[
rPr;s ]$
c
hipk"
lm
n morCrr,am
I
A1-e
Sln
cos0o=cos0-1
cos 90'
:
cos
tan 90'
ft
- tanl
:
0
TC
2
TT
2
rs
not defined
less
frequently and can always be obtained by the reciprocal relations.) If you feel
comfortable with radian measure by now, you can omit the degree measure
entirely.
If an equation can be worked around to a form qrch as sin 0 : * a, cos 0 :
* b, or tan 0 = t c, for instance, where a and b are any of the_numbers 0, 1,
i, tl@, or $12, and c is any of the numbers O,l,11u5, or.f, then we can
obtain all solutions in the manner outlined abovE. For other values of a, b,
and c, a calculator or tables can be used.
;p v ]lil
rhat
-
(The cotangent, secapt, and cosecant are not listed because they occur
-
#
sin 90"
tan 0o : tan 0
[1" 0n" l['hus+
Pq6'r
amd
orur
$,
f
sin0o-sin0-0
e"ll[
L
EXAMPI E
12
Findallvaluesqf 0,forwhich0 <0 <2n,satisfying 2cos2 0 - cos0
- 1:0.
Solution We treat this first as a quadratic equation in cos 0 and factor:
(2cos0+l)(cos0-t;:g
This is true if and only if 2 cos 0 + | :0 or cos 0 : l,
$
is"
fiutioms &rc
lr referemcs
thatis, if and only
if
cos0:-1 or cos0:1
thome
:
Since cos 60' *, the reference angle in the first of these is 60". Since the
cosine is negative in quadraqts II and III, it follows that we are seeking the
angles 120" and 240'.In radians these are 2nl3 and 4n13. The only primaq
0o, or 0 radians. Thus, the primary solution
angle for which cos 0 = 1 is
set for the equation is {A,2n13, 4nl3}.
0:
qgle I"s 30r'.
nd 150' fln
Remark. In many equations involving trigonometric functions the variable
must be treated as a real number, which is the reason for writing the above
answers as the radian measures of the angles. Recall that the values of sin 0,
of
cos 0, and the other trigonometric functions are unchanged whether we consider 0 a real number or the radian measure of an angle. An example of a
mrs
$-lt'.
340
8/ANALYTICAL TRIGoNoMETRY
mixed algebraic and trigonometric equation will perhaps help to make this
, point clearer. The equation
2sinx-x:0
is clearly satisfied when x :0, but by trial and error, using a calculator (or
by some more sophisticated numerical procedure), we can find that another
solution is
x = 1.8955
of an angle, then the degree measure of the angle is approximately 108.6'. Now it would not b,e true thar
x : 108.6'; in fact, it would not even make sense, for we would have
If
we think of this x as the radian measure
2 sin 108.6'
-
108.6'
:
0
which is a totally meaningldbs statement.
EXAMPLE
13
f
Find all primary solutions of sin 2x
x:0.
cos
Solution This time we will omit most discussion and go only through the
would be expected in doing the problem.
sin2x*cosx:0
2sinxcosx+cosx:0
cos x(2 sin
COSX:0
-L
1._
x * 1) :
2sin x-
7T 3n
)'
EXAMPLE
14
l*
2, 71 6.3n 2,
Find all primary solutions of tan
d"
-1
sinx- -12
2
v-
The solution set is
0
7n lln
6'
llni6't.
-
3 cot
a:0.
tan d,-3cotd-0
Solution
,
The reference angle is 60',
tafld"-
3
r^"r-o
tanza-3:0
tan d,- t^rf
or 7Tl3 radians.
Therefore,
2n 4n 5n
N-1, T,
T,T
6
steps that
fI
6/TRrGoNoMErRrcEeuATroNS 341
[o make
Since we multiplied by an unknown expression, namely, tan a, it is essential
that we check to see that this was not zero for the values of s found, which
is clearlythe case here, since tan
0. So the solution set is {z/3,
2n13, 4n13, 5nl3\.
tlus
a: t"l1+
catrculator (or
uhat another
EXAMPLE
r degree meabe true thar
l5
Find all primary solutions of
Solution The fact that
cos n
:2
+
sin x.
x : | - sin2 x suggests that if we square both
it will be easier to work with.
cos2
given equation,
I have
sides
of the
3cos2x:4+4sinx+sin2x
: 4 + 4 sin x + sin2 x
4sin2 x -14 sin -r + 1 :0
(2 sin ..r * 112 : g
sinx: -*
7n lln
^- 6' 6
3(1
the steps
.f
thaT
-
sin2 -t)
We again must check our answers, because squaring does not necessarily lead
to an equivalent equation; it may introduce extraneous roots. So we check in
the original equation.
When x :7n16, we get
r(-*)12+(-il
33
-r=,
So 7nl6 is not a solution.
When x
-
llnl6,
*(+) r 2+(-1)
33
,:,
So
EXAMPLE
16
Solution
x - 1lnl6 is the only primary solution.
Find all primarisolutions of 2 sin 30 - .,6 tan 30 :
2sin30-
.f
sin 39
cos 30
0.
-0
- .fl - 0 (We multiplied by cos 30.)
sin39-0 | .or 30:+ (So we did not multiply by zero.)
sin 3g(2 cos 30
342
8/ANALYTTcAL TRTGoNoMETRY
Now we want all values of 0 lying between 0 and 2n. We must therefore find
all values of 30lying between 0 and 6n. In general, if n0 is involved, in order
to find all values of 0 between 0 and 2n, we find all values of n0 between 0
and 2nn, and then divide by n. We have from sin 30 : 0 that
30
: 0,
TE)
2n, 3n, 4n,
5n
and
0:0, ;,+,
From cos 30 :
,ll
Tc,
+, +
lZ, we have
lln 13n 23n 25n 35n
30: rE
6' 6' 6' 6' 6' 6
TE lln l3n 23n 25n 35n
0- 1g'
18' 1g' 1g' 1g' 1g
In both
cases we found the first two values
then 2n again. The complete solution set is
7E
{0,
EXAMPLE 17
#,
5'
of
30, then added 2n once, and
lln 2n L3nn' 23n 4n 25n 5n 352)
1g 'T' 18 '
1g 'T' 1g 'T' 18
-\
J
Find all real solutions to tan x
-
cot
x:2.
tanx-cotx:2
sinx_cosx_,)
Solution
COSX SlnX
sin2 x- cos2
2x:+,
3n
g)
x- 2sinxcosx
- cos 2x - sin 2x
tan 2x: -1
7n lln L5n
(We "went around"
T, 4 , 4
7n lln 1,5n
T' 8'
twice.)
8
As none of these values of x makes sin x, cos x, or cos 2x eqtoal to zero, we
did not multiply by zero. We must also check to see that we did not lose any
roots when dividing by cos 2x;that is, we must check to see if any of the
roots of cos 2x: 0 are roots of the original equation. Since cos 2x : 0 when
2x : nl2, 3n12, 5n12, 7nf2, and, so when x : nl4, 3n14, 5n14, 7nf 4, we check
and flnd that none of these is a root of the original equation. So we have
found all primary solutions. To get a// solutions, we add arbitrary multiples
EXERCI
6rrRrcoNoMETRrc EeuArroNS 343
herefore flnd
YEd.
d
of 2n to each of these. So the complete solution set is
in order
7n l,ln l5n l9n 23n 27n 3ln 35n 39n 43n 47n
!v ' T'
g' g' g' g' g' g' g' g' g' g'
between 0
ls
II
)
or we could write
\3n18
*
2nn,7nl8
*
Znn,
llnl9 *
2nn, l5nl8
* 2nn: n:0, +1,
+1.
We could continue with examples, each possessing its own special features,
but these illustrate the main techniques. It might be useful to summarize some
of the points that the examples were meant to bring out:
1.
h'once, and
2.
'I
x: *b,
and so on.
Find the radian measure of all primary angles (that is, 0 < x < 2n) satisfying the elementary equations in 1. If all solutions are desired, add arbitrary multiples of 2n to each of these to obtain the complete solution
set.
4.
EXERCISE SET
A
to zero. wE
m( trose an!'
o[
0$rE
!; :0 wfo66
h me chesk
io
cos
3. If in arriving at the elementary equations both sides of an equation were
J
anr-
Use algebraic techniques such as factoring, multiplying by the LCD, and
squaring in conjunction with basic trigonometric identities to simplify so
as to obtain one or more elementary equations of the form sin x : * a,
rr.e liar e
ry muluiptres
squared, or if both sides were multiplied by an expression containing a
variable, it is necessary to check the answers. (In the case of multiplying
by an unknown, it is sufficient to check that the multiplier was not zero.)
If both sides of an equation are divided by an expression containing a
variable, then the values of the variable for which this expression equals
zero must be checked in the original equation to see if they are solutions.
(In general, such division can be avoided by factoring.)
If the final elementary equation(s) is of the form sin nx : * a, cos ,4-r :
+b, and so on, then to get all primary values of x, find all values of n-r
between 0 and 2rut, and divide by n. This amounts to finding the angles
in the first revolution and then adding2n a total of n - 1 separate times
and finally dividing everything by n.
6
Find all primary solutions unless othenvise specifled.
L. 2cos x-J3
3. secx -2
5. cosx-2coszx:0
7. cos2x*cosx:0
9. sin2/-cos2t-l
11. 2cos x- cotx-0
13. 4tan2a:3sec2e
15. cos2 x- sinx*5:0
2, cotx*L:0
4. sin2 x - 1 (Give all solutions.)
6. sin20-cos0:0
8. 2tanrlr-tan$secr/-0
10. sec20:2tan0
12. 2cosz x+ sinx- 1:0'
14. sin2 x- 2(cosx- 1)
16. tan 2x - cos 2x * sec 2x - 0
r
344
8/ANALYTICAL TRIGONOMETRY
x
:
17. 1, -.cos
smx
19.
sin 60
*
21. tan!*
2
sin x
sin 30
cos
x
:
:
0
B 26. sin g * cos 0 :
cos
20.
sin2 4e
-
cos 20
sin 40
+1
-0
- 2 -0
2
:
24.
O
2cos
x-
cotxcscx
-0
| (Give all solutions.)
sin x
:- cos x
28. 1-cosx
1+sinx
30. 13 tan 0 :2 sec g -
\F cos 2x - 0
29. sin 20 - 4sin 0 : 3(2- cos 0)
27.
2 cosz 20 - 3
22. sin2!-sin2x:0
I
23. 2(sinz 20 - cos2 20) + f
25. 2tan2g+sec0+2-0
18.
2x tan 3x +
1
In Problems 3I-36 use a calculator or tables to find all primary solutions correct to three
significant figures.
32. cos2x+2cosx-0
Y. 3sinx-tanlx-0
36. secx-2-L0cosx
31. 12sinzx-sinx-6-0
33. 2sec2x:5tanx
35. 2sinx-cscxf 3:0
In Problems 31 - 3g find all primary solutions without using a calculator or tables.
38.
37. tan3 x*tan2 x- 3tann-3:0
39. 2sin3x*3sin2n-1:0
7
Graphs of the Trigonometric Functions
To obtain the graphs of y : sin x and.),
8sina
0-Zsin2 e-3 -0
:
cos x we make use of the definitions
of the sine and cosine as the ordinate and abscissa, respectively, of a point on
the unit circle (Figure 13). Notice that we have used r to represent arc length
on the circle from (1,0) to P(x) rather than the customary 0. We have used
capital letters to label the axes. Table 1 indicates how sin x and cos x vary as
x varies from 0 to 22. These facts can be verified by visualizing the point P(xl
moving around the circle in a counterclockwise direction, starting at (1,0).
Furthermore, it should be clear from considering Figure 13 that the ordinate
Figure 13
P(x)
(cos x, sin
7/GRAPHS OF THE TRTGONOMETRTC FUNCTTONS
Table
345
I
As x goes from
A to nl2
nl2 to n
n to 3ny2
3nl2 to
2rc
sin x goes from
0to
1to0
0to -1
1
-1to0
cos
x goes from
1to0
0to -1
-1to0
0to
1
and abscissa of P(.r) (that is. sin -x and cos .r) \'ar)- in a uniform way, with no
breaks or sudden changes. as P(-rl mo\-es around the circle.
This information. together with our knosledge of sin.r and cos x for x :
nf 6, nf 4, rl3 and related values in the other quadrants. enables us to draw the
graphs with reasonable accuracy. Since sin -r and cos -r each has period 22,
the graphs repeat every 2n units. So once we know the graphs in the interval
from 0 to 2n, we can extend them indefinitely in either direction. The graphs
are shown in Figures 14 and 15.
Dfus
Figure 14
UilTMS
ist
Figure 15
om
@iltu
ruM
Ey &$
,fro,,,
;[,
(illttr"
lire
The maximum height attained by the sine curve and by the cosine curve is
calledtheamplitude. So, fory: sin xandy: cgs xtheamplitudeis 1.
To obtain the graph of y : tan x we again refer to Figure 13 and use the fact
that tan x : sin x/cos x. Therefore, we consider the ratio of the ordinate of
P(x) to its abscissa as P(x) moves around the unit circle.
When x : 0, tan x : 0ll: 0, and as x increases toward nl2, the ratio of
the ordinate of P(x) to its abscissa steadily increases, taking on the value 1 at
x : n 14. For x near n 12 the ratio becomes very large and gets larger and larger
without limit as x approaches.nl2. At x: nl2 the ratio is not defined. For x
slightly greater than nl2 the ratio is negative, since the ordinate is positive and
the abscissa is negative, but its absolute value is large. At x:3n14 the ratio
346
8/ANALYTICAL TRIGONOMETRY
is
-
I, and at n, it is back to 0. Since
tan(x * n) -
tanx*tann
1-tanxtann
Fis
tanx*0
1-(tanx) '0 =tanx
it follows that tan x has period n. So we can draw its graph between 0 and n
and duplicate this in each succeeding interval of length n. Similarly, we can
extend it to the left. With this information and our knowledge of tan x for
special values of x such as z/6 and nf3,we can draw the graph, as in Figure 16.
The vertical lines at odd multiples of nl2 are asymptotes.
Figure 16
ris
Figure 17
)/-cotx
The grapfu of y
Figure 17.
:
cot x can be similarly obtained and we show ils graph in
ofx is defined as the reciprocal ofthe xcoordinate ofP(x), that
of cos x. Since lcos rl < 1,it follows that lsec xl > t. ttre
reciprocal
is, as the
x are the same. When cos x : l,sec.x : 1, and when
x
cos
of
sec
and
signs
:
This latter occurs at x : nl2,3nf2, -n12, 5nl2x
is
undefined.
.x
0,
sec
cos
gets
cos x gets close to 0; so sec x gets arbitrarily
nf2,
to
x
close
on.
As
and so
with the values of sec x for x : nl6, nl4, nl3.
together
analysis,
this
large. Using
Figure 18. The graph of7: csc x can be
graph
in
the
on,
we
sketch
so
and
similarly obtained (Figure l9).
Both sec x and csc x have a fundamental period of 22. No amplitude is
defined for them. They each have asymptotes; for the secant they are the lines
x:n12, -n12,3n12, and so on, and for the cosecant the lines x:0, ft, -fr,
the
secant
2n, and so on.
7/GRAPHS OF THE TRIGO\O\IETRilC
F-{--\CTIC\S il7
Figure 18
:gnmemC
rly.
U
IV
r"
2n
\,n-e .,eu.
- 5nl2
lf, tam "r icr
I Figr,lne 16
- 3nl2
- 7T12
3n12
5ru1
Figure 19
Figure 20
AmplituJ.
-,
Tc+-x
n12
f
gne,pr
P(xt"
Iad
rirl
E4
Period
-;l
Amplitude
-
1
i Period---l
f Period->
rm
1
Period
::irrrii:
w;'klu'
_\il
a
rtiw"--u
,a[,{r3.r cfllr
:(8
dinme
r$
what we have shown are the basic graphs of the trigonometric functions.
You should become especially familiar"with the fust three. The crucial
facts
about the sine curve are illustrated in Figure 20. This, together
with the knowl_
edge ofthe periodicity, enables us to obtain a rapid skJtcn
of the curve.
For the basic sine curve /: sin x, the fundamentar period is 2n
and the
amplitude is 1. Now we want to consider the effect of introducing
positive
constants aand.b:
the
hs$
I m- -n-
Y- astnbx
we will take one at a time. First consid er y q sin x. This has the
effect of
multiplying every y value of the basic curve by o. So when
the basic curve is
at its maximum height of 1, the new curve will be a
units high. Thus, the ampli-
348
8/ANALYTICAL TRIGONOMETRY
Figure 21
! :
Amplitude
A
sln x
+
tude becomes a, and the period remains unchanged, as shown in Figure 21"
We have exhibited the curve for one period only (also called one cyclQ, since
the extension is obvious.
Next, let us consider./
:
sin 6x. We know that the sine curve completes one
22. But then n goes from 0 to 2nlb. So,we conclude
to
from
Znlb.
This can also be seen in another way. To sa1-.
period
of sin 6x is
that the
that sin x has period 2n means that sin(x + 2n) : sin x. Thus, sin(6x * 2n) :
sin 6x. But sin(bx 1- 2n) : sin 6[x + (2nlb)f. So,
cycle when 6x goes
0
/
sin
a(x
r-\
.+):
sin 6x
Thus, when we add 2nlb to any x, we get the same value as sin 6x. That is.
sin bx has period 2nlb.The effect of the multiplier b in this position, then. is
to alter the period; it is shortened if b > I and lengthened if b < I (Figure 22t.
At'
t'
Figure 22
-)'
.\mplitude
I
-
1
:
sin bx
2nlb
nlb
7rl2b
Remark. When the variable x represents time, the period is the time required
to complete one cycle. The reciprocal of this, called the frequency, gives the
number of cycles (or fraction of a cycle) completed per unit of time. Thus,
Frequency:
rb :
Period
t
This is a widely used concept in electronics.
Now we can put these two changes together to get the graph of y : a sin D-t.
as shown in Figure 23.If , in addition, a constant k is added to the right-hand
side, giving ! : a sin bx * k, this has the effect of shifting the entire graph of
! : a sinbx vertically by k units-upward if k > 0 and downward if k < 0.
We consider next an equation of the form
y-asinb(x-a)
(11
n
7/GRAPHS OF THE TRIGO\O\IETRIC
FU\CTIO\S y9
Figure 23
Amplitude:
a stn bx
a
I
v
f Period
ilr,"0---l
n Figure
Period
1
Period
a
*
b(*
ome
-
- 2nlb---->
a) equal to 0 gives
b(*
- a) x* a-
Tia)-. To say
-
-
I
The sine will complete one cycle when b(*
we conclude
(Dr
Amplitude
11.
crctrg, simm
mpletes
]
arfl
:
-
a) goes from 0 to 2n. Setting
0
0
X-(X
and setting b(*
-
a) equal to 2n gives
b(*-a)-2n
bx- That us"
[bn, them. is
(Figure -l I
x- O:
b
x: 2n
+d'
b
So a complete cycle occurs in the interval from ato (2nlb) * a. This is a distance
of 2nlb, so the period of 2nlb remains unchanged, but the curve is shifted a
units horizontally. If a > 0, the shift is to the right, and if a < 0, it is to the
left. we call a the phase shift, and say that the curve is lal units out of phase
with the curve / : a sin Dx. This is illustrated in Figure 24.
Finally, we note the effect on the graph of ! : a sin Dx if either a or b is
negative. If a < 0, every y value is the negative of what it would have been if
a were positive. For example, in y : -2 sin Jr, every y value is the negative
of the corresponding y value in y :2 sin x. So the effect is to flip the graph
of y:2 sin x about the x axis. lf b < 0, we make use of sin(-0): -sin 0.
irc requeuud
ftr glvm thr
m-
2n
Thu"'.
Figure 24
!:
a sin
b(x
-
(2nlb)
f-dsilrlSru
c right-h*qr*dr
lirc graph str
rd if& < #
1
r Xltn
a)
Period
-
2nlb
+a
350
8/ANALYTICAL TRIGoNoMETRY
For example, we would write
y:
sin(-2x)
/:
as
-sin 2x,
and proceed
as
above.
A srmilar analysis to that for the sine holds for the cosine function (se
Figure 25). Note, however, for the cosine that cos(-0): cos 0, and so the
graph of y : cos(-2x) would be identical to the graph of ./ : cos 2x. with
appropriate modiflcations because of the different period (for the tangent and
cotangent) and lack of amplitude, the other functions, too, could be analyzad
in an analogous way. You will be asked to consider certain of these in Exercise
Set
7.
Figure 25
!:acosb(x-o()
Ampli
(Znlb)
+
u
x
It rfl
1 Period
shift -
E,XAMPLE
18
Sketch the graph of
y
:3
a
sin
-
fx.
Solution The amplitude is 3 and the period rs 2n +
Figure 26
EXAMPLE 19
+
Sketch the graph of
y
-
2nlb
+
-
4TE
(Figure
26).
l
2 sin(3x
-
TE).
Solution We first factor out the 3 to put this in the form of equation
(12):
/n\
sin'f
-,
"r:2
This is therefore a sine curve with amplitude 2, period 2nf3,and phase shifr
z/3 units to the right (Figure 27).
n"t qu
7/GRAPHS OF' THE TRIGONOMETRIC FUNCTIONS
d proceed as
fuilction
3s1
Figure 27
(see
t, and so the
cos ?.r" Wiffi
e
tangent and
il be anallzed
se in Exersise
EXAMPLE
20
Solution
Sketch the graph of
y-
cos12* + (nl3)f.
We first write the equation in a form analogous to equation (12):
./:cos
So the phase shift u
The amplitude is
l,
rQ*I):"or
r?-( .J
is - nl6. The curve is therefore shifted zr/6 units to the left.
and the period is 2nl2: z (Figure 28).
Figure 28
2t:
EXAMPLE 2I
Solution
rnd phase shift
Sketch
! :2tan$x.
There is no amplitude defined for the tangent curve, but the effect of the coefficient 2 is to multiply all y values of the basic curve by 2; in particnlar, when
the basic curve is I unit high, the new curve will be 2 units high. Since the
fundamental period of the basic tangent curve is z, the period of this curve
is z + | :2n. We show two complete cycles in Figure 29.
352
8/ANALYTICAL TRIGoNoMETRY
Figure 29
8
EXERCISE SET
The Invet
7
In Problems l-25 sketch one cycle of each of the curves. Give the period and, where
appropriate, the amplitude and phase shift.
A L. y-stn2x
3. y-2sinx
5. ! :3 cos(zrxl2)
7. !:2sin(-3x)
9. y - -sin(xl2)
11. y- sin2x*
13. y-2sin3x-l
15. !:ttannx
) !: cos 3x
y -2stn3x
Z.
4.
6.
*
8.
10.
! : -2
cos
x
12. !:2 cos x - 3
14. y-3cosnx*2
16. y - 2 cot(xl2)
1
B
:
/
z\
-;)
17.
!:rrr(r-;)
19.
y
19.
!:sin 2(,.;)
20.
!:cos 3(,.;)
22.
y-
24.
/
!:!tan(r,
21. y-2sin(3x+2)
23.
!:1,-2rrn(\ n*-+)
4/
2s. y_ 2+2.or(r,
26'
a'
Figpe
y - 4 sin(nxl3)
y :3 cos( -TEx)
cos(x
3cos(3
-2x)
z\
-;)
)
il,t*i'f'ill3""1.f
dividing by
",trTTE
GTtr
show that a sin x
sin(x + 0)
where 0 is as shown in the accompanying sketch.
* b cos x can be
8/THE INVERSE TRIGONOMETRIC FUNCTIONS 353
b.
Use the technique of part a
to
!:
What is the value of 0 in this
-*J
6,.1-r,lsl
\\ ,lIC. "--L---
of
sin x + \rc cos x
case ?
27.
By the technique of Problem 26,parta, write each of the following as a sine function.
Determine the amplitude, period, and phase shift.
29.
Discuss the effect on the basic secant curve of introducing positive constan ts a and
b to obtain y - a sec bx.
29.
Sketchy-2sec(xl3).
L. y-sinx-cosx
8
analyze and sketch the graph
b. !:3cosx-4sinx
The Inverse Trigonometric Functions
Let us look again al the graph of .r' : sin .t (Figure 30). It is immediately evident
that this is not a one-to-one function. so it has no inverse. However, by a
suitable restriction on the domain of the sine. an inverse can be found. The
standardchoiceistorestrict,.r-sothat -r 2 < x < r 2. Thesinecurvewiththis
domain will be called the principal part of the sine cun,e.
=
Figure 30
L3nl6
Ifx is so restricted, then for each y such that | < y < r, there is exactly one
r such that y: sin x. For exampie, if we take-y : L, we get x : nl6. On ttre
other hand, if y: -t, then x : -i16. The equation :iin x expresses y in
"y
terms of x; we would like to solve this equation for x in terms oly. Unfortunately, we as yet have no way of doing this. with the aid of the graph (or
tables or a calculator), we can find x for any given y, but we have no simpte
equation that expresses x in terms of y. what we do is invent a symbolism,
;::
-
x=sin-1"/
Jrr
which is real,o'x is the inverse sine
ofy." Actually, we should probably read it
"principal inverse sine of y," since it is the principal part of the sine curve
that is used in finding x. Unless otherwise stated, we will in the future understand
that sin- 1 y means the principal value.
as the
There is another symbol in wide use that means the same thing as
itis
sin-1y;
arcsinyrread"arcsineof y." Itsoriginprobablyliesinthelengthof arcon
a unit circle used in the definition of the trigonometric functions. Thus, arcsin
I might be interpreted as "the length of arc on the unit circle for which the sine
354
8/ANALYTICAL TRTGoNoMETRY
ir 1." We know the length in this
case is zr/6, so
.TE
sm6:,
1
7T
and
arcsln
6
1
,
are two ways of viewing the same fact, The first says that the sine of the number
nl6 is t. The secortd says that nl6 is the number whose sine is |.
In what follows we will use both notations for the inverse sine (as well an
ur
inverses of the other trigonometric functions), since both are in wide use.
consider some examples.
kt
EXAMPLE
22
Solution
Find the value of:
a.
sin-1
a.
sin-
1
11
:
b.
sin-'(-r) c. sin-10
nl2, since nl2 is that value of x on the principal part of the
curve whose sine is
sir
1
b.
sin-11-j1
:
-116
It is important
here to note that the answer is not 112/6, though an an_el,.
of llr,6 radians and an angle of - nl6 radians are coterminal. The distinction is most clearly seen by looking at the graph of the sine curve. On tbc
.r axis. - tt16 certainly is not the same point as llnf 6, and even though
sine of each is - l, we choose - nl6 since it falls within the restricted rangr
on -r that defines the principal part of the sine curve.
tb
sin- 1 0
EXAMPLE
23
0
Evaluate:
a.
arcsin
J3
2
Solution
-
a.
arcsin
th :n
23
b. arcsin(#)
b.
arcsin
(#) : -4
-
c.
arcsin(
c.
arcsin(-1)
7T
1)
: -:
2
To plot the graph of the inverse of the principal part of the sine curve, $e
follow our usual procedure with inverses of solving for x and then interchanging
x
and y, giving
-Y
:
sin-1 x
Now, since the roles of x and y have been switched, we must restrict x so that
-1, <x < 1 andy will fall in the range -nl2 <y <n12. The graphis just the
reflection of the principalpart of the sine curve in a 45' line through the origin
(Figure 31). Any value of x between - 1 and 1 corresponds to a unique value
8/THE TNVERSE TRTGONOMETRTC FUNCTTONS
355
Figure 31
n
- Utl'
rrmLttr
-1
Tcrc
\
I
rl2
; Urrel .tS
s-
Ler ;s
1
_nl4
I
_1
1l
(x(
\-"lz SyS "l2l
- 7T12
&E
of y between -nl2 and nl2.When x is positive, y is positivg and when x is
negative, y is negative. Notice that y: sin-1x if and only if sihy: x and
1
-nl2 < y < nl2. So ryhen we wish to evaluate sin- x for a particular x, we ask
what number (or what angle in radians) between -nl2 and nl2has x as its sine.
We consider next the inverse of the tangent function, since it has much in
common with the sine. The principal part (or principal branch) of the graph of
! : tan x is that portion between -nl2 and nl2 (Figtre 32). On this portion,
$,yr#r
m m$Iffi
:
dimrumu-
if any value ofy is specified, a unique value of x is determined. Solving y
for x leads to either
-&ffi.tc
ryfumr
g:tan-r y
ud rmrryrr
or
:
tan
x
x - arctan y
Figure 32
,m
.rr
Again, we interchange the roles of x and )), and write
m"mu
!:
tan-1 Y
or
|:
atctan x
hrrouu*ilqm
and restricty such that -nl2 < y < nl2. The graph of this function is shown in
Figure 33. With our knowledge of special angles, we conclude, for example, that
"milfuwl
$m dfihn:
E@mr
rc mdihro
tan -1 , -fi
4
tan- 1(- 1) :
-! 4
_1
tan
I
G7T
VJ :; J
tan-t
g
-
0
356
siANALYTICAL TRIGoNoMETRY
Figure 33
A consideration of the graph of the cosine function will make it clear thar
some different portion will have to be used as the principal part (Figure 341
because when x lies between -nl2 and nl2, y is always positive and hence do*
not assume all of its possible values. Furthermore, for a giveny there is generalll'
not a unique x determined. We choose instead the portion of the curve lying
between 0 and a as the principal part. Then, for eachy such that 1 < y S 1there is a unique x for which y cos x. Again, we express the dependence of -r
on y by writing
-
:
x-cos'y
_1
x - arccos y
or
1
Interchanging the roles of x and y, we obtain the graph of y : cos - x (Figurt
35). When x is positive, y is positive and between 0 and nl2; when x is negatir-e.
y is positive and between nf 2 and n.
Figure 34
v
Figure 35
ft
[-t(x(
I o<v< t)
2nl3
n12
_l
I:cos'x
n14
- 1-
Llz
u,l1
The inverse cotangent is seldom used, but is defined in much the same way
in Figure 36.
The secant and cosecant present certain problems, and there is no general
agreement on what parts of the curves to invert, that is, which portions to define
as the inverse cosine. Its graph is shown
8/THE TNVERSE TRTGONOMETRTC FUNCTTONS 357
Figure 36
!-cot
--r
i
lhemce Coes
is gemeraltr3
fl-
ndeunce ofr'r
I r' ilFigure
ir negaive-
-;;.;r
o<v<n
I
:
crrrr-e hrmg
-I < l'<
^x
as the principal branches. Fortunately, this lack of agreement is not serious. The
inverse cosecant is seldom used, and since it can always be circumvented. ue
will omit it entirely. The inverse secant is sufficiently useful to warrant consideration. For purposes of later use in calculus, there is some justification for the
choice of principal parts made below. Let us consider first the graph ofy
sec -r
(Figure 37). No single branch of the graph is suitable for defining the inverse
function. While the reason is not now apparent, we select the highlighted portions in Figure 37 to define the inverse function. Thus, when
we take
0 <x < nf2,and wheny < -1, we take TE<x <3nl2.When we interchange
the roles of x and y, we obtain the graph shown in Figure 38.
ft cilear ruhar
(Figure 1S*r
_1
y2l,
-
Figure 37
- 3nl2 - nlz
irlz
TE
3nl2
A
lAl
Figure 38
,'
7q
11
\.
-l::'- 0<y <nl2
[*rr,
[x< -l,n<y<3n12)
--TE12
-1
EXERCISE SET
Etr
A
EBmemaI
Btodefinc
8
In Problems l-32 give the
Fme wlar
2n
-1
1. a. sin- 1(- ,fTtzl
2. a, tan- '(- 1)
3. r. arccot( - 1)
value.
b.
b.
b.
arccos(-V3i2)
arcsec(
sin-1
1
-2)
]
358
8/ANALYTICAL TRIGoNoMETRY
4.
5.
6.
7,
8.
9.
a. arccos I
b. tan-1 I
a. arcsec I
b. sin-1(-)
a. arctan0
b.
cos-11-1;
r
a. arccos0
b. sec-L(-21"f3)
u. cos d lf a: sin-1(-f) b. tar0 if 0 : arccos(-lo)
a. sin x if x : sec-1(-3; b. sec 0 if 0: tan-1 f;
10. a. sin[cos-1(-$)]
b. sec[sin-1(-1n)]
b. cos[tan-11-$]
11.
tan(arccos!)
12. ^.
a. sin(arctan fi)
b. csc[arccos(-zJ]
13.
cos[sin-1(-])l
b. cot[cos-1(-fr)]
14. ^.
sin(a - p) if a : sin-r t and P: cos-1(-:13)
15. cos(a + B) if a : arctan(-l) afi P : sec-1 i
16. tan(u+ $ if a: sin-11-$; and p :cos-1(-p1)
17. sin20 if 0 : cos-l(-t)
18. cos 20 if 0 : sin-1(-+)
19. tan20 if 0 : sin-'(--5-rJ
20. cos[2 cos-l(-!)]
21. tan(2 tan- t t)
22. sin[2 sin - 1( - f)]
1(-f)]
23. cos[2 sin25. cos[] arcsin( - f)l
n.
sin g and cos 0 if 20
24. sin$ cos - 1 fi)
26. tan +[tan - 1( - )]
125
: tan-r(-at)
28. sin(al2), cos(al2), and tan(al2) if a :
29. sin(sin-1 {f + cos-'!;
31. tan(tan-1 l+tan-L 11
1
cos - 3
30.
cos[cos-l S
- sin-l1*fi;]
B 32. tan-t l - tan-1(-$
Hint. Call this u - B and find tan(d - 0.
33. If 0 : arcsin -r, show that xi.Il? : tan g.
y, If 0 : arctan .r-. show that , 1 + 12 .r2 : csc 0 cot 0.
35. Show that the expression ....2 3. -I7l 9 changes to sec g - cos g when the substitution
0: tan-1(,r:3r is made.
X. Show that . E -r : sin 0 under the substitution 0: sec-r(xl2).
o
7
Trigonometric Form of Complex Numbers
f ii*?:11?'#3,,::",:"i..Tl':T'f Br:T,i1ffi::.T#l:11i3::J:#.:
with this the point (a, b) in the plane. As in Figure 39,let r denote the distance
from the origin to the point (a, b), and let 0 denote the angle from the positive
x axis to the line from the origin to the point (a, b). Then we have
r=ffaF
a=rcos0
(13t
b=rsin0
The complex number z
: q I bi can therefore be written in the form
s-r(cos0+isin0)
(14i
L\ \\IT
g/TRIGONOMETRIC FOR\{ OF CO}[FI-E\
\[- \IBERS 359
Figure 39
This is called the trigonometric form (or polar form) of the complex number :.
The form a * bi is referred to as the rectangular form. It should be noted that
the trigonometric form is not unique, because if any integral multiple of l; is
added to 0, the value of z is unchanged. For example, if
- 2('"' \+"t";)
I -3r
then also
/ 7n i sin
/-/ ('o'T
+
--t)
4?
2(,",
+ *isin +)
and so on, where integral multiples of 2n are added to the angle. Normally, we
choose 0 to be between 0 and2n, but there are exceptions.
The number r is called the modulus of z and is often denoted by lzl. Recall that
for real number x, the absolute value lxl can be interpreted geometrically as the
distance on the number line between the origin and the point representing r.
Similarly, lzl represents the distance in the plane between the origin and the
point representing z.Infact,lzl is sometimes referred to as the absolute value
of z. The angle 0 is called the argument of z.
t-xrl
b
T)-
zubstiirutiom
EXAMPLE
21
Find the trigonometric form of the complex number
,
:
modulus and argument of z?
Solution
J3
f
i. What are the
The number \fr f i is in rectangular form a + bi,wrth a: ,13 and b r-vlr+1-z,and0- TEl6 (Figure 40). Therefore, by equation l4l,
(
ryrnbers amd
,[*,rt";)
- z("o
\
ffe associate
z
the distanse
the positire
'
The modulus of z ts 2 and its (primary) argument is nl6.
)
I
Figure 40
tt3n
fl["**
t. So
360
8/ANA LYTICAL TRIGC}NOM ETRY
One of the advantages of the trigonometric form of complex numbers is that
products, quotients, powers, and roots are particularly easy to calculate with
numbers in this form. Consider first the product of two such numbers' say
:
zr: rz(cos 02 + i sin 0r). On multiplication.
:
making use of the fact that i2
- I and arranging terms, we have
ZtZz: rrrr[(cos 0, cos 0, - sin g, sin 0r) * i(sin 0, cos 0, + cos 0, sin 0r)]
zt
rr(cos 0r
+ i sin 0r) and
By the addition formulas for the cosine and the sine, this can be written in the
form
zrzz= rrrr[cos(O, +
e) + isin(01 + 0)f
(l5l
So the modulus of the product is the product of the moduli, and the argument of
the product is the sum of the arguments of the two numbers.
EXAMPLE
25
THEORT
Find the product of the two comPlex numbers
zL
Solution
DE MOII
/
ir\
- 2(.o' i+ i sin,I)
and
zz
:: (.o, + - rt,
;)
\
By equation (15), the product is
z,.zz:2 3
*i sin(;.;)]
[cos(:.;)
: o ("o, i* t'n\)
This can be put in rectangular form by evaluating the trigonometric functions:
,' l):0 * 6i:6i
Ztzz:6(0 +
In a similar w&y, we can prove that tf z,
Z:A
EXAMPLE
26
-
0r)
0, then
+ isin(0r -
0r)]
(15
Find the quotient z, f z, rf
zL- 4t",
Solution
[cos(g1
*
+.i sin?)
zr
z2
and
zz:2(.",
i+ t ri";)
::['"'(+ -;)*isin(+ il]
- z(.", \+-
"";)
EXA}TP
r
g/TRTGONOMETRTC FOR\I OF CO\[PLEX
fiers lS -;ilirr
hulam w*":r-T.
In rectangular form this
hiplraail'lrtr-
ritten ::-
l
-
is
Z:rG*,+) :1+i"3
mhem,* su',&
ffr s,n r,
\U\{BERS 361
It is in raising to a power that the trigonometric form has the grearesr adr artage. We will state without proof the following important result due rr-r rhi
French mathematician Abraham De Moivre (1667-1754):
rruF
::
4ummud
I
MOME'S
THEOREM
DE
Let z= r(cos0+isin0) be the trigonometric form of any complex number.
Then for any natural number z
7"
= r"(cos n0 + i sin n0)
Thus, the modulus of zn is the nth power of the modulus of z, and the argument
is n times the argument of z.That the result is plausible follows from a consideration of a few powers of z. By equation (15),
$)
z2
: r' r[cos(O + 0) + i sin(0 + 0)]
: r'(cos 20 + i sin 20) s
Similarly, applying equation (15) again,
23
;fumcnlcrf,$
:
:
z2
.z:
.rlcos(2O
12
+
0)
+
isin(20 + 0)f
,'(cos 30 + i sin 39)
This process could be continued, and for any given power of z,De Moivre's
theorem would be confirmed. However, this does not co4stitute a proof of the
theorem. It can be proved using a technique called mqthematical induction.
which we will study later in this book.
t,
j
6r,'i
EXAMPLE
27
Expand (1
+
i)u.
Solution With the aid of a sketch (Figure
41) we determine that
r:
Writingz:l*i,wehave
z6:(r + il6
*I
-
/
= Gl46 (cos
Vt"' i+i sin;l'
6n _, 6r\
q +i sin
4)
_.
-8t",+.i sin +)
In rectangular form the answer is
(1
+i)u
-8[0+i(-1)]_ -8i
Ji
and A
:; {
362
r
8/ANALYTICAL TRIGoNoMETRY
Figure 41
Remark. You might wish to compare this solution with the work involved in
dding this problem by expanding by the binomial theorem and simplifying the
result. You will find that the method we used is much simpler.
Finally, we consider the problem of taking roots of cornplex numbers. Again
: r(cos 0 * i sin fl, and suppose we wish to find <li.W" first recall that if
any integral multiple of 2n is added to 0, the same value of z results. That is.
for any integer k,
let z
z
:
rlcos(O
+ 2kn) + i sin(0 + 2kn)l
Making use of De Moivre's theorem we can prove the result
;1,' = rr i,[cos
(ry)
. i sin
(l7t
W)
Furthermore. there are exactly z distinct roots that can be obtained by taking
k = 0, l, 2o . . ., n - l.These z roots are equally spaced on a circle of radius r1''-
No.:. In working with the trigonometric form of complex numbers, the
angle 0 may be expressed in either degrees or radians. Ifdegrees are used, then
equation (17) must be written
,, n -,,,n[.",
EXAMPLE
28
Solution
Find all cube roots of
Let z
: -8 : -8
-
* i sin (,.*,
(*+*)
8.
+ 0i. Then r :
8
and0
:
r(Figure42). So byequation(17t.
fi - z, z- 8ra [co, W) . i sin ej1
and the three distinct roots correspond to k :0, k : 1, and k
and (, be the roots corresponding to these values offt. Then
(o
:
2G".
)]
:
i * i sin N
(
(t : 2f"r- n*2n *
i sin
:2.
Let (0, (r.
r(i., +) :r + ;.'/3
z*22\
):
-
2(cos 7t + i sin n)
: -2
EXER.(
e/TRTGONOMETRTC FORM OF COMPLEX
(z:z(.o,
*#+ir,"$3)
-z(.",
+.i
NUMBERS 363
sinT)
-2G +)--1 -ir3
These roots are shown graphically in Figure 43.
nnohed im
ffiins &c
fs"
Figure 42
AgEsrE
nll thar ff
i That ln
Figure 43
1
+
ir13
x
* l*r$i
bf
(z
t-rn*
-1- iT
frrn'*-
bcrr tfuE
tr6d-
tt?,*ilm
EXERCISE SET
A
9
In Problems l-10 find the trigonometric form of the complex number. Draw a sketch in
each case.
1. l-i
3. 15-i
5. -4
2,
4.
6.
8.
7. -z - ziJl
9. 5
rim
I
tr-L
In Problems
1r.
cil
(m,- t'o-
/
z(co'
t+i^15
-8i
+Ji - qiJ,
10. i-l
ll-20 find the rectangular form of the complex number.
4n
4z\
f + isin!)
/ 3n 3z\
13. t(."r;+isinr)
rs. u("orA*
6
\
,.,n
It)
6)
- / 7n
17. Ur(.or?+isinZ/
':)
19.
2+2i
3(cos 150'
*
I sin
150')
/ ln + isin,3z\
t2. *f"'
;
)
14.7(cosz+isinz)
16. / 5n 5z\
'('totT+isinr/
tB.
5(cos o
20.
$(cos 315"
+ i sin o)
*
i sin 315")
364
8/ANALYTIcAL TRIGoNoMETRY
In Problems 21-27 find z, . z,
and zrf z, using equations (15) and (16). When possible
write answers in rectangular form.
a
calculator,
tables
or
without using
2t.
zr
r\
/ n
: 2(cos
22. zr :
U+
S(cos
z
*
i sin;1,
i sin
zz
z), ,,
--
/ 5n+ isin
..5r\
: 3(cos
O1
4
+(orT. ,t^T)
/nn\/3n3n\ zz : 2(cos
i+ istn 4 )
)+ i sin)).
A. z, : J2 + iJ2, zz: 4',11 - 4iJ,
25. z, : ,13 - i, zz: 2 + 2i,,,t3
26. zr : 3(cos 70' + i sin 70"), zz:5(cos 50'+ i sin 50')
n. zt : 4(cos 100" + i sin 100"), zz: 6(cos70' + i sin 70')
23. zr : 4[cos
In Problems 28-34 find the indicated power using De Moivre's theorem. Express answers
in rectangular form.
[,(...|.,,r,f)].
240)]s
+
30.
28.
, sin
[2(cos 240"
32.
34.
B
[r l)'o
t-O- i"Dl'
[,(.,f *,.,,f;)]'
31. (J3 - 0'33. (2 - 2iJr4
In problems 35-40 flnd all the roots indicated. Express answers in rectangular form
wheneter this can be done without tables or a calculator.
35.
Cube roos of
8i
3j. SixthrootsorL
39.
Square roots
In Problems
of
4l-M
- l6i
45.
36.
Fourth roots of
;r.
-l - '€
-?.,?A
40. ";;;rootsor
Fifth roots of - 16'18 +
16i
frnd the complete solution set.
41. x6 + 64 -_ 0
43. x6 : 64
42.
44.
xa
x3
+ 256 :0
+ 27i:0
Prove equation (17).
First wiite z : r[cos(0 + 2kn) + i sin(0 + 2kn)f, and let ( : p(cos d + ,sin d)
unknown nthrooiof z. By De Moivre's theorem determine p and $ such that
Hint.
be an
rt_-
10
zs.
Polar Coordinates
In all of our graphing so far we have used the rectangular coordinate system.
For certain g.upt irrg problems, however, an alternative system, called the
polar coordinateiystem, has definite advantages. In this system points in the
lo/PoLAR cooRDIN,
a
TES 365
planearelocatedbygivingradialdistancesfromafixedpoint,calledthepole,
extends horiirrJ ungt., from a i*ia.uV called the polar axis. The polar axis
1',r
pesffi3l{g
,"tt"ffytotherightfromthepole'u'd*tmakeitintothepositivehalfof
Now let 0 be any angle
a number line, with the pole correspondin g to zero'
in Figure 44'
with the pole as its vertei and the polar axis as its initial side, as
along the terfor a positive number r, if we measure r units from the pole
the ordered pair
minal side of 0, aunique point P is determined, and we call
units from
(r,0) the polar coordinates of P' lf r is negative' we measure lrl
45
ifi. i"f" Jn the ray diretied opposite to the terminal side of 0' In Figure
we iilustrate the points (2, nl6) and
(-2,
Figure 44
Figure 45
nl6)'
ffi;ffi$rxm;er$
rr'
Pole
(-2, A
Polar axis
Inthemannerdescribed,theorderedpair(r,0)determinesauniquepointP
gghr
many
inihe plane. On the other hand, if we are given P there are infinitelyhaving
angles
many
infinitely
are
there
for it, since
;;i* "i folar coordinates For
example, 12,nl6), (2,13n16), {2, -llnl6), and
terminal side through P.
ic,r:um
;
(-2,7n16) represent the same point' (Verify this')
system so
If we ,op"ii*por. a rectangular coordinate system on a polar
the polar
with
x
axis
positive
the
and
pole
tt ai tt e origin coin"ides with ihe
and
rectangular
between
relationships
axis, as in Figure 46, we can determine
and
system
rectangular
the
p
y)
in
polar coordiiates. Let have coordinates lx,
(r, 0) in the polar system. The following relationships hold true:
i
I
\J
;
rtu
tan
#-.'-;$r
d
fli
Y:rsin
Figure 46
ints
um
rM
(x*
12:x2+ v'
$:fCOS ol
srrc,h dxe[
IIE rystem*
catled rhe
0:/x
eJ
'1
(18)
(1e)
366
8/ANALYTTcAL rRrcoNoMErRY
The next two examples illustrate how we can use these equations to change
from one coordinate system to the other.
EXAMPLE
29
a.
b.
Change(..,8,_-1)topolarcoordinates(r, g),wherer>0and
Change (- J2, 3nl4) torectangular coordinates.
0<0 <2;
Solution a. From equations (18) we find that
-1
tanU:..:
.u/ 3
-2_A
b.
Since x > 0 and,r' < 0, we knox' that I is in quadrant
The polar coordinates are therefore (2,llnl6).
From equations (19)
IV, so 0 :
lh
b
:
- -, : cos;: -vr(
#)
/
1\
,sin7- -vrftr) : -1
- -\ r.'3;
.,
1
So the rectangutrar coordinates are (1,
EXAMPLE
30
a.
b.
Solution
a.
-
1).
Find the rectangular equation of the curve whose polar equation il
r:
2 cos 0.
Find the polar equation of the curve whose rectangular equation
)'2 :
rs
2s'
For r + 0 rve replacqcos
0
_by xf
rto
L\ \\X PL
obtain
,':2;
12
Now we use the fact that
12
:
-2x
x2
+ y2 to get the
desired rectangultr
equation:
x2 +
!2:2x
which we recognize as being a circle. The fact that the origin is on th*
circle assures us that no further points on the graph are obtained if r 1;-,
10,'POLAR
r - 0 is the pole, which coincides with the
since the only point for which
.i to change
COORDINATES 367
origin.
b.
From equations (19) we obtain
cos o
-':::j ;::2r
r(r sin2 ,
r-0 | r
}cos0csc20
(0
+
Zkn)
The solutio rr r : 0 is contained in the second equation. When 0 - ii l. the
equation r - 2 cos 0 csc2 g gives r -- 0, so r - 2 cos 0 cscz 0 contains all
points on the curve. Thus, the answer is
r :2 cos A csc2 0
To graph a polar equation we can make a table of values by assigning
values to 0 and calculating r. We typically assign to 0 the radian measure of
special angles we have studied, and we use our knowledge of the behavior of
the functions as 0 varies from 0 to 22. As an alternative, the graph is sometimes more easily obtained by changing to the corresponding rectangular equation. However, some of the more interesting curves in polar coordinates have
corrcsponds to
complicated rectangular equations. The equation r cos 0
in rectangular coordinates, so we know that its graph is a vertical line.
In this case it is clearly easier to work with the rectangular equation. On the
other hand, the equation r2:4cos20 corresponds to xa +2x2y2 + y4
4x2 + 4y2: 0. (Verify this). If you attempt to draw the graph from this rectangular equation, you will soon be convinced that the polar equation is easier
to work with. In the next example we show its graph.
:2
x:2
-
E,XAMPLE
31
Draw the graph of the equati on
12
-
4 cos 20.
0 < 0 < z, since cos 20 will complete one cycle in
this interval. Writing the equation in the form r: t2Vcos 20,we see that
values of 0 for which cos 20 < 0 must be exciuded. Thus we cannot have
nl4 < 0 < 3nl4.We make a table of values as follows. Notice that we choose
values for 0 so that 20 is an angle whose functions we know. The values of r
are rounded to two decimal places.
Solution It is sufficient to consider
lln I 12
:
--l
,1,f'F:
11 1
15
-;]u
l!l
_i.
The graph is shown in Figure 47. The curve is called a lemniscate.
368
8/ANALYTICAL TRIGoNoMETRY
Figure 47
lltr
7T
t2
n
7t
0
I
X:
oilt{
Called a r.,
of heal
We now present a summary of some standard polar curves, showing their
equations, their graphs, and their names.
Straight Lines
2
Line through pole making an
rcos 0 - a
Same
angle a with polar axis
asx:
a
rsin 0 -
Same
b
6y -
b
Circles
r-Zacos0
T:A
Circle'of radius
center at pole
a,
Circle of radius a,
center on polar axis,
passing through pole
x0 POLAR
Limagons:
r-a+
b cos 0
(a-
b, 0)
a>
A:b
r
-
a(L
+
tklr
in, = r,
a<b
b
Limagon without loop
(If a > 2b, the "dimple"
on the left is absent.)
cos 0)
Called a cardioid because
of heartlike shape
ring
Lemniscate
z
Rose Curvesz
12
r
COORDT\{TES 369
:
-
Limagon with loop
o2 cos 20
acosn0
-I
-,I
&
=[
n
even
(n : 2)
n odd
When n rs even, there are 2n petals;when
, {e*
trit"
p6ilR
r
(n:
3)
is odd, there are ru petals.
You may have noticed that each of the equations we have given involves
the cosine function only and that the
are all symmetriJ to the polar
"urn"Jthat if cgs
axis, since cos(-0): cos 0. It can be shown
0 is replaced by iin 0,
the effect is to rotate the graph in a countercloctwise direction by 90". More
generally, the graph of r : f(0 * a) is the sanre as that of r :
/(0) rotated
through an angle a. Since cos{0 - nl21 - sin g, this confirms that replacing
'-
cos 0 by sin 0 rotates the graph rf 2 radians, or 90". Similarly, since cos(g
-cos
0 and cos(0
-
3n12,1:
-sin
0, replacing cos g
")
by -cos 0 or by _sin
J
0
370
8/ANALYTICAL TRIGoNoMETRY
In the case of cos n0 being teth" rotation is 90"fn, with similar results for the other
rotates the graph 180' or 270o, respectively.
placed
by
sln
i0,
por example, the graph of the equation 12 : 2a sin 20 is the lemniscate
rz :2acos 20 rotated tluough an angle of 45". We conclude this section with
an example of a type of problem often encountered in calculus'
cases.
EXE RC
H,XAMPLE
32
r: 3 cos 0 and r: 1 + cos 0 on the same polar coordinate
the area inside the first curve and outside the second. Also
shade
and
system,
find all points of intersection of the curves.
Draw the curves
a
Solution The first curve is a circle of radius ! centered at @, 0); the second curve is
the
in
one
described
is
the
area
cardioid, as shown in Figure 48. The shaded
problem. To find the points of intersection we set the two r values equal to
one another and solve for
0:
3cos0:1*cos0
2cos0:l
1
cos0- 2
7T 5n
e- 1,7
Figure 48
at(1, nl3) and (|,5n 3tis probably better to give the equivalent representation (2, -rrl3) for the
We flnd r to be
It
second point.
* in each
case. So the curves intersect
ny s&mg the two equations simultaneously we found two points of intersectlon. Buifrom the giaph we see that the curves also intersect at the pole'
Why did this not show up in our solution? The answer lies in the fact that
rrr
hqimg rB*
b ffistr
mflffirrumr8
fiffi]
PoL{R
c(loRDI\{TEs 371
the curves pass through the pole for different values ol t. The circle g.'o
through ttre pote wtren g:n12, whereas the cardioid -errs throrytr rire poie
when 0 : zr. Nevertheless, the pole is a point of intersection'
lm'ttliff;
EXERCISE SET IO
AlnProblemsland2plotthepointqinapolarcoordinatesystem.
uu&iml,mmr
d_
mr8
l$
/
3z\
e' (-2,/
''
e
Ed iEE fu
qIBe-
1^('19 bt-;) c(-,+) d('il
2. a.
liCI
e'
3.
(4,n)
/
b.
(-r,) c. t,-?)
d. (o
il
5z\
[-'' - r/
Give flve different sets of polar coordinates for the point (2,2n13). lnclude at least
one set for which r < 0 and one for which 0 < 0.
In problems 4 and 5 give the corresponding rectangular coordinates for the given polar
coordinates.
b(-*';)
4.a('':)
/
dk')
c(^')
:)z\
e' (.-t,-;J
7z\ h.
/
s. a. (r",r.-Zl
/ -9n\
e. (Vt,Z/
fr.lq
\ 6/
c. /-,0,
\
-T)
4/
In problems 6 an|T glvethepglar coordinates (forwhich
r>0
d.
and
@,3n)
0<0<2n)cote-
sponding tq the points with the given rectangular coordinates'
c. (0,5)
d. (-4,0)
6. a. (1,.F)
e. (-3, -J3)
q. G2,2)
7. a. (4, -4)
e. (4,try, -J6)
b. l-2.f,2) c. (0, -2) d. (; +)
In Problems 8-18 change to the equivalent polar equation'
,a
*
t'ff. )m
n,
--'"
h for *ffi
of u:lm-
fu
hEt
ptnrue
lirn'lt
9. x+Y:O
8. x?+y2:9
10. 2x - 3Y: !
12. y+l:0
14. x'-Y':4
16. y2 :2x + |
18.
,tnT -2:
11' x:3
13. *'+Y'-4y:o
15. x!:2
l'1}. (x * 1)2 * -i'2 :
x
1
372
8/ANALYTICAL TRIGoNoMETRY
In Problems 19-30 change to an equivalent rectangular equation.
TE
19. r-4
20. 0--
21. rcos0:2
?3. 2rcos0-3rsin0-5
?5. r:3 sec 0
27. r- -2cos0
22. rsin0- -1
24. r :2 sin 0
26. r : 4 csc 0
28. 0 _ tan-t z
29. va_
I
30.
l-cos0
3
r2 sin 20
:
I
In Problems 3l-52 identify the curve and draw its graph.
3L. r :3
33. 0:.
r:
536.
r:4 cos 0
r:I*sin0
r - 4 - 5 sin 0
r:2cos20
r - -cos 30
12
- cos2 0 - sinz 0
r :2 cos 6 -
-)
34. rcos 0:2
3n
4
35. r- -3csc0
37. r: | - cos 0
39. r - 5 - 4 cos 0
41. (' : 9 sin 20
41". r-sin30
45. r:3 sin20
47. r :2 * sin 0
/
\
B
a)
J-at-
38.
40.
42.
44,
6.
48.
3z\
tt
1
49. r-2cosl0--l
so. r- 2+2.".(r.;)
51. r - 3(sin 0 - 1l
53. Write in polar forrn.
52. 12:8singcosg
a.
rr-1.,
_
4/
4{_Tl
* l.r}3 r
54. Write in rectangutrar fonn.
4
L' r- L-zsino
b'
Draw the graphs in Problerns
55.
b. *, - y,
55
-
4x
coS 0
+
0
r sec' 0 - sin 0 - 3 cos
0
-61.
4
r:2rrn'l
57. / -
-
2+cos0
sin 0
Hint. Write in the form r : a cos(O - u)
58. l: 3cos0 -4sin0 (SeehinttoProblem5T.)
0
59.
"z r=2cosi
OO.
,:Jqsirr0
61. r:3cos59
62. Bychangingtorectangularcoordinates,showthatthegraph
of r:4sec210121 isa
parabola. Draw the graph.
Problpms 63-65 involve threq spiral curves. Dtaw them for the indicated values of a.
63.
@.
Spirpl of'Archirnedes,
Hyperbolic spiral, r :
r: a0 (a:
al9 (a:2)
l)
1I/PARAMETRTC EQUATIONS
65.
66.
Logarithmic spiral, r : eoo (a - l)
Graph the bifolium r : a sin 20 cos 0 for a -
373
1.
In Problems 67-71 show the two curves on the same coordinate
system, indicate the
specified area by crosshatching, and flnd all points ofintersection.
67. Outside r :2 and inside r : 2(1 + cos 0).
68. Outside r: I * sin 0andinsider:3 sin0.
69. Inside r : 3 cos 0 and inside r :2 - cos 0.
70. Inside r : -4 cos 0 and outside r : 4 *4cos
71. Inside r : 5 cos 0 and outside r : 2 + cos 0.
11
0.
Parametric Equations
As we have seen, some curves are more conveniently represented by polar
equations than by rectangular equations. There is, however, still another way
of representing curves by equations that in some cases is more natural and is
better suited to analysis of the curves than either rectangular or polar equations. In this third way,a rectangular coordinate system is used, but x and y
are each written separately as functions of an auxiliary variable, called a parameter.ly assigning permissible values to this parameter, points (x, y) on the graph
can be found. For example, consider the equations
[x:t-l
:2t
LY
,."
where I ian be any real number. Here the letter r is the parameter, and the two
equations together are called parametric equations. To draw the graph of the
curve represented by these equations, we can construct a table as shown below
and then plot the point (x, y) on a rectangular coordinate system (Figure 49).
The graph appears to be a straight line. Note that the parameter r does not
appear on the graph. We use it to f,nd x and y but plot the points (x, y) only.
Figure 49
ntg'-r
mda-
ma
37
4
8/ANALYTICAL TRIGoNoMETRY
is desirable to find the rectangular equation of a curve whose
parametric equations are given. In our example, we could solve for t from the
:
first equation, getting t: x I 1, and substitute into the second, getting y
In
the
examples
line'
is
straight
graph
a
2x + i,which confirms the fact that the
that follow, we will see some other techniques for eliminating the parameter.
but it should be noted that it is not always possible to do so. It is important
to be aware also that the parametric equations may impose certain restrictions
on the graph that are not present in the rectangular equation. We will illustrate
this in some of the examPles.
In general, parametric equations for a plane curve C are of the type
Sometimes
it
(x:
f(t)
tY
s(o
:
varies over some prescribed interval I (which may be all of R). The
totality of points (x, y) obtained by letting , vary over I is the curve C. When
C is described in this way, we say it is represented parametrically. The letter r
is commonly used as a parameter, since in applications it often represents time.
However, other letters may be used. The next three examples further illustrate
where
,
these concepts.
EXAMPI-.,E 33
Let C be the curve defined parametrically by
(x-2r
\;:;
-23r<2
Eliminate the parameter and draw the graph of
Solution
From the first equation we get
t:
C.
xl2, and on substituting into the second, we
have
x2
:4y
which is the equation of a parabola. We note, however, that with the given
domain for t, the graph extends only from (- 4,4) to (4, 4\, as shown in Figure
50.
Figure 50
EXAMPLE
34
Show that
[*-2cost
tr-2sin t
is a parametric representation of a circle.
o<t<2n
EXAU
r1/PARAMErRrcEeuATroNs 375
rwhosg
rmrk
[8
-!i
Solution To eliminate the parameter
*'
,:
:
4
cos2
t + 4 sin2 t
:
4(cos2
t -f sin2 t) : 4
:
rflmpnff
or x2 + y2 4,which is the equation of a circle of radius 2, centered at the origin.
is instructive to observe how the circle is traced out as , varies over [0,2n].
(*, y) as a moving point whose position is deterWe can think of the point P(r)
mined by r. We see from the given equations that P(0) (2,0), P(nl2): (0,2),
P(n): (-2,0), P(3n12): (0, -2), and P(2n): (2,0). Thus, the circle is traced
out in a counterclockwise direction, starting from (2,0).
In this case we can see a geometric interpretation of the parameter r. As shown
in Figure 51, it is the radian measure of the angle in standard position with terminal side 0P.
rlmttr-
It
portefi[
rictious
hrstru":r
fL Tbs
+ y'
we square both sides of each equation and then add:
:
:
Figure 51
- Whsm
lsmcr r
h
titrBg-
hsmu.re
EXAMPLE
35
Discuss the curve C defined parametrically by
J.-1+cos2r
[l'
d-w,e
- sin r
7E
2
TT
-2
Solution Observefirstthatforrinthegivenintervalwehave0< x < 2and -
1
<
y<
1.
To aid in identilying the curve, we eliminate the parameter, making use of the
identity cos 2, : | - 2 sinz t. Since / : sin ,, we get
E
x:
CIT{mr
rFrymmu
I
f cos2t:1
+ (1 - 2sin2
t):2-2y2
of,
y2:-l@-2)
This is the equation of a parabola with vertex at (2,0), opening to the left. But
because of the restrictions on x and y imposed by the parametric equations,
the graph of C consists only of that portion of the parabola shown in Figure 52.
Note how the curve is traced out as , varies from -nl2to nl2.
Figure 52
376
8/ANALYTICAL TRIGoNoMETRY
As we have seen in Examples 33 and 35, the graph of a curYe defined parametrically may be only aportion of the gtaph of the corresponding tectangulat
equation. Restrictions on x and y may come about because of the interval for
r (as in Example 33) or by the equations themselves (as in Example 35). The
important thing is tirat you must take such restrictions into account when using
the rectangular equation to draw the graph.
The next example illustrates how parametric equations sometimes arise
naturally when describing the motion of a moving object.
EXAMPLE
36
Solution
A projectile is flred at an angle of inclination e (where 0 < a < nl2) at an initiatr
spied-of uo feet per second. Find parametric equations for the path taken by the
piojectile. Neglect air resistance, and assume the ground is level'
Choose axes as shown in Figure 53, and let (x, 1) be the coordinates of the pro'
jectile after , seconds. If there were no gravity, the projectile would simply continue along the straight line ofits initial velocity vector and be at a distance ro;
on it at time r. But gravity causes the t'coordinate to be diminished by tgt2 after
f seconds. So from the figure, we see. using right triangle methods, that
[= -
(
r'or cos 7
Lj'-r'orslnz'- ist'
)
These, then, are parametric equations for the path. We must have t 0, and
t is also limited b1 u'hen the projectile strikes the ground. You will be asked t"
find this upper limit on r in Exercise Set 11. Also, you will be asked to find the
rectan-sular equation olthe path. as well as the maximum hor2ontal and vertica
distances for the projectile.
l\
Figure 53
l..-r----->l
Finally, we derive parametric equations for an important curve known as the
iou can think of a cycloid as the path traced out by a point on the circumference of a wheel as the wheel rolls in a straight line along a level path. Ftrr
example, a point on a bicycle tire would trace out a cycloid as the bicycle roltri
along. The iesulting curve is shown in Figure 54, where a is the radius of tha
wheel. Note that the curve is periodic, with period 2na.
To get the equations, refer to Figure 55, where we show a typical position ol
the wheel, with the point P that traces out the cycloid having originally beer
coincident with the origin. Then, since the length of the arc PC is equal to th*
cycloid.
1|IPARAMETRTC EQUATTONS
hod pare-
377
Figure 54
rtangunm-r
fierral fon
n
35t The
- 4rra
frEm urilqg
Iffi
erise
:4a
iniufle'l
rcm
b]-tM
- 2ra
) iro
4na
Figure 55
f the prclnpily c&m-
length of the segment Oe (why?), we have
knm r*r
W after
lrt
x -Oe
- m - fC - trE - ad -
asin
@
and
y:K,:AC-m-o-ocos@
So we have
> 0. emd
:
{f* :
:ar,ked to
l.y
r find thr
o(6
-
sin
@)
a(t _ cos {)
-oc<d<oo
for the parametric equations.
A fact of historical as well as practical importance is that the cycloid is the
"curve of quickest descent," in the following sense. Let A and B be two points
not on a vertical line. As shown in Figure 56, suppose a wire joins ,4 and I
and a bead is placed on the wire at .4. Among all possible shapes into which
such a wire can be bent. the one that causes the bead to reach B in the least
time is part of an arch of an inverted cycloid, with the origin at ,4. This is rather
surprising, since one might think a straight line would be best, since this is the
d veutilcai
shortest distance. The curve of quickest,descent is called a brachistochrone. The
mt as
discovery that the solution to the brabhistochrone problem is the cycloid was
made independently by a number of seventeenth-century mathematicians,
including Johann Bernoulli and Blaise Pascal.
In the exercises you will be asted.to find the rectangular equation of the cycloid. You will discover that graphing thp curve directly from the parametric
equations is easier than using the reitangular equation.
&e
r the cir-
nth For
p*
ro,fl3m
m of thu
uifiom of
tBy
hffi
rk
rI to
Figure 56
378
8/ANALyrrcAL rRrGoNoMErRy
EXERCISE SET
Review Exe
11
A In Problems
1-6 draw the curve represented by the parametric equations by assigning
values to the pararneter to determine points (x, y).
l. (x:2t-l
{
["u:1*2
3. (x:2-t
5.
-
lt:2,
: sin2 t
[y:cost
2. (x:t2
{
[y:r-l
4. fx:sint
--oo<f<co
6.
fx
-
tl:
(x:
cos
-co<r<oo
t
,lt -
'1
Ly:t-Z
In Problems 7 -20 draw the curve represented parametrically by first noting any restrictions and then eliminating the parameter to obtain the rectangular equation.
8. (x:3-4t
--'l!:3t-l
10. (x:2cosr
-.-' 0<t<2n
{
' ly::rirrt
12. 1fx:sinr-t
ft
ft
2="-2
[y:cos2l-1 --<r<14, (x: e'
ly:r"
16. fx: ht
U: ,
18. lx:2tant
<
0<t<-- -" -2
[y:.ott
20.
_LasaL
{x:ztan0-3
2-"-2
[y:3secg+Z
7. lx:2t-l
U:1*2
9. [x:cosr-l 0<t<2n
i
[y:sinr+2ll. [x:cos2t U<t<ft
<
iY:cort
13. fx: sec t
ft
n
-- -2
2
[y:tanr
15. (x: e'
[-t': e-'
17. [.t:r]
{
, -:t_<t<:r{.t':t'
le.
o<0<2n
{'t:lsin0-l
[1:4cos0*2
7L
B 21. Find the rectangular
22.
equation of the path of the projectile in Example 36, and
identify the curve.
For the projectile in Example 36 find: (a) the time at which the projectile strikes the
ground, (b) the maximum horizontal distance covered by the projectile, and (c) the
maximum height the projectile attains.
23. Find the rectangular equation of the cycloid.
24. Draw the gra-ph of one arch of the cycloid with a :
I
2.Take { in the intervalf0,2n},
and use a calculator to obtain points.
25. Draw the curve defined parumetrically by n - cos3 0, y sin3 0, for 0 < 0 < 2r.
Find the rectangular equation of this curve (the four cusp hypocycloid)"
26. With the aid of a calculator, make a table of values for r in the interval [0, 2nf, and
draw the graph of the curve defined by
f* -
t, :
f / sin r
sin r - r cos,
cos r
(This is called an involute of a circle.)
RE\-IEr,t E\ERCISE
sET 379
Review Exercise Set
A
rmig[Imm'S
1.
Evaluate each of the following without the use of tables or a calcularor:
2.
Express each of the following in terms of sin 0 or cos g.
I
a.
tan(7n112)
a. .-(? - ,)
3.
T
';
.
-,
m*
I
!t!
sin(23n112)
b.
c.
* n)
e' *'\7len* \
',
cos
c.
,nn(o
h
il.
rft#m
h!*s"
lglun
rlm
-
o)
r::H![]*x
b.
+
+ p,
T,si?iild
'".(il
a. sin(a B)
cos(a
c. tat(a - B)
g. lf tan 20 :- _! and nl2 < 20 < z,B)
flnd sin g and cos 0.
9. If tan 0: f and n <0 <3nl2,findsin(012), cos(012), artd,tan(012).
10.
Evaluate:
11.
Evaluatb:
a. cos[arcsin(-$)] b.
a. tan[2cos-1(-a)] b.
sin[2
sin-1(-:r)]
cos(]tan-1
4.f)
In Problems 12-25 prove that the given equations are identities.
12. csc0-cos 0cotg-sing
i,14.
hlrfu
sec(4n
cos{t3z l3r
Evaluate:
13.
'cos
n'trm
(z/B) d.
a. arccos(-|) b. tan-r(-r,,6) c. arcsin(uTl21
d. cos-11-l,; e. tan-1(-i1
4. Ifsec 6 :f, and sin o < o, o < o 32n, find
a. cos20
b. tan20
c. cosf0
5. Ifsin 0 : f and tan 0 <0, 0 < 0 < Zn,fmd:
a. sin20
b. cos20
c. sin $0
r mmflH*
m
d. .or/e - 1)
\ 2/
b.
0+cot0:coSg
A
1+csc0
16. "'!.'^ !3*:
eos 20
thn0
tan] r sec0-1 2tan0
lU.
rv'
sec0+1' tune -sec0+l
20. sin0*cos 0cotg:csc0
zz. r+-3n2e
secg+1-sec0
t^* e
aa'
24. L!
1-.unro:sec2o
cote
15.
L*
-'
-:srn2e
tan0fcot0
cos0
secO_ tana
g
_l*sinO
tan 0* cos 0
17.
sin
19.
tan g(co s 2e
sec 6
+ l) _ sin 20
21.. sin20*(cosg-sin
0lr-l
23. cot7-tan7:2cot,20
)F
25'
srn 20
1-cos2e:coto
In Problems 26-31 find all primary solutions.
26. sinx-2sin2x:0
28. sin2o*cosg:o
30. 2cos23t_ 3cos3t- 2:0
:
27.3tan3x-tanx-0
29. coszo-sing:o
31. 2cos2t:r-sinr
380
s/ANALYTICAL TRIGoNoMETRY
In Problems 32-34 sketch the $raphs'
b. Y:3cos2x
32. a. y-2sinx*l
33. a.
34.
y:ztanff
b. Y:2cosT*'
b.
a. y -rir(1, .;)
35. If
/
,, :2(cos
flnd
zr'
Y
-2 cos("'-,
I 4n+isinrT
. 4z\
5r\ and zz: 3(cos,
i"i"l)
i+
5n
zrand zrfzr.Express ans\lers in rectangular form'
In Probleins 36 and 37 expand using De \{oivre's theorem'
37' 1z+zi,flln
36. (t - i)8
38. Write in Polar form'
a.
-t2
+
8,r
: 16
b.
-tr -.1'r
::Y*r+y,
39. Write in rectangular form.
a'
.t
f-_
b'
=- --'
l--icostl
In Problems J0 and
'll
r1
cos20:4
dras' the graph'
{). a- r:}--1 cos0 b. 12:-4cos20
b. r:2-Zsin0
41. a. rrcosl0:0
In Problems 42-45 draw the
42.
44.
I*:,--,
graph, making use of the rectangular equation'
ti:,;,
In Problems 47 -52
sing
50.
tan-1
? csc2 0
-
lt
lt
-i<o<'
-co<r<oo
] + tan-t(-z)
prove the identities'
cos2g--1 _tan.
tLAfra
sec 0 csc 0
tr:'une
{l:t+,'
r
!1"
4g. Drrr
r*2-tanz0-1sin2 w'
sec'0
cosu -sin20
sec0-1
49. -- ^-:sec0+1
{x:1+sec20
45. lx:2-e2'
0<t<2n
o<t<2n
Jx:l-2sinr
lir:3cosr
B 46. Evaluate:
;. los[tan-' fi - sin-1(-$)] b.
47.
43.
-l<r<oo
-
2 cot0 csc e - |
zcos 0 csc 0 - tan 0 - cot
sin 8o - sin 69
cosg0+cos60-vdnl
0
p^
52'
cos o
1-sin0 =tan0+sec0
eosz1
RE\-IE\\ EXERCISE
SET 381
In Problems 53-56 find all primary solutions.
55.
tan22x*3sec2x*3:0
tan20 :2 cos 0
57.
Draw the graph of:
53.
/n\
\r/
a. !:3-Zcosl)-2xl
54. cos20ser0*secu:l
56. 1 + sin f : cos r
Hint. Square both sides-
b. y:J3cosx-sinx
58. Findalleighthroots of 1.
59. Find all cube roots of -64i.
60. Find the complete solution set to the equation x6 + I : 0.
61. Draw the curves r :3 - 4 cos g and r:2J3tl - cos 0) on the same polar
r)
62.
*:(atb)cos0-a*.(f)e
Draw the epicycloid for which
I
r[.
ttllt,,,
i'i
t*
([< r
:offi
J#
c+.
ordinate system, and crosshatch the area that lies outside the flrst curve and inside
the second. Find all points of intersection.
An epicycloid is a curye generated by a point on a circle that rolls externally around
a fixed circle. If the fixed circle has radius a and the rolling circle has radius b, it
can be shown that parametric equations of the epicycloid are
and y :(a+b)sin0-bsin ft:!\,
a:4 andb:
\b
l.Take 0 from0to2n.
/
Cumulative Review Exercise Set
1.
2.
III (Chapters 7 and 8)
In each of the following a right triangle ABC is given, with right angle at c. Find all
missing sides and angles without the use of a calculator or tables.
a. A:30", b:12
c. A:45", c:16
e. b:6O, c :12
b. B:60", a:4
d. a:10, c:20
Without using a calculator or tables find all six trigonometric functions of each of
the following. Leave answers in exact form.
a. 4nl3 b. -3n14 c. 17n16 d. 75' e.
f. l7nll2 g. -22.5' h. (2n - 1)n, ri an integer
3. Evaluate without a calculator or tables:
a. sin[2 sin-1(-t)]
b. cos[cos-1(-1s,) + ran-1(-3)]
c. tan(l)0, where sec A : ) and sin 0 < 0
d.
/n\
cscIf -
0].
990.
wherecot g :landcos0<0
\4)
4.
Prove the identities:
_
a.
5.
sec0-cos0+txt9 _,_-n , (sin0-cos0)2 l-tan?
b.
tand*seco
cos20 l+tano
Let
ZL:+(.o,
+ .i
sin
+) and Zz:2 (.", 2+, r,r;)
Find:
a. -:SmU
ZrZz b. ztlzz c. z!
6,
7.
d.
The square roots ofz,
Express answers in rectangular form.
Points z4 and B are on level ground on a line with the base of a tower. From the top
of the tower the angles of depression of A and.B, respectively , are 24.6" and 31.2. . If
points ,{ and .B are known to be 20 meters apart, find the height of the tower.
Evaluate without using a calculator or tables:
a. cos[2 sin-1(-.1)] b. sin[2 arctan(-3)]
c. tan|fcos-1(-fi)] d. cos]a, wherecsc o: -iandn<u<3n12
8. Find all primary solutions:
a. 2 sin x * f tan x : 0 b. 2 sin2 lx - 2 cos2 x : l
9. A chord on a circle ofradius 6 inches is 10 inches long. Find:
a. The central angle formed by the radii to the end points ofthe chord
b. The arc length cut oflby the chord
c. The area ofthe sector formed by this arc and the radii
d. The area of the triangle fbrmed by the chord and the radii
10. [n each of the following,
triangle ABC is a right triangle, with right angle at C.
Find all unknown sides and angles.
a. A:16.3e, b:4.68 b. B:5240', c:171.5
c. a:13.7, b:21.3 d. b:50, c:130
11. a. A curve on a railroad track is in the form of a circular arc which is 1.5 kilometers
long. Ifthe central angle corresponding to this arc is 72", find the radius ofthe
b.
circle.
A girl is riding a bicycle with
28 inch diameter tires which are rotating at the
rate of 200 revolutions per minute. Approximately what is the speed of the
bicycle in miles per hour?
sides and angles of the triangles ABC. Also, find their areas.
12. Find all unknown
a. A:32.4', B:47.8', c:25.3
b. B:28.3., b:23.5, c:39.2
c. A:98"24', b:124.3, c:89.76 d. a:32.5, b:26.8, c:18.6
383
384
CUMULATIVE REVIEW EXERCISE SET
III
(CHAPTERS 7 AND
8)
13. Prove the identities:
sin 0
+ sin. 0- . -) cot0
a. sec0-1'rec0+l-'vv..
14.
Find all primary solutions:
a. sinx*sin2x-0
15. Find all solutions to the equation
L6. Evaluate:
a.
b.
c.
17.
b.
sin[cos-1(-f)
*
Draw the graphs
a.
-
+
64i
x
b. cos 4x - 0.
-
2 tan(xl})
cos 2x
tan-1(-Pr )]
arctatf
cos(u+2fl), where seca:
arctan(-2)
x3
stn 2x cos
(1 + cos 2x) cos21x121
f,
csc fl
<
0,
cotB:2, secf < 0
:
y:r +2sin(;-,
b.
r:+*'(**i)
18.
An airplane leaves an airport at 2:O0 p.tu. and flies at a heading of 148" at an average cruising speed of 240 miles per hour. Another plane leaves the same airport at
2:15 p.rra. with a headingof 25'and cruises at an average speed of 400 miles per
hour. If they fly at the same altitude, how far apart arc they at 3:30 p.rr't.?
L9.
Prove the identities:
a.
20.
2cos3 0
*sin20:2cos0
l+sin0
A
Prove the identities:
sin0
1-cos0
L. cot0srn2e-cos20:I
21.
22.
23.
Draw the graphs:
a. y - ]tan1zx12)
b. ! : ] sin-'(Z*) + nl3
?7.
b.
cos x
sln x
=++th3':"..2,
rig in the
of Mexico is viewed
from points A and B
Gulf
offshore oil drilling
on the shore, with the point on the shore nearest the rig lying between A and BIn the triangle formed by the rig and the points ,4 and B, the angle at A is 76"25 and the angle at B is 68'54'. If A and.B are 6 kilometers apart, find the distance from,
each of these points to the drilling rig. How far is the rig from the nearest point
on shore?
Show the curves r : 2 cos 0 and 12: Z.,6 sin 20 on the same polar coordinate
system. Crosshatch the area outside the flrst curve and inside the second, and find
all points of intersection.
Prove that the area of the parallelogram having the vectors V and W as adjacenl
sides is lVl lWl sin 0, where 0 is the angle between V and W.
In the accompanying flgure, triangle ABC is in the horizontal plane and line CD is
vertical. If side c and angles a, B, ard @ are known, derive the following formula
for the length ft of side CD:
24. An
26.
e
2
Observer ,4 is stationed at a point due west of a monument, and measures the angle
of elevation of the top of the monument to be 45'. Observer B is 100 feet due south
of observer A, ard.B measures the angle of elevation of the top of the monument
to be 30". Without using a calculator or tables, find the height of the monument.
Find the complete solution set of each of the following:
a. .r,6sinx:cosx-l
25.
/u 7t\ sina-1
b. tanl
l-\2 4) cosn
o_csinutan0
ttsin(a+p)
CUMULATIVE REVIEW EXERCISE SET trII ICHAPTERS -
If c:21.34, a:
I
$en,r -r,-
I alrpnot"i .a:
I mile= pr:
L:
28.
I
each of the following:
0
I. , ) sin2 ,
[, -2sin20
b' [. :et*e-t
IY :et-e
,rg-ffi
fm[IU]ffii:
[TEflt
tJ&ffJ
A eru
-*
iis*6 1
frIffi:::,,M
tre$[ ;ullrr.
mCI{rc":,"u"]ii
t ad :ns
n d,_ra.m:rr
[ire Cf s
g fbrcrru,iiu
-
56.42'. find ri" A,s,.-
nrj ,:i-:
The pilot of a small alrllafre flew from to'wn
to town -B and returned the follorrins
{ay. _The bearing of B from i4 is N 42.3' E, and its distance from A ir-+sz^.,rr*.
on the trip flrom A to B, a 33.4 mile per hour wind was blowing from 270.: on the
return.trip the wind wab ft'om 305' aizg.l miles per hour. If ihe pilot cruised at
182 miles per hour airspeed, find hbr heading on each leg of
the i.if. ato find her
total time in the air.
a'
du€ yc,r-ii.
and O
38s
D
29. Graph
tk
fr:43"16.,
]
AI A.n "l-*'ff-
$
18'37',
A\D !
t
0<0<n
oO<t<Q
AN-32
ANswERs
ro
SELECTED PRoBLEMS
Review Exercise Seto page 305
5. A:45"'a:sJl,o:suD
3. A:30',c:8,b:4J5
l. B:30',c:t0,a:514
7. A:30',r: 60",b:7J1 9. B:57",a:7.08,b:10.90
l1.. a: lll.34, A:4',7.51, B:42.49" 13. A:63.8"a:30.06, D:14.79
15. a. 4nl3 h. 1nl4 c. nll2 d. 3n e. 8nl9 17. a. 4n x 12.5'1 h. (2401n)' x 76.39"
h. -1 i. -2 i.-J312
re. t. ! b. -+ c. 1l$ d.0 e. -J, r.2 s.t
21. b : 10.72, c :20.82' C: 118.9'
23. Solution1: A:64.49",8:83.51", a:13.62;solution2: A:51.51',B:96'49"a:ll'82
25. A:100.2",8:52.2',C:27.6' 27. Nosolution 29. A:37'36",8:43'05", C:99'59"
31. Solution l: B:104.2t",C:50.59", b:34.38;Solution2: B:25'39",C:129'41",b:15'21
33. 61.2 feet 35. 103.12 feet, area : 4,914.91square feet 37. lvrl : 15.0, lvr + Vrl :
39. lRl : 80.6 pounds, 0: 53"38' 41. 22.86 feet 43' 526'46 miles
45. 8.48 miles from first tower, 7.93 miles from second 47. 646 feet, S 10.6" E
25.9
Chapter 8
Exercise Set 1, page 314
sing- -Zrlltycos 0 - _1,csc0- -312'f1,sec0 - -3,tan 0:2t{2,cot0-lpp
sin0- -#,cos 0:*,csc0- -#,sec0-+,tafi0_ -+,cot 0: -*
\ sin0- -rlBtO.cos 0:i,cscg- _ 4ltlls,sec 0:4,tan0- -J15,cotg- -U^,lB
: tll,cot 0 lltl3
7. 0 - 4nl3; sin 0 - -15 2. cos 0 - -!, csc 0 - -z|",ft,sec 0 - -2,t3n0
g
g
ztlr,cot
0
tan
3,
0
csc
0
cos
sin
0
9.
-j.
== -Z.rlitl"
- - UztD=
- - -3l2tlr,sec (€.
(0,-1)
d.
(c.
1.
0t
(0,
b.
1)
+, - ,ll lz)
11. a.
eJ3p,+)
1.
3.
13.
I
IX
r9
?1
d.
a.
v
3,1
3?
a. nl6 b. nl4 c. n 3 d. nl2 e. 0
. a. sin 0.3 : 0.2955, cos 0.3 : 0.9553, tan 0.3 : 0.3093
b. sin 2.718 :0.4110, cos 2.718 : -0.9116, tan2-'ll8:
-0.4509
c. sin 14.03 :0.9943, cos 14.03 : 0'1070' tan 14.03 :9.2955
cos( -3.625) : -0.8854, tan(-3'625): -05249
d. sin(-3.625):0.a64S,
-0 c. nl4,5nt4
d' 0, z e. nl2,3nl2 f' 0,n g' 3nl2 h' z i' 3nl4'7n14
19. a. nl2 b.
21. a. p(g)iseither(0, 1)or(0, -1),sotangisundefined b. P(g)iseither(1,0)or(-1,0),socotOisundefined
c. piei iseither(0, f) o.iO, -ti,sosecgisundefined d. P(0)iseither(1,0)or(-1,0),socsc0isundefined
23. IlkeR,tan0:ylx:/c,with(x,y)ontheunitcircleif x :tl"tTlz'arldv:t luTi'''Sdif P(0):tttJTTP'
15.
11
klJl
+i.2), then
tan0:
k.
Exercise Set 2, page 321
1. cos 0 : -f , csc 0: -2, sec 0: -*, tan0 :1,cot|:1
3. sing:f,cos 0: -l,csc0:i,secg: -3, cot0: -Z
5. sin 0 : _llJi,csc 0: -16, sec0 -- JS1Z,tan0: -i,cot0: -2
7. cos0 = -Z0fi,cscg:3, sec0: -3l2J1,tan0: -llzJr,cot0: -2Jl
Exercise Set 3, page 327
3. a. -d6 + Jztrt- b' -(..6 + ,fzttt
nllltq 't. a. (r,6 + "Dltq b. (",6 + Jluq e. o. -#
"l
(-\tr
+ 4,f2)P 13' a' ttlrtJs b' 3slr7J5 ls' i. -3t
tr. a. QJL| -z)le b.
cos(nl2) sin 0: cos 0
cos
0
sin(z/2)
+
e):
sin(nlz
a.
+
17,
b. cos(z/2 + 0): cos(nl2) cos 0 - sin(z/2) sin 0 : -sin 0
r.
s.
\Jz +
J6l4,tJ1-
t.,|z
Jatn
-.'Gy+ t.
-t.-6 +
b., -*
b. -ffi
ANSWERS
19. a,
b.
sin(z
cos(z
-* 0)::
0)
-
sin n cos 0
coszcos 0
31. 0,n
33. a. sin acosB
b.
cos
oc
cos z sin
0:
To
SELECTED PROBLEMS
Af{-33
sin 0
+ sinz sin0 : -cos0
cos 1l + cos d sinB cos ?
cos Bcos 7
sin a sinf cos y
-
+
-
cos d cosB sin y
sin a cosf sinl
- sin a sin B sin y
- cos d sin f siny
Exercise Set 4, page 332
1. sin20: -#,cos20: -*
3. sin20:ffi,cos2e:# 5. sin20: -f,cos20:t
7. sin20:2xJT-7,cos20:t-2x2 9. sin0:"tjtZ,cosg: -.,fq: il. sin e: -i,cosg:t
B. t JI -1112 b. {z a ,1t12 c. uD + Jttz d. {I -J12
15. sin(a/2):3lJ1.:,cos(al2): -21\ry, fi. sin(al2): -Jttt,cos(al2):,f613 19. p(0):(i,+)
21. a. -sin 0 b. -sin 0 c. cos 0 d. -cos 0 e. cos 0
39. sin 20 : 2xlQ + x2), cos 20 : (t _ x21l1t + x21
41. sin30:3 sin 0 -4sin3 0, cos 30:4cos3 0- 3 cos 0
Exercise Set 5, page 336
1. a. 2+rll b. 2-J3 3. + 5.
. tan 2a - #, tan 2fr : -#, tan(af 27 - -1,
I l. a. t'612 b. -,l6tz
13. a. O2
7
(e
sQlp
- 4 9. tan20 - -+, tan(012) : 3
b. - rl2tZ 15. a". 2 sin 4x cos x b.
+
tan(Bl2)
j(sin 8x
*
sin 2x)
Exercise Set 6, page 343
1. {nl6,1tnl6} 3. {nl3,5nl3} s. {nl3,nl2,3nl2,snl3} 7. {nl3,n,snl3} s.
11.
19.
23.
27.
31.
37.
{nl2,3nl2}
{nl6,nlz,5nl6,3nl2} 13. {nl3,2nl3,4nl3,5nl3\ 15. nos-olution 17. {nlz,3nl2}
{0,2nl9,nl3,4n19,2nl3,8nl9,n,r0nl9,4nl3,1,4nl9,5nl3,t6nl9} 21. {O,n12}
{n124, 1tnl24, 13n124,23n124,25n124,35n124,37n124,47n124} 25. nosolution
{2nl9,nl4,5nl9,3nl4,8nl9,rtnl9,5nl4, l4nl9,7nl4, l7nl9} 29. no solution
{0.848,2.29,3.87,5.s5} 33. {0.464,r.Lt,3.6t,4.2s} 3s. {0.285,2.36}
{nl3,2nl3,3nl4,4nl3,snl3,7nl4} 39. {nl6,snl6,3nl2}
Exercise Set 7, page 352
1.
Amplitude
-
1,
period
-
z
3.
Amplitude
:
2, perio d
-
2n
5.
Amplitude
-
3,
period
-
4
7.
Amplitude
:
2,pertod
-
2nl3
9.
Amplitude
* l, period :4n
11.
Amplitude
:
1,
13.
Amplitude
-
2, penod
-
2nl3
15. Period -
period - n
1
\
t.it\
AN.34
ANSwERS
To
SELECTED PROBLEMS
17. Amplitude
-
1,
period
-
2n, phase shift
21. Amplitude
-
2,penod
-
2n13, phase
-
nl3
shift --
-?
19.
Amplitude
-
1,
period
:
n, phase shift
: -nl4
23.
Amplitude
-
2, penod
:
2, phase shift
:i
\
9.
v
2
11.
2\.
23.
25.
27.
25.
Amplitude
-
2,pertod
-
?m 3. phase
shift
-
29.
nl6
31.
33.
Jf.
37.
27,
a. y :.@ sin(x - z/41: amplitud, : Jl,period : 2n, phase slttft : nl4
b, y:5 sin(-x * 0). where 6 : 2.5 radians as shown;amplitude: 5, period
39"
41"
:2n,phase
shift: -0 x -2.5
45.
+'
,Bn
-
Exert
29.
Period
Exercise Set
:
6n
8,
1.
page 357
1. a. -nl3
b. 5nl6 3. a. 3nl4 b. nl2 5. r. 0 b. -nl6
1r. a. t b. i 13. a. zJitz b. -+
s. a. -2A13 b. i
le. -#g
21.
+ 23.+
2s.
7. a. nl2 b.
ls. 2l"E
ryrm n. sing:-!,cos0:f 29.*
3r.
17.
7nl6
-X
1+
5.
7"
11.
Exercise Set
1.
0
9,
page 363
21"
-7nl4,r: O;l - - 'fiTcos(7nla) * isin(77114)1
i
3.
0: llnl6,r:2;.,/3 - i -
2[cos(llnl6)
* i sin(ltnl6))
31.
ANSWERS TO SELECTED PROBLEMS
5. 0:fi,r:4;
-4:4(cos n*
9, 0:0,r:5;
5:5(cos
0
+
z
7. 0 -
i sinzr)
4n13, t'
:4; -2 -
2irfr
-
4lcos(4n13)
AN.35
+ i sin(nl3)f
sin 0)
11. -t - i"E 13. -si 15. zJz - zi 17. l - i tg. etJtlzl + eilz)
21. zF2: 6Lcos(3n;2) + i sin(31112)): -6i; zrlzr: !lcos(-z) + i sin(-z)l : -l
23. zp2:8[cos(52i4) +, sin(5nl4)] : -+1D - aia;z1lzr:)lsor1-nl4) + isin(_nl4)f : O - iO
25. zlzr:$[ser113n6) -isin( t3r16)]:4\4t4i:zrlzr:t[cos(32/2) +isin(3ltt2)f : _ it2
27. zF2 -24(cos 170' + i sin 170'); zrlzr: l(cos 30' + i sin 30') : (\E/3) + (r/3)
29. 129lcos(9n12) + i sin(9n12)f :729lcos(nl2) + i stn(nlz)f :729i
31. 256lcos(4ht3)+isin(44itl3)):256lcos(2n13) +,sin(22l3)l: -128 + tZtiJS
33. 256lcos(20n13) + lsin(202l3)l :2s6lcos(2n13) + iiin(2n13)): -t28 + nei15
35. (o : 2lcos(nl6) + I sin(n/6)l : Jr + it (r : 2Lcos(5nl6) + i sin(szl6)l : - !5 * i; (, : 2lcos(3n12) + i sin(3n12)) : - 2i
37. (o:1(cos0+ isin0):1;(r:1[9o_s(z/3) + i sin(z/3y] :_i_+iJ1.lz;(z;rlcos(2n13)+isin(2rll3)): -\+i,fZ1Z;
(a: l(cos z + lsin n): -1; (+: 1[cos(4213) + isin(4ft13)] : -] - iJllZ; (s: 1[cos(52l3; +;sii1Sz7:i]
I - iJ31z
39. (o:4lcos(3n14)+ i sin(32l4)l : -zuD+zi.,,D,h:4lcos(7n14)+isin(7n|4)f :zuD-ziJl
41. {\6 +;, 2i, -,,4 + i, -.lE - i,-2i,,,8 - i} 43. 1z,t + irz, -t + iJl, -2. -l - iy6, r -;nri1
45. By the hint we have p'(cos n$ + i sin nd) : r[cos(0 + 2kn) + l sin(0 * 2kn)]. So p" : r, p : yltn, and n0 : 0 + 2kn.
Therefore,O:(0+2kn)ln.Bytakingk:0,1,2,...,n-l,wegetallthedistinctanswers.
Exercise Set 10, page 371
2: ku i
l.
3.
',,
(2, 8nl3), (-2, 5nl3),
(2, - 4nl3), (-2, - 7rl3),
(2, t4nl3)
It
rit2
5.
t.
F]
l9r
L. (2, 2) b. Q ury, - 3) c. (- s ,12, s Ol d. ( - 4, o) e. (2,2)
a. go, Tnla) b. (4, 5nl6) c. (2,3n12) d. (I,5nl3) e. (z.vT,7n16) 9. 0 :
rcos0:3
13. r:4sin0
2t. x:2
23. 2x-3y-5
11.
Tr
31.
33.
15.
25.
12
-r:3
4lsin20 17. r - -2
27. x2+y2+2x:0
35.
0 19. xz + !2 :
29. yr:l+2x
37 .
cas
3nl4
16
cardioid
Afd-36
39.
ANSwERS
To
SELECTED PRoBLEMS
43.
41. lemniscate
limaEon
4-leaf rose
3-leaf rose
(,,?)
(r,?)
47.
1:
49.
limaEon
51.
circle
:
o,T)
cardioid
o,+)
s
(',?)
53. L. r -
55"
16 csc2 20, together
with r
:
0
b. r :4
r:l-cos0
cos 0 sec20
57. r-Jrcos(0 -
7T14)
circle
cardioid
11. {
:
(r,+)
59.
0l0
nl6
rrl4
nl3
r I 2
1.93
1.85
1.73
7rl2
t.4r
2nl3 3nl4 5nl6 Tc
7nl6
5n
14 4nl3
L 0.77 0"52 0 -0.52 -0.77 -1
3nl2
5n
l3
7
nl4
-1.41, -1.73 -1.85
Use symmetry
llnl6
-
1.93
15.
65.
63.
61. 5-leaf rose
I9.
.'i"',
{X
t" 4;t
\J.T/
(,,9)
67.
69.
(2, nl2), (2, 3nl2)
t|,
7T
3t, (* . 5n 3)
71. (2, nl3), (|,5n13)
23,
Exercise Set 11, page 378
-1
',- =
Review
0
-s -3 -1 |
23
1.
a"
3
4
3.
7,
11.
a"
a.
a.
I
ANSWERS TO SELECTED PROBLEMS
3.
'
t
1
2
x
I
l.
0
AN-37
. lllf
,*
d,i
.r.>
v
u,
7. x -
2y
* 7:
9. (x+l)'+(y*2)r:1
0
i {r
_41
w4
rl
lx
Il. x -2y, - I,lrl < 1, ll,l <
1
13. x2-y2:1,x21
-
:r!m iffi,
.L.lp,:r
15. x!:1,x>0,.1 >0
tg.
(,
l1)'
g ,(y-2)'_
1,6
17, !2 :
21.
x3
! : x tan a -d;ly!,
parabola
LUO
25,
x2l3
+ !213 *
|
Review Exercise Set, pag e 379
b. i7 - 46 c. J'TA d. -VzEE
-V
1-./3
4
4
3. a, 2nl3 b" - 7Tl3 c. *7113 d. ft €.
5. a. _#
b. _*
7. a. -H b. Sf c. t* 9. sin(?12) : 5lJ 34,-TEl4
cos(l12) _ _ 3lr[U, ian(L12) _ _
t
Ll..a. -+ b. 2lrfi 27. {0,n, nl6,5nl6,7nf 6, llnl6} 29.
nl6,5nl6}
r. a. tii
{inl2,
c.
31.
2l"E
{7n16,
|lnl6,
ft12}
AN.38
33.
ANSwERS
To
SELECTED PRoBLEMS
a.
b.
Amplitude
-
2, perrod
:
1l
4
17
35. zt' Zz: 6[cos(l3nl6)* i sin(L3nl6)f : 6[cos (nl6) * i sin(z l6)): 3 J1 + 3ii zrlzz - f [cos (- nlz)* i sin( - 7El2)1 - -2i13
37. 256[cos$nl3)* i sin(4n13)7 : -128 - l28iJ3 39. il. 5x2 - 4y', + 24x * 16:0 b. x2 - !2 :4
41. L,
(,*)
43.
23.
45. (y - l)'
''2:x-2
53.
{n13,2n13, 4n13, 5n13,
57,
a.
59.
{4i, -2.18
Amplitude
-
-
21
nl2,3n12} 55.
2, perrod
2i,2J3
-
:
2i}
TE,
{n12, 3nf 2, n
phase shift n /6
61.
(3 +
zJ3,
TT
2),
2*
x < 2, y >
l6]1
\
/n x):
\
2({cos.x-- jsin x):2sin(;-2 sin (, - ,),
\2
plitude - 2, period * 2n, phase shift : nl3
b.
5nl6), (3
: -(x -
+ zJa,7nl6),
29.
pole
ChI
Exer
I,
15"
2!.
Cumulative Review Exercise Set
III
(Chapters 7 and 8), page 383
1. a. B-60,c:816,o-4rl3
b. A-30",c-g,b:4",13 c. B_ 45",a:b:8J,
d. A-30",8:60o, b:LO.r6 e. A- B-45",a-60
3. a. -# b. ;+ ". -tlrl5 d. -..,6
5. a. zp2: 8[cos(32l2) + isin(32l2)] : -81 b. zrfz2:2lcos(7n16\ + isin(7r.16)): -.J5 c. z!: l6lcos(2n13) + , sin(21tl3)l : -a + aif
d. (o : 2lcos(2n13) + , sin(27rl3)l : - I + i,fr; (, :2lcos(5n13) + , sin(52l3)l : t - inll
7. a. Z b. -+ c. f d. tt,t5
9. a. 1.97 radians b. 11.82 inches c. 35.46 square inches d, 16.58 square inches
i
23.
ANSWERS TO SELECTED PROBLEMS
11.
17.
a, r : 1.194 kilometers b. 16.66 miles per hour
a. Amplitude - 2, period : 47r, phase shift - )LILI
-
11'
J
15.
{4i, -2111
b.
-
2i,
Amplitude
2$ -
- j,
2i\
period
:
2, phase shift
v
21.
a.
Period
-
2
b.
a.
25.
(rr, ;), (,, ;)
29.
b.
{0, anl3}
a.
- -L
The given equation is equivalent to
* : * sin 2(y or nf 12 < y <
23.
AN.39
7rl3),
with -7112 < 2(y
7nll2.
-
7rl3)
< nl2,
,,,r1
{n 6. 5n,6,7n16, llnl6}
27. h _ 1I.76,0 _ 35021'
-Xy-2),0<x <2,0<y<2
b. x2-y':4rx>'2
Chapter 9
Exercise Set 1, page 396
t. (4,2) 3. (-1, -3) s. (4,3) 7. (#,?) e. (4,_2)
ts. (2,-1,4) 17.(-#,-*) *.(##,t#)
21. If both sides of the second equation
are multiplied
8x - 6y - 3
-8x+6y-4
0
-7
(_5,6)
13. (_2,3,4)
by - t and then 5 is added to both sides, the result
the first equation. So the equations are equivalent. The solution ret
23.
11.
i,
This is impossible, so the equations are inconsistent.
lf -,)'
I
/
3c
- 5\ ,. n}.
)
the same as