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CHAPTER 2
2.1 RADIAN
AND DEGREE
MEASURE
2.2 T HE
TRIGONOMETRIC
FUNCTIONS AND THE UNIT CIRCLE 2.3 TRIGONOMETRIC
FUNCTIONS AND RIGHT TRIANGLES 2.4 T RIGONOMETRIC
FUNCTIONS OF ANY
ANGLE
2.5 G RAPHS
OF SINE AND
COSINE FUNCTIONS
2.6 OTHER
TRIGONOMETRIC
GRAPHS
2.7 INVERSE
TRIGONOMETRIC
FUNCTIONS
2.8
TRIGONOMETRY
ApPLICATIONS OF
TRIGONOMETRY
2.1
Angles
RADIAN AND DEGREE MEASURE
I
Radian Measure /
Degree Measure I
Applications
Angles
FIGURE
2.1
As derived from the Greek language, the word trigonometry means
"measurement of triangles." Initially, trigonometry dealt with relationships
among the sides and angles of triangles. As such, it was used in the develop­
ment of astronomy. navigation, and surveying.
With the advent of calculus in the 17th century and a resulting expansion
of knowledge in the physical sciences, a different perspective arose--one that
viewed the classic trigonometric relationships as functions with the set of real
numbers as their domains. Consequently, the applications of trigonometry
expanded to include a vast number of physical phenomena involving rotations
or vibrations. These include sound waves, light rays, planetary orbits, vibrating
strings, pendulums. and orbits of atomic particles. Our approach to trigonome­
try incorporates both perspectives, starting with angles and their measure.
An angle is determined by rotating a ray (half-line) about its endpoint.
The starting position of the ray is called the initial side of the angle, and the
position after rotation is called the terminal side, as shown in Figure 2.1. The
endpoint of the ray is called the vertex of the angle.
105
106
CHAPTER 2
TRIGONOMETRY
In a coordinate system, an angle is in standard position if its vertex is the
origin and its initial side coincides with the positive x-axis, as shown in
Figure 2.2. Positive angles are generated by counterclockwise rotation, and
negative angles by clockwise rotation, as shown in Figures 2.3 and 2.4. To
label angles in trigonometry, we use the Greek letters a (alpha), f3 (beta), and
(J (theta), as well as uppercase letters A, B, and C. In Figure 2.4, note that the
angles a and f3 have the same initial and terminal sides. Such angles are
coterminal.
y
y
)'
Positive angle (counterclockwise) ----.:....-4------..;~:c
--....lIo:-~--:c
Initial side
--,.-".I----:C
Negative angle (clockwise) Coterrninal Angles
FIGURE
2.2
FIGURE
2.3
FIGURE
2.4
Radian Measure
The measure of an angle is determined by the amount of rotation from the
initial to the terminal side. One way to measure angles is in radians. This type
of measure is needed in calculus. To define a radian we use a central angle
of a circle, one whose vertex is the center of the circle, as shown in Figure 2.5.
Central Angle
FIGURE
2.5
e
DEFINITION OF A RADIAN
One radian is the measure of a central angle (J that subtends (intercepts)
an arc s equal in length to the radius r of the circle (see figure).
arc length '" radius when
e
1 radian
SECTION 2.1
2 radians
3 radians
-f...J===:::::::..~:---+L-
RADIAN AND DEGREE MEASURE
107
Because the circumference of a circle is 21Tr, it follows that a central angle
of one full revolution (counterclockwise) corresponds to an arc length of
s = 21Tr. Moreover, because each radian intercepts an arc of length r, we
conclude that one full revolution corresponds to an angle of
21Tr
r
6 radians
21T radians.
Note that since 21T = 6.28, there are a little more than six radius lengths in a
full circle, as shown in Figure 2.6.
In general, the radian measure of a central angle () is obtained by dividing
the arc length s by r. That is,
5 radians
FIGURE 2.6
s
- = (),
Radian measure
r
where () is measured in radians. Because the units of measure for sand r are
the same, this ratio is unitless--it is simply a real number. The radian measures
of several common angles are shown in Figure 2.7.
e 12I (21T)
rr
e
2
t
Quadrant II
fJ
I
4 (21T)
6
1T
=2
I
8 (21T)
I
e = :2 (21T)
1T
4
= 1T
I
1T
6 (21T) = 3
fJ
=
21T
Quadrant I
Radian Measure for Several Common Angles
1T
1T
=
1T 0<9<2
2<9<1T
FIGURE 2.7
e=
rr - - - - - - i - - - - - < - fJ
Quadrant III
Quadrant IV
31T
1T<9<2
31T
2 < 9 < 21T
3rr
8=­
2
FIGURE 2.8
= 0
Recall that the four quadrants in a coordinate system are numbered coun­
terclockwise as 1, II, III, and IV. Figure 2.8 shows which angles between 0 and
2'1T lie in each of the four quadrants.
You can find an angle that is coterminal to a given angle ()by adding or
subtracting 2'1T (one revolution), as demonstrated in Example I (Note that a
given angle has many coterminal angles. For instance, ()
1T/6 is coterrninal
with both I 31T/6 and -111T/6.)
108
CHAPTER
2
TRIGONOMETRY
Sketching and Finding Coterminal Angles
EXAMPLE 1
A. To find an angle that is cotenninal to the positive angle (} = 13'TT/6. you
can subtract 2'TT to obtain
13 'TT
6
Thus, the terminal side of (} lies in Quadrant l. Its sketch is shown in
Figure 2.9(a).
B. To find an angle that is cotenninal to the positive angle (}
3'TT/4, you can
subtract 2'TT to obtain
5'TT
4'
as shown in Figure 2.9(b).
REMARK The phrase "the terminal
side of fJ lies in a quadrant" can be
abbreviated by simply saying that "0
lies in a quadrant." The terminal sides
of the quadrant angles 0, 'TT/2, 'TT, and
3'TT/2 do not lie within any of the four
quadrants.
-27T/3, you
C. To find an angle that is coterminal to the negative angle (}
can add 27T to obtain
27T
4'TT
(} = - 3
+ 27T = ­
4'
as shown in Figure 2.9(c).
1317
e
6
6
1'--'---4:""'-'-'_0
417
e =317
­
4
17
'TI-L----=:>i--_- 0 3
l ' - - - - J - -_ _ O
517 4
311 (a) FIGURE
31'
}1!
T
(b)
2
(el
T
......... ... 2.9
Figure 2.10 shows several common angles with their radian measures.
Note that we classify angles between 0 and 7T/2 radians as acute and angles
between 7T/2 and 7T as obtuse.
~ ~
Acute angle:
between 0 and
17
2
FIGURE 2.10
17
8=­
2
Right angle:
quarter revolution
8
Obtuse angle:
17
between 2 and
17
Straight angle:
half revolution
217
Full revolution
SECTION 2.1
109
RADIAN AND DEGREE MEASURE
Two positive angles a and f3 are said to be complementary (or comple­
ments of each other) if their sum is 1T/2, as shown in Figure 2.II(a). For
example, 1T/6 and 1T/3 are complementary angles because
1T
1T
-+-=
1T
632
(a) Complementary angles
Two positive angles are supplementary (or supplements of each other) if their
sum is 1T, as shown in Figure 2.] ](b). For example, 21T/3 and 1T/3 are
supplementary angles because
21T
-
3
+
1T
== 1T.
3
EXAMPLE 2 (b) Supplementary angles
FIGURE
Complementary. Supplementary. and Coterminal
Angles
A. The complement of (J = 1T/12 is
2.11
1T
]2
1T
2
61T ]2
1T
51T
]2
=
12' as shown in Figure 2.l2(a). B. The supplement of (J = 51T/6 is
51T
1T 1T
6=6'
as shown in Figure 2.12(b)
C. In radian measure, a coterminal angle is found by adding or subtracting
21T. For
171T/6, you subtract 21Tto obtain
(J
171T
6
as shown in Figure 2.12(c). Thus, 171T/6 and 51T/6 are coterminal.
7T 7T
7T
2
-
2
2
57T
12
7T
+"':"''''::''1--'--0­
7T
12
6
37T
(b)
(a)
FIGURE
2.12
(c)
2
... " .. " ... ".
110
CHAPTER Z
y 90° =
.--,----­
TRIGONOMETRY
Degree Measure
~ (360°)
~(3600)
60°
45° = ~(360o)
30° =
12 (360°)
0°
--t-----l<---'----i-''---x
360"
330°
315
300°
0
A second way to measure angles is in terms of degrees. A measure of one
degree (10) is equivalent to 1/360 of a complete revolution about the vertex.
To measure angles in degrees, it is convenient to mark degrees on the circum­
ference of a circle as shown in Figure 2.13. Thus, a full revolution (counter
clockwise) corresponds to 360°, a half revolution to 180°, and a quarter
revolution to 90°.
Because 21T radians is the measure of an angle of one complete revolution,
degrees and radians are related by equations
3600
21T
rad
and 180 =
0
1T
rad.
From the latter equation, we obtain
FIGURE 2.13
o
I = -
1T
180
rad
and
I rad
which lead to the following conversion rules.
CONVERSIONS: DEGREES +-+ RADIANS
REMARK Note that when no units of
angle measure are specified, radian
measure is implied. For instance, if we
write e = 1T or e = 2. we mean
e 1T radians or e 2 radians.
1T rad
I. To convert degrees to radians, multiply degrees by 1800 '
2. To convert radians to degrees, multiply radians by
1800
1T
rad
.
To apply these two conversion rules, use the basic relationship
180°.
EXAMPLE 3
rad
=
Converting from Degrees to Radians
(l35~)C;or~)
A. 135°
1T
Multiply by 1T/180
31T
= -rad
4
B. 540°
= (540.deg-)(
1T rad )
180.Geg
Multiply by 1T/180
= 31T rad
C. -
270°
(-270 .deg-)(
31T
--rad
2
1T rad )
180.eeg
Multiply by 1T/180
....... .. SECTION 2.1
Techno{091J Note _ _ _ _ _ __
Calculators and graphing utilities have
both
and radian modes, \Vhen
you are using a calculator or graphing
utility, be sure you use the correct
mode, Most graphing utilities allow
you to convert directly from degrees
to radians and from radians to degrees,
Try using a graphing utility to convert
0
135 to radians and
TI/2 radians to
degrees,
EXA M PLE
4
Converting from Radians to Degrees
~rad = (-~~)C~~) A. lIt
RADIAN AND DEGREE MEASURE
Mulrip!\' !J.,' J 80!"
= -90° B. 9; rad = (9;
~)C~O~)
= 810°
c.
2
rad = (2fttdJC~~)
360
-deg
Mulripl,' b,' 180/"
1T
........ .. "" 144.59° With calculators it is convenient to use decimal degrees to denote fractional
parts of degrees. Historically, however, fractional parts of degrees were ex­
pressed in minutes and seconds, using the prime (') and double prime (")
notations, respectively, That is,
I'
=
one minute
I (1°)
60
I
I" = one second = -(I ')
60
Consequently, an angle of 64 degrees, 32 minutes, and 47 seconds is repre­
sented by 8 64° 32' 47",
Many calculators have special keys for converting an angle in degrees,
minutes, and seconds (DO M' S") into decimal degree form, and conversely, If
your calculator does not have these special keys, you can use the techniques
demonstrated in the next example to make the conversions,
EXAMPLE 5
Converting an Angle from DO :W S" to Decimal Form
Convert 152° 15' 29" to decimal degree form,
SOLUTION
152° 15' 29"
1520 + ( -15)0 + (29)°
-­
60
3600
"" 1520 + 0,25 0 + 0,00806°
152,25806°,
&
••••••
II •
112
CHAPTER
2
TRIGONOMETRY
Applications
The radian measure formula, e = sir, can be used to measure arc length
along a circle. Specifically, for a circle of radius r, a central angle e subtends
an arc of length s given by
s =
where
re,
Lel1gTh
IIf
c;rdi/or
<iII
e is measured in radians.
EXA M PLE 6
Finding Arc Length
A circle has a radius of 4 inches. Find the length of the arc cut off (subtended)
by a central angle of 240°, as shown in Figure 2.14.
SOLUTION
re,
To use the formula s =
240°
we must first convert 240° to radian measure.
= (240 deg) ( 7T rad) = -47T rad
180 deg
3
Then, using a radius of r = 4 inches, we find the arc length to be
s =
re
167T
3
= 16.76 inches.
FIGURE 2.14
Note that the units for
units.
re are determined by the units for r because e has no
........ .
The formula for the length of a circular arc can be used to analyze the
motion of a particle moving at a constant speed along a circular path. Assume
the particle is moving at a constant speed along a circular path (of radius r).
If s is the length of the arc traveled in time t, then we say that the speed of the
particle is
distance
Speed = - - ­
time
s
Moreover, if e is the angle (in radian measure) corresponding to the arc length
s, then the angular speed of the particle is
e
Angular speed = -.
t
SECTION 2.1
113
RADIAN AND DEGREE MEASURE
Finding the Speed of an Object
EXAMPLE 7
The second hand on a clock is 4 inches long, as shown in Figure 2.] 5. Find
the speed of the tip of this second hand.
SOLUTION
The time required for the second hand to make one full revolution is
60 seconds
I minute.
The distance traveled by the tip of the second hand in one revolution is
s = 211'(radius)
211'(4)
= 811' inches.
Therefore, the speed of the tip of the second hand is
FIGURE
2.15
s
811' inches
.
= 0.419 m./sec.
Speed = - =
r
60 seconds
....... ..
Finding Angular Speed and Linear Speed
EXAMPLE 8
A lawn roller that is 30 inches in diameter makes one revolution every
~ second, as shown in Figure 2.16.
A. Find the angular speed of the roller in radians per second.
B. How fast is the roller moving across the lawn?
SOLUTION
A. Because there are 211' radians in one revolution. it follows that the angular
speed is
e
Angular speed = ­
r
211' rad
= 2.411' rad/sec.
B. Because the diameter is 30 inches. r
=
15 and s
=
211'r
=
3011' inches.
Thus,
Speed
=­
s
t
3011' in.
FIGURE
2.16
= 3611' in./sec
= 113.1 in./sec. ....... .. 114
CHAPTER
2
TRIGONOMETRY
If you have a programmable calculator, try entering an "angle-drawing program."
Then use the program to draw several different angles. The following program is
for the TI-81. Programs for other calculators are given in Appendix B.
AN ANGLE­
DRAWING
PROGRAM
:Disp "ENTER,MODE"
:Disp "C,RADIAN"
:Disp "I,DEGREE"
:Inp'..lt M
:Disp "ENTER,ANGLE"
:Input T
: O--'I'rnin
: abs T--Trr,ax
: .15-+Tstep
: cos T--A
: sin '-.;:'--B
: Para;-:-,
: 1--S
:If M=l
: fi,!: /2.80-+':'
:Rad
: If T<C
:-1--+5
: " ( .25-. 04T') cos
"--Xn
: ,. S (.25+. 04T) sin T"-+Y l l
:Cl~Draw
:AII-Off
-1.sl---'---+-+-J."'+--+--'----ll.5
: 1. 5--Xmin
: 1. 5--Xmax
: l--Xscl
:-l--Y;-:-,in
: l--Ymax
: l--Yscl
: DispGraph
:=-.ine(C,O,A,B)
:Pause
: F:mction
:End
To run this program, enter 0 for radian mode or 1 for degree mode. Then enter
any angle (the angle can be negative or larger than 360°). After the angle is dis­
played, press ENTER to clear the display. The figure shows an angle of -750°, as
drawn by this program.
WARM-UP
The following warm-up exercises involve skills that were covered in earlier ~tions,; You
will use these skills in the exercise set for ,this section.
In Exercises I-to, solve for x.
3.
51T
1T;: -
6
45
x
180
2. 790
+x
4.2'17" -
1T
5.-=-
7. x
1T
180
x
9. -
60
20 3
4
= 720 +
= 180
+ 135
1. x
8. ­
5'17"
X
= 3"
240
180
180
330
'IT
x
6.-=
10. _x_
3600
X
x
1T
0.0125
SECTION 2. I
115
RADIAN AND DEGREE MEASURE
SECTION 2.1 . EXERCISES
........................................................................................................................
In Exercises 1-4. determine the quadrant in which the terminal
side of the angle lies. (The angle is given in radians.)
L (a)
7r.
5
r.
5'
(b) -
13. (a)
(b)
~_450
II r.
9
2. (a)
r.
12
(b)
3. (a)
I
(b) -2
4. (a) 5.63
In Exercises 13-16. determine two coterminal angles (one pos­
itive and one negative) for the angle. Give your results in
degrees.
(b) -2.25
14. (a)
/ ~o
,~e" -I,"
(bJe=L
In Exercises 5 and 6, detennine the quadrant in which the
terminal side of the angle lies.
S. (a) 1300
6. (a) .- 260"
15. (a)
(b) 28SO
e=
(b)
740
c
/~
(b) -3.4 0
.
,
In Exercises 7-10. sketch the angle in standard position.
7.
5r.
(a) ­
(b)
(b)
, 5[J
.
7r.
1
" \
4
5r.
2
In Exercises 17-20, find (if possible) the positive angle comple­
ment and the positive angle supplement of the angle.
9. (a) 30°
(b) 150"
10. (a) 405 0
(b) -480°
In Exercises 11 and 12, determine two coterminal angles (one
positive and one negative) for the angle. Give your results in
radians.
11. (a)
(b)
12. (a)
-420°
.(~'.\
(b) 3
8. (a)
e=
16. (a)
4
2r.
(b)
f""
21'
e=-­
15
~. r. 17. (a) 3
18. (a) J
3r.
(b) ­
19. (a) 18°
(b) 115 0
20. (a) 79°
(b) 150c
4
(b) 2
In Exercises 21-24. express the angle in degree measure. (Do
not use a calculator.)
3r.
21. (a) 2
22. (a)
23. (a) (b) ­
7r.
12
7r.
:3
II r.
24. ( a ) ­
6
(b) ­
7r.
6
r.
9
II r.
(b)
30
34r.
15
(b)­
116
CHAPTER 2
TRIGONOMETRY
In Exercises 25-28, express the angle in radian measure as a
multiple of 71'. (Do not use a calculator.)
25. (a)
26. (al
27. (a)
28. (a)
30°
315 0
-20°
- 2700
(b) 1500
(b) 1200
(b) -240°
(b) 144°
In Exercises 29-32, convert the angle from degrees to radian
measure. Express your result to three decimal places.
29. (a) 115°
30. (al - 216.35°
0
31. (al 532
32. (al -0.83°
(b) 87.40
(b) -48.27°
(b) 0.54°
(b) 345 0
In Exercises 43-46, find the radian measure of the central angle
of a circle of radius r that intercepts an arc of length s.
Radius 43. 15 inches
44. 16 feet
45. 14.5 centimeters
46. 80 kilometers
Arc
4 inches
10 feet
25 centimeters
160 kilometers
In Exercises 47-50, on the circle of radius r find the length of
the arc intercepted by the central angle e.
Radius 47. 15 inches
48. 9 feet
49. 6 meters 50. 40 centimeters
Central Angle
180°
60°
2 radians
371'
4' radians
In Exercises 33-36, convert the angle from radian to degree
measure. Express your result to three decimal places.
33. (al ­
71' 7
1571'
34. ( a ) 8
35. (aJ -4.271'
36. (al -2
571'
(b) II
(b)
6.571'
(b) 4.8
(b) -0.S7
In Exercises 37 and 38, convert the angle measurement to
decimal form.
37. (a) 245° 10'
38. (a) -13SO 36"
(b) 2° 12'
(b) -408° 16' 2S"
In Exercises 39-42. convert the angle measurement to DO M' S"
form.
39. (aJ 240.6°
40. (aJ - 345.12°
41. (a) 2.S
42. (a) ~-0.35S
(bJ
145.8°
(b) 0.45
(b) -3.58
(b) 0.7865
Distance Between Cities In Exercises 51-54, find the distance
between the two cities. Assume that the earth is a sphere of
radius 4000 miles and that the cities are on the same meridian
(one city is due north of the other).
Latitude
--_._­
51. Dallas
Omaha
52. San Francisco
Seattle
53. Miami
Erie
54. Johannesburg, South Africa
Jerusalem, Israel
32°
41°
37°
47°
25°
42°
26°
31°
47' 9" N
15' 42" N
46' 39" N
36' 32" N
46' 37" N
7' IS" N
10' S
47' N
55. Difference in Latitudes Assuming that the earth is a
sphere of radius 4000 miles, what is the difference in lati­
tude of two cities, one of which is 325 miles due north of
the other?
56. Difference in Latitudes Assuming that the earth is a
sphere of radius 4000 miles, what is the difference in lati­
tude of two cities, one of which is 500 miles due north of
the other?
SECTION 2.1
57. Instrumentation
The pointer on a voltmeter is 2 inches
long (see figure). Find the angle through which the pointer
rotates when it moves inch on the scale.
1
RADIAN AND DEGREE MEASURE
117
59. Angular Speed A car is moving at the rate of 50 miles per
hour. and the diameter of each wheel is 2.5 feel. (a) Find the
number of revolutions per minute of the rotating wheels.
(b) Find the angular speed of the wheels in radians per
minute.
60. Angular Speed A truck is moving at the rate of 50 miles
per hour. and the diameter of each wheel is 3 feel. (a) Find
the number of revolutions per minute the wheels are rotat­
ing. (b) Find the angular speed of the wheels in radians per
minute.
61. Angular Speed A 2-inch-diameter pulley on an electric
• •
FtGURE FOR 57
58. Electric Hoist
An electric hoist is used to lift a piece of
equipment (see figure). The diameter of the drum on the
hoist is 8 inches and the equipment must be raised I foot.
Find the number of degrees through which the drum must
rotate.
motor that runs at 1700 revolutions per minute is connected
by a belt to a 4-inch-diameter pulley on a saw arbor. (a)
Find the angular speed (in radians per minute) of each
pulley. (b) Find the rotational speed (in rpm) of the saw.
62. Angular Speed How long will it take a pulley rotating at
12 radians per second to make 100 revolutions?
63. Circular Saw Speed The circular blade on a saw has a
diameter of 7.5 inches and the blade rotates at 2400 revolu­
tions per minute (see figure). (a) Find the angular speed in
radians per second. (b) Find the speed of the saw teeth (in
feet per second) as they contact the wood being cuI.
l--
7.5 in.
FIGURE FOR 63
64. Speed of a Bicycle
The radii of the sprocket assemblies
and the wheel of the bicycle in the figure are 4 inches.
2 inches. and 13 inches. respectively. If the cyclist is pedal­
ing at the rate of I revolution per second. find the speed of
the bicycle in (a) feet per second and (b) miles per hour.
1 ft
I
T
FIGURE FOR 58
-1 1--2 in.
FIGURE FOR 64
118
CHAPTER 2
TRIGONOMETRY
ClRCL.E
Tdgurwmetric Functions ! Domain and Period of
Td~ fU!ictiom with a Calculator
The Crill Circle
The Unit Circle
}' The two historical perspectives of trigonometry incorporate different methods
for introducing the trigonometric functions. Our first introduction to these
functions is based on the unit circle. Consider the unit circle given by
(0, 1) (~1,
0)
(1, 0)
---<t----T------t--x
(0,
Unit Circle: x~
FIGURE
1)
as shown in Figure 2.17. Imagine that the real number line is wrapped around
this circle, with positive numbers corresponding to a counterclockwise wrap­
ping and negative numbers corresponding to a clockwise wrapping, as shown
in Figure 2.18.
+ y'
2.17
y
(J, 0)
-j-----..----+- x
-I
-2
(a) Positive numbers
FIGURE
-2
(b) Negative numbers
2.18
As the real number line is wrapped around the unit circle, each real
number t will correspond with a point (x, y) on the circle. For example, the real
number 0 corresponds to the point (I, 0). Moreover, because the unit circle has
a circumference of 21T, the real number 21T will also correspond to the point
(1, 0),
SECTION 2.2
THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE
119
The Trigonometric Functions
From the preceding discussion, it follows that the coordinates x and yare two
functions of the real variable t. These coordinates are used to define the six
trigonometric functions of t.
sine
cosine
tangent
cosecant
secant
cotangent
These six functions are normally abbreviated as sin, esc, cos, sec, tan, and cot,
respectively.
DEFINITION OF TRIGONOMETRIC FUNCTIONS
Let t be a real number at (x, y) the point on the unit circle corresponding
to t.
sin t = Y
esc t
cost=x
I
sect=-,x*-O
x
y
(0, I)
v
tan t = '--,
x
(I, 0)
x*-O
cot t =
y*-O
y
x
,
y*-O
Y
~------~----~~_x
(-I, 0)
'I
As an aid to memorizing these definitions, note that the functions in the
second column are the reciprocals of the corresponding functions in the first column.
(0,
Unit Circle Divided into 8 Equal Arcs
FIGURE
2.19
°
y
( _~2'
v'23)
(0, 1)
~--.,.---~
(~
3 ~)
(_v'2'2
2'
v'23)
(-;,D
(1, O)
~~----~------~~X
(0, -I)
Unit Circle Divided into 12 Equal Arcs
FIGURE
2.20
In the definition of the trigonometric functions note that the tangent or
secant is not defined when x = 0. For instance, because t = n/2 corresponds
(0, I), it follows that tan( n/2) and sec( n/2) are undefined. Simi­
to (x, y)
O. For instance,
lar/y, the cotangent or cosecant is not defined when y
because t = 0 comesponds to (x, y) = (1,0), cot and esc 0 are undefined.
In Figure 2.19, the unit circle has been divided into eight equal arcs,
corresponding to t-values of
n n 3n
0,
4' 2' 4'
5n 3n 7n
n,
4' 2' 4' and 2n.
Similarly, in Figure 2.20, the unit circle has been divided into 12 equal arcs,
corresponding to t-values of
n n n 2n 5n
7n 4n 3n 5n I In
0, 6' 3' 2' 3' 6' n, 6' 3' 2' 3' 6' and 2n.
Using the (x, y) coordinates in Figures 2.19 and 2.20, you can easily
evaluate the trigonometric functions for common t-values. This procedure is
demonstrated in Examples I, 2, and 3.
120
CHAPTER
2
TRIGONOMETRY
E\aJuating Trigonometric Functions of Real Xumbers
EXAMPLE 1
DISCOVEr{y
Use a graphing utility to draw the
unit circle as follows.
Evaluate the six trigonometric functions at the following real numbers.
A. t
I. Set the utility 10 radian and
parametric mode.
2. Let 0
T :$ 6.3. with
= 0.314
3. Let -1.5 :$ X :$ 1.5 and
-- 1.5 :$ Y :$ 1.5.
4. Let XIT
cos T and Yn =
sin T.
5. Activate the graphing utility.
The screen should look like
that shown in the figure. Use the
trace key to move the cursor
around the unit circle. The
T-values represent the measure of
in radians. What do the X-and
Y-values represent? Explain your
reasoning.
e
1T
="6
51T
= '4
B. t
SOLUTION
= 1T/6 corresponds to the first-quadrant point
(V3)2, 1/2), you can write the following. A. Because t
1T sin 6
1T
cos - = x =
6
1T
tan -
6
1T
csc 6
y=-
Y
x
2
v3
2
=2
2
v3
1T
sec 6
1/2
1T
cot 6
v3/2
(x, y) =
2v3
3
v3
51T/4 corresponds to the third-quadrant point (x, y)
you can write the following.
B. Because t
( - \12/2, - \12/2),
\12 51T
sin - = y 4
51T
cos 4
1--+---+---+---11.5
51T
tan -
4
x
2
=
\12 2
\12/2
y
=:..
x
51T
csc- =
4
sec
cot
2
-\12
51T
4
51T
4
....... .. Evaluating Trigonometric Functions of Real Numbers
EXAMPLE 2
Evaluate the six trigonometric functions at the following real numbers.
A. t =
0
B. t = 7T
SOLUTION
A. t
0 corresponds to the point (x, y)
sin 0
y
cos 0
x
v
tan 0 = =x
0
=
1
=0
= (L 0) on the unit circle.
csc 0 is undefined
sec 0 cot 0 is undefined
SECTION 2.2
B. t
=
1T
THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE
corresponds to the point (x, .v)
sin
1T
Y
cos
1T
= X
tan 1T
=0
v
-I
=- = -
0
x
Ex AMPLE 3
= 0
1
= (- I,
121
0) on the unit circle.
CSC 1T
is undefined
sec
1T
= -I
cot
1T
is undefined
.. " " ... "
"
Evaluating Trigonometric Functions of Real "iumbers
Evaluate the six trigonometric functions at the following real numbers.
1T
A. 1=
B. t =
3
51T
SOLUTION
A. Moving clockwise around the unit circle, you can see that
corresponds to the point (x, y)
(1/2, v3/2)
Sin(
-~)
cos(
-~) =x
2
tan (
-~) = ~ =
-V3
v3 y=
2
sec(
r
~)
csc(
-~)
r./3
I
2
=2
1T) =
cot ( - '3
X
Y
B. Moving counterclockwise around the unit circle one and a quarter revolu­
51T/2 corresponds to the point (x. y) = (0. I).
tions, you can see that f
.
sm
tan
(0. -1)
-1 ;::; x ;::; 1
FIGURE 2.21
2
y= 5r.
cos -~ = x
2
y
~---+---+---
51T x
0
5r.
v
= '- is undefined
2
x
51T
2
5r.
csc
~
sec
2
cot
~
is undefined
5r.
2
x
- = 0
y
.. " . " ...
"
-1;::; y ;::; 1
Domain and Period of Sine and Cosine
The domain of the sine and cosine functions is the set of all real numbers. To
determine the range of these two functions. consider the unit circle shown in
y and cos f = x. Moreover.
Figure 2.21. Because r = 1. it follows that sin t
122
CHAPTER 2
TRIGONOMETRY
because (x, y) is on the unit circle, you know that -] S Y S J and
-1 S x S 1, and it follows that the values of the sine and cosine also range
between I and I. That is,
-ls),SI
I S sin t S
-Isxsi S cos t S I. and
Suppose you add 211' to each value of t in the interval [0, 211' J. thus
completing a second revolution around the unit circle, as shown in Figure 2.22.
The values of sin(t + 211') and cos(t + 211') correspond to those of sin t and
cos t. Similar results can be obtained for repeated revolutions (positive or
negative) on the unit circle, This leads to the general result
sin(t + 21Tn)
In Figure 2.22, positive
multiples of 21T were added to the t­
values. You could just as well have
added negative multiples. For instance,
1T/4
21T and 1T/4
41T are also
coterrninal to 1T/4.
= sin t
and
cos(t + 21Tn) = cos t
for any integer n and real number t. Functions that behave in such a repetitive
(or cyclic) manner are called periodic.
t :=
y
37T 37T
t
:=
t
4' 4 +
:=
7T 7T
2
2'
27T, ...
+
7T
27T,
2 + 47T,
7T 7T
4' 4 +
7T 27T,
57T 57T
:=
4+
47T, ... 7T, 37T, •..
-----+-----+-----+--~x
t := 0,
t
...
4'
4
+ 21T, ...
77T 77T
4' 4 +
27T, 47T, ...
77T 21T, 4 + 47T, ... Repeated Revolutions on the Unit Circle
FIGURE
2.22
DEFINITION OF A PERIODIC FUNCTION
A functionJis periodic if there exists a positive real number c such that
J(t
+
c)
=
J(t)
for all t in the domain off The least number c for which J is periodic is
called the period off
From this definition it follows that the sine and cosine functions are
periodic and have a period of 211'. The other four trigonometric functions are
also periodic, and we will say more about this in Section 2.6.
SECTION 2.2
THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE
Csing thE: hTioct In
EXAMPLE 4-
A. Because
2.luate
S~ne
123
and
1317
17
6 = 217 + 6' you have
1317 = sin ( 217
sin (5
B. Because
cos( -
717
2
-417
7;) =
. 17
17)
+ "6
Sin -
6
=
I
2
.
17
+ -, you have
2
cos( -417
+
~)
= cos
~=
0,
Recall from Section 1.7 that a function f is even if
f( -t) = f(t)
and is odd if
f( -t) = - f(t).
Of the six trigonometric functions, two are even and four are odd, as stated in
the following theorem.
EVEN AND 000 TRIGONOMETRIC FUNCTIONS
The cosine and secant functions are even.
cos( -t) = cos t
sec( -t)
sec t
The sine, cosecant, tangent, and cotangent functions are odd.
sine - t)
-sin t
tan( - t) = -tan t
csc( -t) = -csc t
cot( -t) = -cot t
Evaluating Trig Functions with a Calculator
DISCOVERY
Set your graphing utility in
degree mode and enter tan 9(),
What happens" Why" Now set
your graphing utility to radian
mode and enter lan(3.14159; 2!.
Explain the calculator's anS\\er.
From the arc length formula s = r8, with r = I, you can see that each real
number t measures a central angle (in radians). That is, t = r8
1(8) = 8
radians. Thus, when you are evaluating trigonometric functions. it doesn'r
make any difference whether you consider t to be a real number or an angle
given in radians.
A scientific calculator can be used to obtain decimal approximations of the
values of the trigonometric functions, Before doing this, however. you must be
sure that the calculator is set to the correct mode: degrees or radians. Here are
two examples.
124
CHAPTER 2
TRIGONOMETRY
Degree mode:
cos 28
Radian mode:
tan ­
12
j)i.\pjrl'"
11'
Most calculators do not have keys for the cosecant, secant, and cotangent
functions. To evaluate these functions, use the reciprocal key with the func­
tions sine, cosine, and tangent. Here is an example.
11'
esc
8
\
= . ( / ) = 2.613.
Sin 11'
EXA[v'!PLE 5 8
Using a Calculator to Evaluate Trigonometric
Functions
Use a calculator to evaluate the following. (Round to three decimal places.)
A. sin(-76.4)"
B. cot 1.5
c. sec(5° 40' 12")
SOLUTION
Rounded to 3
FunCTion
Mode
Degree
Radian
Degree
A. sine 76.4t
B. cot 1.5
c. sec(5° 40' 12")
Note that 5° 40' 12"
= 5.67°.
Decimal Places
-0.97196\ 0006
0.0709148443
1.004916618
-0.972
0.071
1.005
.........
w
Suppose you are tutoring a student who is just learning trigonometry. Your stu­
dent was asked to evaluate the cosine of 2 radians and, using a calculator, ob­
tained the following.
You BE THE INSTRUCTOR Keyscrokes
0.999390827
You know that 2 radians lies in the second quadrant. You also know that this
implies that the cosine of 2 radians should be negative. What did your student do
wrong?
SECTION 2.2
WARM-UP
THE TRIGONOMETRIC FUNCTIONS AND THE UNIT CIRCLE
125
.The following wann-up exercises involve skills that were covered in earlier sections. You
,Will use theSe skills in the exercise set for this section.
.fu EXerclses,1 and 2; simplify the expression.
•
1.
I
2·
~r:
-V3
V2
2.
2
~ r:
-v2
2
2
In Exercises 3 and 4, find an angle () in the interval [0, 21TJ that is coterminal with the given
angle.
3 81T
• 3
4.
4
In Exercises 5 and 6, convert the angle to radian measure.
5. 30·
6. 13Y
In Exercises 7 and 8, convert the angle to degree measure.
7.
7T
3" radians
8. -
31T
2
radians
9. Determine the circumference of a circle with radius I.
10. Determine the arc length of a semicircle with radius I.
SECTION 2.2 . EXERCISES
...•.•••.....••.......................................
.....••.......•••..................................... .....•...... In Exercises 1-8, find the point (x, y) on the unit circle that
corresponds to the real number I (see Figures 2.19 and 2.20).
1. [ ==
3. t
1T
4
51T
6
41T
5. 1==3
37T
7. 1 = 2
1T
2. [
4.
6.
3
51T
I
4
1=
8. [
=
111T
6
1T
In Exercises 9-16. evaluate the sine. cosine. and tangent of the
real number.
9.
11.
13.
1T
4
51T
t =
4
I IT.
I
15. [
6
41T
3
1T
4
10. t ==
[
12.
14.
16.
5'77
I =
6
21T
I
3
7rr
I
4
126
CHAPTER 2
TRIGONOMETRY
In Exercises 17-22, evaluate (if possible) the six trigonometric
functions of the real number.
37T 27T
17. 1 = -
IS. 1=
7T 19. 1=-
37T
20. 1 = ­
4
2
2
47T 21. 1=
3
117T
22. 1=
3
6
In Exercises 45 and 46, use the accompanying figure and a
straightedge to approximate the values of the trigonometric
functions. In Exercises 47 and 48, approximate the solutions of
the equations. Use 0 :5 I :5 27T.
45. (a) sin 5
(b) cos 2
46. (a) sin 0.75
(b) cos 2.5
47. (a) sin I = 0.25
48. (a) sin I = -0.75
(b) cos I = 0.75
(b) cos I = -0.25
1.75
2.00
In Exercises 23-30, evaluate the trigonometric function using
its period as an aid.
23. sin 37T
24. cos 37T 87T
25. cos3
26. sin ­
\
2.25
0.8
\
2.50
97T .1.0
o.
0.4
2.75
4
137T)
2S. sin ( -6
29. sin( _ 9;)
30. cos( _ 8;)
1.25
1
1.00
'
/
1
t
0.75
/
0.50
0.25
0.2
3.00
197T 27. cos - 6
1.50
1
I
3.25
I
I
0.2
!
!
0.4 0.6
I:::J 6.25
0.8 1.0
6.00
3.50
1
1
/
In Exercises 31-36, use the value of the trigonometric function
to evaluate the indicated functions.
31. sin t =
/
4.00
t
4.50
(a) sin I
-i
(a) cos (-I)
(b) sec (-I)
49. Harmonic MOlioll
(b) sec(-I)
35. sin t = ~
(a) sin( 7T - I)
(a) cos( 7T - r)
(b) sin(1
+ 7T)
(b) cos(r
+ 7T)
In Exercises 37-44, use a calculator to evaluate the trigonomet­
ric function. (Set your calculator in the correct mode and round
your answer to four decimal places.)
41. tan I 10.5°
43. csc 0.8
The displacement from equilibrium of
an oscillating weight suspended by a spring is
y(l) =
I = ~
4
39. cos 34.2°
5.00
FIGURE FOR 45-48
34. cos I = - ~
7T
4.75
(b) csc I
(a) cos I
37. sin -
5.50
(b) CSc(-I)
32. sin( -I) = ~
33. COS(-I) =
5.75
5.25
4.25
(a) sin(-I)
36. co,
3.75
3S. tan 7T
where v is the displacement in feet and I is the time in seconds. Find the displacement when (a) I = 0, (b) I = j, and (c) I = ~. 50. Harmonic Malian
The displacement from equilibrium of
an oscillating weight suspended by a spring and subject to
the damping effect of friction is
y(t) =
40. cot I
42. sec 54.9°
44. sin( -0.9)
j cos 61, je
I
cos 61,
where y is the displacement in feet and I is the time in
seconds. Find the displacement when (a) I = 0, (b) I = j,
and (c) I = ~.
5 E C TI 0 N 2.3
51. Electric Circuits The initial current and charge in the
electrical circuit shown in the accompanying figure is zero.
The current when J 00 volts is applied to the circuit is
given by
TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES
127
52. Use the unit circle to verify that the cosine and secant
functions are even and the sine. cosecant tangent, and
cotangent functions are odd.
5e- 21 sin t
I
if the resistance, inductance, and capacitance are 80 ohms,
20 henrys, and 0.01 farads, respectively. Approximate the
0.7 seconds after the voltage is applied.
current t
c
R
o------c
E(t)
FIGURE FOR 51
2.3 TRIGONOMETRIC FUNCTIONS AND
RIGHT TRIANGLES
Trigonometric Functions of an Acute Angle I Trigonometric
Identities I Applications Involving Right Triangles
Trigonometric Functions of an Acute Angle
second look at the trigonometric functions is from a right triangle per­
spective. In Figure 2.23, the three sides of the triangle are the hypotenuse, the
opposite side (the side opposite the angle 8). and the adjacent side (the side
adjacent to the angle 8). Using the lengths of these three sides. you can form
six ratios that define the six trigonometric functions of the acute angle 8.
In the following definition it is important to see that 0° < 8 < 90° and
that for such angJes the value of each of the six trigonometric functions is
positive.
OUf
.8
Adjacent side
FIGURE
2.23
128
CHAPTER
2
TRIGONOMETRY
RIGHT TRIANG.LE DEFINITION OF TRIGONOMETRIC
FUNCTIONS
Let e be an acute angle of a right triangle. Then the six trigonometric
functions of the angle e are defined as follows.
sin
e
csc
hyp
cos 8 hyp
sec8=­
adj
hyp tan 8 cot
adj hyp
e =opp
­
e =adj­
opp
The abbreviations opp, adj, and hyp represent the lengths of the three
sides of a right triangle.
the length of the side opposite 8
the length of the side adjacent to
the length of the hypotenuse
opp
adj
hyp
e
Find the values of the six trigonometric functions of
shown in Figure 2.24.
e in
the right triangle
SOLUTION
By the Pythagorean Theorem, (hyp)2 = (opp)!
\125
hyp
Thus, adj
SJ
;y
§'
C'
.:$>"<.
4
sin 8
cos
'\8
e
\
3
FIGURE 2.24
tan 8 =
3, opp
= 4,
4
hyp
-
5
3
hyp
adj
5
(adj)2, it follows that
= 5.
and hyp
= 5.
csc
e=
hyp
opp
=
5
4
sec
e = ­ hyp
adj
=
5
3
cot 8
+
-
adj
opp
3
4
"' ,. "
:;
..
".
" '"
~
!:!,;;""'~~'r:::£iiJ.£
In Example I, you were given the lengths of the sides of the right triangle,
but not the angle 8. A much more common problem in trigonometry is to be
asked to find the trigonometric functions for a given acute angle e. To do this,
construct a right triangle having e as one of its angles.
5 E C T JON 2.3
129
TR tGONOMETR IC FUNCTIONS AND RIGHT TRIANGLES
E\aluating Trigonometric Functions of 45°
EXAMPLE 2
Find the values of sin 45°, cos 45°, and tan 45°.
SOLUTION
Construct a right triangle having 45° as one of its acute angles, as shown in
Figure 2.25. Choose the length of the adjacent side to be 1. From geometry,
you know that the other acute angle is also 45°. Hence, the triangle is isosceles
and the length of the opposlle side is also I. Using the Pythagorean Theorem,
you can find the length of the hypotenuse to be
VI" +
hyp
=
12
vi
Finally, you can write the following.
opp
hyp
cos 45°
45°
tan 450
V2 1
= V2 = 2 adj
1
V2
hyp
V2
2
= opp = adj FIGURE 2.25
Ex AMPLE 3
,. ,. ,. II II ,. .. ,. /I
Evaluating Trigonometric Functions of 30° and 60°
Use the equilateral triangle shown in Figure 2.26 to find the values of sin 60°,
cos 60°, sin 30°, and cos 30°.
SOLUTION
Try using the Pythagorean Theorem to verify the lengtlE> of the sides given in
Figure 2.26. For (J = 60°, you have adj
I, opp = V3, and hyp = 2, which
implies that
FIGURE 2.26
opp -= V3
hyp
2
For (J
DISCQVERY
Set your graphing utility in
degree mode and choose un anglc
\. Now evaluate costr) and
sinl90
xL Whut do you
oh,cn c·: Repeat this experiment
II
ith ,inC,) and
COS\I.)O
.11.
=
30°, you have adj
. 30° = -opp
Sin
and
hyp
I
2
= -
cos 60°
adj
hyp
I
2
=­
V3,opp = 1, and hyp = 2, which implies that
and
°
adj
cos 30 = hyp
V3
=.
2
.. ,. .. II .. ,. .. /I ..
Because the angles 30°, 45°, and 60° (7T/6, 7T/4, and 7T/3) occur fre­
quently in trigonometry, we suggest that you learn to construct the triangles
shown in Figures 2.25 and 2.26.
130
CHAPTER 2
TRIGONOMETRY
SINE, COSINE, AND TANGENT OF SPECIAL ANGLES
7T
0
sin 30 = sin 6
•
Sin
45
7T
2'
cos 30°
= Sin -7T4 = V2
2'
0
•
sin 60"
.
7T
Sin -
3
cos 6
cos 45° = cos V3
= -
V3
Ti'
= 2'
7T
V2
4
2
tan - = ­
6
3
tan
7T
I
cos - = -.
2'
3
4
7T
tan 60° = tan -
2
3
,r:;
v3
Trigonometric Identities
In the preceding list, note that sin 30° = ~ = cos 60", This occurs because 30°
and 60° are complementary angles, In genera!, it can be shown that cofunctions
of complementary angles are equal. That is, if () is an acute angle. then the
following relationships are true.
sin(90° - ())
tan(90° - 8)
sec(90° - 8)
=
cos(90"
8)
())
cot (90°
csc(90° - 8)
cos 8
cot (J
csc (J
= sin
8
tan () sec 8 For instance, because 10° and 80° are complementary angles, it follows that
cot 80°.
sin 10c = cos 80° and tan 10°
FUNDAMENTAL TRIGONOMETRIC IDENTITIES
Reciprocal Identities
sin 8
csc
Tecnno(OiJlJ Note _ _ _ _ _ __
Use your calculator to confiml several
of the trigonometric identities at the
right for various values of 8, For
instance, calculate (sin
(cos 0,))' and observe that the vulue
is l.
(J
csc
sec 8
(J
J
= - ­
sin
cos 8
(J
tan 8
cos 8
I
cot
=-­
(J
I
J
cot 8 = - ­
tan 8
=-­
sec 8
Quotient Identities
sin ()
tan 8 = - ­
cos (J
CO!
8=
cos 8
sin 8
Pythagorean Identities
sin" 8
REMARK
+ cos 2 8 =
We use sin'
(J
J
+ tan:: 8 = sec:: 8
to represent (sin 8)', cos'
(J
I
+ cot2 8
to represent (cos
(J)',
csc" 8
-
and so on,
SECTION 2.3
TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES
4
EXAMPLE
131
Applying Trigonometric Identities
Let 6 be the acute angle such that sin 6 = 0.6. Use trigonometric identities
find the values of the following.
10
B. tan 6
A. cos 6
SOLUTION
+ cos" 6
(0.6)2 + cos" 6
A. sin 2 6
cos" () = I - (0.6)"
cos 6
=
YO.64
=
=
0.64
0.8.
B. Now, knowing the values of sin () and cos (), you can find the value of
tan ().
tan () 0.8
sin 6 = 0.6
FIGURE 2.27
sin 6
0.6
cos 6
0.8
0.75.
Try using the definitions of cos () and tan 6, and the triangle shown in
Figure 2.27, to check these results. • .••.••••
EXAMPLE 5
Applying Trigonometric Identities
Let () be an acute angle such that tan ()
the values of the following.
A. cot 6
3. Use trigonometric identities to find
B. sec ()
SOLUTION
A. cot 6
3
tan ()
Rt'Ciprocai /dell IiI.'
3
B. sec 2
tan" 6 +
sec 2 ()
31 + I
sec" () = 10
sec () =
()
Pyrlwgoreon idcntifY
VIo
tan 6
=3
FIGURE 2.28
Try using the definitions of cot () and sec 6, and the triangle shown in
Figure 2.28, to check these results.
.. .......
132
CHAPTER
2
TRIGONOMETRY
Applications Involving Right Triangles
Many applications of trigonometry involve a process called solving right
triangles. In this type of application, you are usually given two sides of a right
triangle and asked to find one of its acute angles. or you are given one side and
one of the acute angles and asked to find one of the other sides.
EXAMPLE 6
Using Trigonometr) to Solve a Right Triangle
A surveyor is standing 50 feet from the base of a large tree, as shown in
Figure 2.29. The surveyor measures the angle of elevation to the top of the tree
as 71.5°. How tall is the tree?
SOLUTION
From Figure 2.29, you can see that
tan 71S = opp
ad]
y
x
where x
50 and y is the height of the tree. Thus. you can determine the
height of the tree to be
y
x tan 71.5° = 50(2.9&&68) = 149.4 feet. EXAMPLE 7
......... Using Trigonometry to Solve a Right Triangle
A person is standing 200 yards from a river. Rather than walking directly to
the river, the person walks 400 yards along a straight path to the river's edge.
Find the acute angle 0 between this path and the river's edge, as indicated in
Figure 2.30.
SOLUTION
From Figure 2.30, you can see that the sine of the angle 0 is
FIGURE 2.30
sin 0 = opp
hyp
200 400
2
Now, you can recognize that 0 = 30°. ...... .. . SECTION 2.3
TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES
133
In Example 7, you were able to recognize that the acute angle that satisfies
the equation sin e = ~ is e = 30°. Suppose, however, that you were given the
equation sin e == 0.6 and asked to find the acute angle e. Because
.
sm -30°
I
= -2 = 0 .5000
and
.
sm 45° =
I
v2
=
0.7071,
you can figure that e lies somewhere between 30° and 45°. A more precise
value of can be found using the inverse key on a calculator. (Consult your
calculator manual to see how this key works on your own calculator.) For most
calculators, one of the following keystroke sequences will work.
e
Degree mode: .61IN~ )sinl
Degree mode:
~di [sir!) .6 ~~:ri~J
Thus, you can conclude that if sin
EXAMPLE 8
DI51'/0I' 3686989765
e
Disl'/a\' 3686989765
0,6, then
e
Using Trigonometry to Solve a Right Triangle
I
40 ft
1
A 40-foot flagpole casts a 30-foot shadow, as shown in Figure 2,31. Find
the angle of elevation of the sun. Angle of
elevation ~\.
e
FIGURE
SOLUTION
I
/
1
.... · - - - 30 ft
e,
Figure 2.31 shows that the opposite and adjacent sides are known, Thus,
tan
---1
40
30'
=­
With a calculator in degree mode you can obtain
2.31
COMPARING
DEFINITIONS
OF
TRIGONOMETRIC
FUNCTIONS
e = -opp
adj
e=
53.13°.
....... .. In Section 2.2 and in this section, we presented two different definitions of
trigonometric functions. One was the "unit circle" definition and the other was
the "right triangle" definition. Write a short paper that compares the two
definitions. Then use both definitions to find the values of the six trigonometric
functions at
30°. For this value of which definition do you prefer" For
= 37T, which do you prefer and why?
e
e
e,
134
CHAPTER
2
TRIGONOMETRY
WARM-UP The following warm-up exercises involve skills that were covered in earlier sections. You
will use these skills in the exercise set for this section.
In Exercises 1-4, find the distance between the points.
1. (3, 8), 0, 4)
2. (5, 2), (2, -7)
3. (-4,0), (2, 8)
4. (-3, -3). (0, 0)
In Exercises 5-10, perfonn the indicated operation(s). (Round your result to two decimal
places.)
5. 0.300 x 4.125
7. 6. 7.30
2.40 9. 19,500
0.007
SECTION
2.3 .
.......................
~II
X·
43.50
3740
8.­
28.0
151.5 10.
(10.5)(3401)
1240
EXERCISES
••• "."''' .....
\ln'''.'' ..
''~ll'
..
~.Cl
.... O •••
,,~
0."1:"" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In Exercises 1-8. find the exact values of the six trigonometric
functions of the angle e. (Use the Pythagorean Theorem to find
the third side of the triangle.)
5.
1.
15
2
8
8.
7. 4.
3.
5
4
8
5
5
SECTION 2.3
In Exercises 9-16. sketch a right triangle corresponding to the
trigonometric function of the acute angle (J Use the Pythago­
rean Theorem to determine the third side and then find the other
five trigonometric functions of O.
9. sin
11. sec
(J
== ~
e
10. cot (J == 5
12. cos 0 ~
14. csc 0 = J:}
:2
13. tan (J == 3
IS. cot 0
16. sin
e ;;,
TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES
In Exercises :25-34. use a calculator to evaluate each function.
Round your result to four decimal places. (Be sure the calcula­
tor is in the correct mode.)
(b) cos 80 0
25. (a) sin 10°
26. (a) tan 23.5°
27. (a) sin 16.35°
(b) cot 66S
(b) csc 16.35°
28. (a) cos 16° 18'
29. (a) sec 42° 12'
(b) sin 73° 56'
(b)
17. sin 60° ==
2-
cos 60°
1
3.
(J (J
== 5.
33. (a) csc I
(b) sec
34.
(a)
1
tan 2
(b)
(n.2 I)
cot( '!!.
.2
~)
sin 30°
(d) cot 60°
v3
tan 30°
In Exercises 35-40. find the value of (J in degrees
(0° < (J < 90°) and radians (0 < e < 7T/2) without using a
calculator.
3
(b) cot 60°
(d)
cot 30°
3v2
sec (J == - ­
35. {al sin
4
(a) sin (J
(c) tan (J
20. sec
16
(b) cos 0.75
(b)
(a) csc 30°
(c) cos 30e
19. esc
32. (a) sec 0.75
=2
(a) tan 60°
(c) cos 30e
18. sin 30° == :2
r.
(b) tan
16 V1
csc 48° 7'
(b) sec 4° 50' IS"
30. (aJ cos 4° 50' IS"
r. 31. (a) cot In Exercises 17-20. use the given values and the trigonometric
identities to evaluate the trigonometric functions.
135
(b) cos
e
(d) sec(90°
tan 0
(a) cos e
(c) cot(90° - 0)
e)
2\16
cot e
(d) sin (J
37.
(b)
esc (J (a)
sec
39.
(a)
(b)
I
36. (a) cos
2
2
I
esc
e
sin
v2
e
2
(b)
40. (a) cot
3
2
e I
tan e = v3
cos e 38. (a)
2V3
e v2
tan
(b)
e= 2
e
(bl cot
(b)
e
sec
(b)
e v3 3
e v2
In Exercises 21-24. evaluate the trigonometric functions.
21. (a) cos 60°
22. (a) csc 30°
23. (a) cot 45°
7T
24. (al sin -
3
(b) Lan ­
(b) sin
7T
In
Exercises
4
(00
< e < 90°) and radians
7T
verse key on a calculator.
4
(b) cos 4Y
41.
(b) esc 45°
43.
(a)
(b)
(a)
(b)
41--44.
sin e == 0,8191
cos (j 0.0175
tan
tan
e=
e
find
the
(0
<
value of e in degrees
< 7T/2) by using the in­
(J
42. (a) cos
(b)
cos
U920
44. (a) sin
0.4663
(b)
cos
=
(J
e
(J
0.9848
0,8746
== 0.3746
e
0.3746
136
CHAPTER 2
TRIGONOMETRY
46. Solve for x.
45. Solve for y.
53. Height A 6-foot person standing 12 feet from a streetlight
casts an 8-foot shadow (see figure). What is the height of
the streetlight~
y
10
T
100
x
48. Solve for
47. Solve for x. r.
FIGURE FOR 53
25
r
30
x
50. Solve for x.
49. Solve for r.
54. Height A 6-foot man walked from the base of a broad­
casting tower directly toward the tip of the shadow cast by
the tower. When he was J 32 feet from the tower, his
shadow started to appear beyond the tower's shadow. What
is the height of the tower if the man's shadow is 3 feet long 0
~30
10
x
55. Length A 20-foot ladder leaning against the side of a
house makes a 75° angle with the ground (see figure). How
far up the side of the house does the ladder reach')
51. Solve for y. 12
52. Solve for
r.
y
50"
20
FIGURE FOR 55
SECTION 2.3
56. Width of a River A biologist wants to know the width w
of a river in order to properly set instruments for studying
the pollutants in the water. From point A. the biologist
walks downstream J00 feet and sights to point C. From this
sighting. it is determined that
SOc (see figure). How
wide is the river?
e
137
TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES
59. Machine Shop Calculations A steel plate has the form of
a quarter circle with a radius of 24 inches. Two ·inch holes
are to be drilled in the plate positioned as shown in the
figure. Find the coordinates of the center of each hole.
i
y
24
20
FIGURE FOR 56
20
24
x
FIGURE FOR 59
57. Distance From a J50·foot observation tower on the coast,
a Coast Guard officer sights a boat in difficulty. The angle
of depression of the boat is 4° (see figure). How far is the
boat from the shoreline 0
FIGURE FOR 57
60. Machine Shop Calculations A tapered shaft has a diame­
ter of 2 inches at the small end and is 6 inches long (see
figure). If the taper is 3°. find the diameter d of the large end
of the shaft.
58. Angle of ElevatIOn A ramp 17 ~ feet in length rises to a
loading platform that is 3 feet off the ground (see figure).
Find the angle that the ramp makes with the ground.
e
*
t-------.---~
FIGURE FOR 58
FIGURE FOR 60
6 in.
~-.~------i
138
CHAPTER 2
TRIGONOMETRY
61. Trigollomerric Funcrions by Acrual Measuremenr
Use a
compass to sketch a quarter of a circle of radius 10 centime­
ters. Using a protractor, construct an angle of 25° in stan­
dard position (see figure). Drop a perpendicular line from
the point of intersection of the tenninal side of the angle
and the arc of the circle. By actual measurement. calculate
the coordinates (x, y) of the point of intersection and use
these measurements to approximate the six trigonometric
functions of a 25° angle.
)'
i
69. A 3D-foot ladder leaning against the side of a house is 4 feet
from the house at the base (see figure).
(a) How far up the side of the house does the ladder reach?
Express your answer accurate to two decimal places.
(b) Use the fact that cos e 4/30 to find the angle that the
ladder makes with the ground. Express your answer in
degrees accurate to two decimal places.
(e) Use the tangent of the angle found in part (b) to answer
part (a) again.
(d) Why do your answers to parts (a) and (c) differ
slightly?
10
FIGURE FOR 61
62. Trigonometric Functions by Actual Measurement
Repeat
Exercise 61 using an angle of 75°.
In Exercises 63-68, determine whether the statement is true or
false, and gi ve a reason for your answer.
63. sin 60° csc 60° = 1
64. sec 30°
csc 60°
65. sin 4SO + cos 45° =
66. coe 10° 10°
67. sin 60°
sin 30°
= sin 2°
68. tan[ (0.8)2J
tan 2(0.8)
FIGURE FOR 69
SECTION 2.4
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
139
2.4
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
..........................................................................................
Trigonometric Functions of Any Angle
! Reference Angles
Trigonometric Functions of Any Angle
In Section 2.3, you learned to evaluate trigonometric functions of an acute
angle. In this section you wi1llearn to evaluate trigonometric functions of any
angle.
DEFINITION OF TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
Let e be an angle in standard position with (x, y) any point (except the
origin) on the terminal side of e and r = Vx" + y" (see figure).
e
y
cos
e=
x
-
tan
e=
'-,
x
Sin
csc
r
r
V
x*,O
e=
r
-,
y*,O
\'
sec
e = xr
cot
e
x
y
x*,O
y*,O
)'
r
=
vx? + /
REMARK
Because r =
cannol be zero, it follows that the sine and cosine
functions are defined for any real value of 8. However, if x = 0, the tangent and secant
of (] are undefined. For example, the tangent of 90° is undefined. Try calculating the
tangent of 90° with your calculator. Similarly, if y
0, the cotangent and cosecant of
8 are undefined.
-
140
CHAPTER
2
TRIGONOMETRY
y
(x, y)
y
If 0 is an acute angle, then these definitions coincide with those given in
the previous section. To see this, note in Figure 2.32 that for an acute angle
x
adj, y := opp, and r = hypo
e,
Evaluating Trigonometric Functions
EXAMPLE 1
Let ( 3, 4) be a point on the terminal side of
tangent of O.
\6
x
e.
Find the sine, cosine, and
SOLUTION
FIGURE 2.32
Referring to Figure 2.33, you can see that x
y
= Y(-3)"
r =
+
4"
:=
-
\I2s
3. y = 4. and
S.
Thus, you can write the following.
e
Y
4
r
5
cos 0
x
r
sin
tan
-3
-2
FIGURE 2.33
-]
e=
~
:=
x
-3
3
5
5
4
-3
-­
4
....... .. 3
The signs of trigonometric function values in the four quadrants can be
determined easily from the definitions of the functions. For instance, because
cos 0 =
x
r'
it follows that cos () is positive where x > 0, which is in Quadrants I and IV.
(Remember, r is always positive.) In a similar manner you can verify the
results shown in Figure 2.34.
y
y
rT
i<9<1T
Quadrant II
sin 8:
cos 8; ­
tan 8; ­
1T
0<9<2 Quadrant I x<o y> 0., sin 8: .....
cos 8:
-+
x>o
>0
,_ Y
tan 8: + ;c
Quadrant III
sin e: ­
cos 8:
lan
e:
~.
Quadrant IV
sin e:
cos 8: +
tan e: ­
;c
x < o.
y<o
"
~
'. x > 0
y<O
3rT
1T<8<T
Signs of Trigonometric Functions
FIGURE 2.34
3rT
T<8<21T
SECTION 2.4
141
Evaluating Trigonometric Functions
EXAMPLE 2
Given tan
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
e = 1and cos
0 > 0, find sin
e and sec
0.
SOLUTION
Note that 0 lies in Quadrant IV because that is the only quadrant in which the
tangent is negative and the cosine is positive. Moreover, using
v
5
4
tane=~=--
x
and the fact that y is negative in Quadrant IV, you can let y = - 5 and x = 4.
Hence, r
V 16 + 25 v4l and you have
.
SIn
-5
e = ~v
- - - = -0.7809
v4l r
EXA M PLE
3
and
sece=­
r
- - = 1.6008.
4
x
....... .. Trigonometric Functions of Quadrant Angles
Evaluate the sine function at the four quadrant angles 0, rr/2, rr, and 3rr/2.
SOLUTION
To begin, choose a point on the terminal side of each angle, as shown in
Figure 2.35. For each of the four given points, r
I, and you have
y
1T
2
(0, I)
sin 0 x
(-1, 0)
(l, 0)
0=0
r
I
. rr
y
=2
r
SIn -
0
1T
y
.
y
sm rr = r
31T
2
FIGURE
2.35
(0, -I)
. 3rr
sm2
~ =
r
IX. \
I
-=1 i
ix. I)
1
o
- = 0
I
I. Ol
10. I)
It.r; =!
I
_I
I
=
1.
Ix. I)
= 10.
1.0\
-II
Trying using Figure 2.35 to evaluate some of the other trigonometric functions
at the four quadrant angles and check them on your calculator.
•••••••••
Reference Angles
The values of the trigonometric functions of angles greater than 90° (or less
than 0°) can be determined from their values at corresponding acute angles
called reference angles.
142
CHAPTER 2
TRIGONOMETRY
Quadrant II
DEFINITION OF REFERENCE ANGLES
Reference
angle: e'
e' ==
e' ==
Let (J be an angle in standard position. Its reference angle is the acute
angle (J' formed by the terminal side of (J and the horizontal axis.
e ,
!
11 -
180
0
e (radians) e (degrees) Figure 2.36 shows the reference angles for
Find the reference angle
Quadrant III e'
e'
ee-
in Quadrants II, III, and IV.
Finding Reference Angles
EXAMPLE 4
Reference angle: e' (J
11 (radians) 0
180 (degrees) for each of the following.
(J'
B. (J
= 2.3
C. (J
=
135°
SOLUTION
A. Because
axis is
//'
(e
Reference
angle: e'
Quadrant IV
e' ==
e' ==
211
360°
(J
= 300° lies in Quadrant IV, the angle it makes with the x­
B. Because (J
2.3 lies between 1T/2 = Ic5708 and 1T = 3.1416, it follows
that (J is in Quadrant II and its reference angle is
e (radians)
- e (degrees)
(J'
= 1T - 2.3
0.8416.
Rat/wns
C. Because (J = 135 is coterminal with 225°, it lies in Quadrant III. Hence,
the reference angle is
0
FIGURE 2.36
(J' Degree.",.
Figure 2.37 shows each angle
(J
and its reference angle
(J'.
y
y
y
0
e' ==
(a)
e in Quadrant IV
FIGURE 2.37 225 and -135
are coterminal.
11
(b)
e in Quadrant II
(c)
e in Quadrant III
....... .. 0
SECTION 2.4
y
TRIGONOMETRIC FUNCTIONS Of ANY ANGLE
143
To see how a reference angle is used to evaluate a trigonometric function,
consider the point (x, y) on the terminal side of 8. as shown in Figure 2.38. By
definition, you know that
(.x; y)
f
='~
sin 8
r
v
tan 8
and
= :...
x
For the right triangle with acute angle 8' and sides of lengths Ix i and Iy I. you
have opp sin 8'
= opp =
------~~~-~_4--------X
opp
FIGURE 2.38
1)'1,
adj
Ixl
r
hyp
tan 8'
and
_ opp _ Iyl
­
- adj - Ix
I
Thus. it follows that sin 8 and sin 8' are equal, except possibly in sign. The
same is true for tan 8 and tan 8' and for the other four trigonometric functions.
In all cases, the sign of the function value can be determined by the quadrant
in which lies.
e
EVALUATING TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
To find the value of a trigonometric function of any angle
e.
1. Determine the function value for the associated reference angle e'.
2. Depending on the quadrant in which e lies. prefix the appropriate sign to the
function value.
By using reference angles and the special angles discussed in the previous
section. you can greatly extend your scope of exact trigonometric values.
Table 2.1 lists the function values for selected reference angles. For instance,
knowing the function values of 30° means that you know the function values
of all angles for which 30° is a reference angle.
TABLE 2.1
i
(degrees)
0°
30°
0
-
e
0
I
2
cos H
J
e
0
(j
45° I 60°
90°
180 0
270°
!
H (radians)
sin
tan
I
_ .-~
i
1T
1T
6
4
1T
1T
3
2
-
v2 v3
i
!
v3 v2
I
2
-v3
3
31T
1T
I
v3
I
0
0
undef.
I
J
0
0
undef.
144
CHAPTER 2
TRIGONOMETRY
5
EXA M PLE
Trigonometric Functions of ;\onacute
Evaluate the following.
A. cos
471'
1171'
"3
C. csc-­
4
SOLUTION
A. Because e = 471'/3 lies in Quadrant III, the reference angle is e' =
(471'/3) - 71' = 71'/3, as shown in Figure 2.39(a). Moreover, the cosine is
negative in Quadrant III, so that
71'
1
(- )cos - = - .
3
2
471'
cos -
3
B. Because -210° + 360°
150°, it follows that -2100 is coterminal with
the second-quadrant angle 150°. Therefore, the reference angle is
1800 - 1500
30°, as shown in Figure 2.39(b). Finally, because the
tangent is negative in Quadrant II, you have
e
V3
KCfi'reii.
3
C. Because (I 171'/4) - 271'
371'/4, it follows that 1171'/4 is coterminal with
the second-quadrant angle 371'/4. Therefore, the reference angle is e'
71' - (371'/4)
71'/4. as shown in Figure 2.39(c). Because the cosecant is
positive in Quadrant II, you have
csc
1171'
4
=
71'
(+ )csc -
4
1
sin 71'/4
= v2.
Check the above values with your calculator.
y
y
e =41T
­
3
e=
e'
111T
4
e' =!!.
3
(a) (el
(b)
FIGURE 2.39
The fundamental trigonometric identities listed in the previous section
(for an acute angle 8) are also valid when is any angle.
e
SECTION 2.4
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
t'~ing
EXAMPLE 6
145
Identities to Evaluate Trigonometric Functions
*' Find the following.
Let 8 be an angle in Quadrant II such that sin 8
B. tan 8
A. cos 8
SOLUTION
A. Because sin 8
obtain
*, use the Pythagorean Identity sin" 8 + cos" 8
=
1 to
(:31)2 + cos" 8 =
I
8
9
9
cos" 8 = 1 - - = -.
< 0 in Quadrant II, use the negative root Because cos 8
cos 8
Vs
2\/2
= - ---;:: = - - - .
Y9
3
B. Using the result from part A and the trigonometric identity tan 8
sin 81cos 8, you obtain
tan 8
3
= --- =
2Y2 Y2
4
.........
3
Scientific calculators can be used to approximate the values of trigono­
metric functions of any angle, as demonstrated in Example 7.
EXA M PLE
7
Evaluating Trigonometric Functions with a Calculator
Use a calculator to approximate the following values. (Round your answers to
three decimal places.)
TechnololJlJ Note _ _ _ _ _ __
utilities
Remember that most
do not have keys for the cosecant.
,ecant. and cotangent functions. To
evaluate the,e functions. you need to
with the 'ine.
use the reciprocal
cosine and u:.ngenl. For instance.
In Example '",Al. you can lind the
cotangent of .+10° by
I tan-POlin
A. cot 410 0
B. sin(-7)
C. tan
141T
5
SOLUTION
FunClion
A. cot 4100
B. sin( -7)
141T
C. tan
5
Mode
Display
Rounded 10 3 Decimal Places -
Degree
Radian
0.8390996312
-0.6569865987
0.839
-0.657
Radian
-0.726542528
-0.727
.........
146
CHAPTER 2
TRIGONOMETRY
At this point, you have completed your introduction to basic trigonometry.
You have measured angles in both degrees and radians. You have studied the
definitions of the six trigonometric functions from a right triangle perspective
and as functions of real numbers. In your remaining study of trigonometry. you
will continue to rely on both perspectives.
For your convenience we have included in the endpapers of this text a
summary of basic trigonometry.
There are many different techniques that people use to memorize (or reconstruct)
trigonometric formu las. Here is one that we like to use for the sines and cosines
of common angles.
MEMORIZATION
AIDS
Ii
sin fj
cos fj
OC
30"
Va
\./]
2
::
V3
Y4
2
45<
60°
90"
vi]
..
")
2
2
V2
VI
::
va
2
::
2
Write a paragraph describing the pattern indicated by this table. Discuss other
memory aids for trigonometric formulas.
WARM-UP
The following warm-up exercises involve skills that were covered in earlier sections. You
will use these skills in the exercise set for this section.
In Exercises 1-6. evaluate the trigonometric function from memory.
1. sin 30°
3. cos
7T
4
7T
5. sec6
2. tan 45°
7T
4. cot­
3
7T
6. esc ­
4
In Exercises 7-10. use the given trigonometric function of an acute angJe (} to find the values
of the remaining trigonometric functions.
7. tan
e
9. sin (}
8. cos (}
10. sec
e=
3
SECTION 2.4
SECTION
2.4 .
147
EXERCISES
In Exercises J-4, determine the exact values of the six trigono­
metric functions of the angle e.
1. (a)
TRIGONOMETRIC FUNCTIONS Of ANY ANGLE
(b)
y
y
•
In Exercises 5-8, the point is on the terminal side of an angle
in standard position. Determine the exact values of the SIX
trigonometric functions of the angle.
5.
(a)
0,24)
(b) (7.
24)
6.
7.
8
(a)
(8. 15)
(-4. 10)
(-5. -2)
(b) (-9.
-40)
(a)
(a)
(b) (3. -5) (b) (
~. 3) (8, -15)
2. (a)
y
(b)
In Exercises 9-12. use the two similar triangles in the figure to
find (a) the unknown sides of the triangles and (b) the six
trlgonometric functions of the angles Cl'1 and 0'>
y
(-12, -5)
(1, -I)
FIGURE FOR 9-12
3.
(a)
(b)
y
(-\13,
y
!
.
/
------~+_~--~x
9.
B~~. .
1)
----+---..r-----_ x
UI
10. b l
= 3. b l
12.('1
l1.ul=l.c I
12.b,
4,
=
U,
=
13.(',
9
= 26
2.b,=5
4.a2=4.b 2
10
(-2, -2)
In Exercises 13-16. determine the quadrant in which
(b)
(-2, 4)
e < 0 and cos e < 0
e > 0 and cos e < 0
sin e > 0 and cos e > 0
sin e < 0 and cos e > 0
SIn e > 0 and tan e < 0
cos e > 0 and tan e < 0
sec e > 0 and cot e < 0
esc e < 0 and tan e
0
13. (a) sin
(b) sin
J
B
-----~r_~----x
14. (a)
(b)
IS. (a)
(b)
16. (a)
(b)
e lies.
148
CHAPTER 2
TRIGONOMETRY
In Exercises 17-26. find the values (if possible) of the six
trigonometric functions of e using the given functional value
and constraint.
40. (a)
7T
41. (a) Functional Value
17. sin
e
18. cos
e= 4
5
19. tan 0
20. cos e == ~
21. sec 0 = -2
tane<O
sin
e is undefined
sin e == 0
tan e is undefined
e>
44. (a) 0
7T
37T
2
2
22. cot
-sos
23.
sec 0
24.
0
-145 0
31. (a) 0
=
32. (a) e =
33. (a) 0
27T
3
77T 4
3.5
ll7T 34. (a) 0 == -3-
e'.
and sketch 0 (b) 0:= 127"
(b) e:= 226°
-72°
(b) 0
(b) 0
=
-239°
77T
(b) 0
35. (a) 225
36. (a) 300°
37. (a) 750° 39. (a)
:3
107T (b) ­
3
6
177T
3
45. (a) sin 10°
46. (a) sec 225 0
47. (a) cost - 110°)
48. (a) csc 330°
49. (a) tan 240 0
50. (a) cot 1.35
7T 51. (a) tan 9
52. (a) sine -0.65)
(b) esc 10
0
(b) sec 135 0 (b) cos 250 0
(b) csc 150°
(b) cot 210°
(b) tan 1.35
107T
(b) tan­
9
(b) sin 5.63
6
(b) 0
9
(b) 0:= 5.8
77T
(b) 0
10 0
(b) 330
(b) 510°
0
137T
In Exercises 45-52, use a calculator to evaluate the trigonomet­
ric functions to four decimal places. (Be sure the calculator is
set in the correct mode.)
e
In Exercises 53-58, find two values of that satisfy the equa­
tion. Give your answers in degrees (00 S (] < 360°) and radi­
ans (0 s e < 27T). Do not use a calculator.
53. (a) sin 0
= ~
54. (a) cos (]
(b) -225°
0
47T
(b)
87T
In Exercises 35-44, evaluate the sine, cosine, and tangent of
each angle without using a calculator. 38. (a) -405
117T 4
2
-I
In Exercises 27-34. find the reference angle
and 0' in standard position.
27. (a) e 203°
28. (a) 0 = 309
29. (a) e:= - 245°
(b) ­
2
6
7T
OS27T
7T
25. The terminal side of 0 is in Quadrant III and lies on the
line y == 2.x.
26. The terminal side of e is in Quadrant IV and lies on the
line 4x + 3y := O.
30. (a) 0
42. (a)
43. (a)
(b) ­
6
7T Constraint
e lies in Quadrant II
e lies in Quadrant III
sin e < 0 ~
57T
(b) 4
57T 7T
'4 (b)
120°
27T
(b) -
3
55. (a) csc 0 =
V2
2
2Y3
3
56. (a) sec (]
2
57. (a) tan (] = I
58. (a) sin 0
Y3
2
(b) sin (] == -~V2
(b) cos (]
(b) cot (]
2
-I
(b) sec 0 = -2
(b) cot (]
(b) sin (] =
-
Y3
Y3
2
SECTION 2.4
In Exercises 59 and 60. use a caiculator to approximate two
values of 0(0°
0 < 360°) that satisfy the equation, Round to
two decimal places,
59. (a) sin 0
60. (a) cos 0
(b) sin
0,8191
= 0.8746
(b)
0
-0,2589
e=
-02419
cos
TRIGONOMETRIC FUNCTIONS OF ANY ANGLE
149
72. Sales A company that produces a seasonal product fore­
casts monthly sales over the next two years to be
23, I + 0,4421 + 4.3 sin
S
1TI
'6'
where S is measured in thousands of units and I IS the time
in months, with I = I representing January 1991. Predict
sales for the following months.
(a) February 1991
(b) February 1992
(c) September 1991
(d) September 1992
In Exercises 61- 64, use a calculator to approximate two values
of 0(0 ::; 0 < 21T) that satisfy the equation, Round to three
decimal places,
(b) cos
61. (al cos 0 = 0,9848
62. (a) sin 0
63. (a) tan
e=
64. (a) cot 0
0.DI75
(b) sin
1.192
5,671
(b) tan
e -05890
0 = -0,6691
e = -8,144
(b) cot 0
1.280
73. Dislance An airplane flying at an altitude of 5 miles is on
a flight path that passes directly over an observer (see
figure), If 0 is the angle of elevation from the observer to
the plane. find the distance from the observer to the plane
when (a) 0 = 30°. (b) 0 = 75 c • and (c) 0 = 90"'
In Exercises 65-68, use the value of the given trigonometric
function and trigonometric identities to find the required trigo­
nometric function of the angle in the specified quadrant
e
Given FUl1crioll
e 5
66. tan e = ~
67. csc e -2
68. sec e = :;9
65. sin
.<
Quadrant
Find
IV
cos
III
IV
III
e
sec e
cot e
tan e
5mi
FIGURE FOR 73
In Exercises 69 and 70. evaluate the expression without using
a calculator.
69. sin" 2 -" cos" 2
71. Average Temperature The average daily temperature T
(in degrees Fahrenheit) for a city is
T
= 45 - 23
COS[~(I
365
74. Consider an angle in standard position with r = 10 cm. as
shown in the figure. Write a short paragraph describing the
changes in the magnitudes of T, Y. sin
cos
and tan
as 0 increases continuously from Oc to 90°,
e,
y
32)J.
where 1 is the time in days with I = I corresponding to
January I, Find the average daily temperature on the fol­
lowing days,
(b) July 4 (t = 185)
(a) January I
(C) October 18 (I = 291)
(x, y)
__
~=-
__ ______ x
~
FIGURE FOR 74
e.
e.
150
CHAPTER
2
TRIGONOMETRY
,o... . .
.:,;.,:;.
GRAPHS OF SINE AND COSINE FUNCTIONS
hsk Sine and Cosine Curves I Key Points on Basic Sine and Cosine
ne~
Amplitude and Period of Sine and Cosine Curves / Translations
.f' :;ine and Cosine Curves
Basic Sine and Cosine Curves
f,
~.
In this section you will study techniques for sketching the graphs of the sine
and cosine functions. The graph of the sine function is called a sine curve. In
Figure 2.40, the solid portion of the graph represents one period of the func­
tion and is called one cycle of the sine curve. The gray portion of the graph
indicates that the basic sine wave repeats indefinitely to the right and left. The
graph of the cosine function is shown in Figure 2.41. To produce these graphs
with a graphing utility, make sure you have set the mode to radians.
Recall from Section 2.2 that the domain of the sine and cosine functions
is the set of all real numbers. Moreover, the range of each function is the
interval [-1. I]. and each function has a period of 27T. Do you see how this
information is consistent with the basic graphs given in Figures 2.40 and 2.41 ?
Note from Figures 2.40 and 2.41 that the sine graph is symmetric with
respect to the origin, whereas the cosine graph is symmetric with respect to the
y-axis. These properties of symmetry follow from the fact that the sine func­
tion is odd whereas the cosine function is even.
Range:
-1
y.:51
FIGURE 2.40
Period: 21T
2.41
Period: 21T
FIGURE
SECTION 2.5
GRAPHS OF SINE AND COSINE FUNCTIONS
151
Key Points on Basic Sine and Cosine Curves
To construct the graphs of the basic sine and cosine functions by hand, it helps
to note five key points in one period of each graph: the intercepts, maximum
points. and minimum points. For the sine function, the key points are
J ~ICr,'cpl
\liJ,ximC:111
~ I: l (' r,-' L' \' t
\! j :)I1',';~lln
(0,0),
(~, 1).
(7T. 0),
e -I).
\
7T
2 '
(27T,0).
For the cosine function. the key points are
\!iJ,il11111ll
\IlIli L1~1!li
Inlc;'L'ef'!
(~2' °),
(0, I L
(7T, - J),
1:-1 1 ;f'.. .-,:':.'!
e;, Q).
.1"
(27T, J ).
Note how the x-coordinates of these points divide the period of sin x and cos x
into four equal parts, as indicated in Figure 2.42.
}'
}'
I
t
Maximum Intercept Minimum Intercept Maximum
Inte cept Maximum Intercept Minimum Intercept
Full
period
Three-quarter
Half
period
period
Period: 211'
Period: 211'
FIGURE 2.42
EXAMPLE 1
Tecnnu(oiJlJ Note _______
Sketch the graph of y
When using a graphing utility to graph
trigonometric functions. pay special
attention to the viewing rectangle you
use. For instance. try graphing
\' "" [sin(30x 1]/30 on the standard
viewing rectangle. What do you
observe') Use the zoom feature to find
a
rectangle that displavs a
good view of the graph.
2 sin x on the interval
l 7T, 47T].
SOLUTION
The y-values for the key points of y = 2 sin x have twice the magnitude of
those of y
sin x. Thus, the key points for y
2 sin x are
(0.0),
(~'2).
(7T.O),
e;.-2).
and
(27T,O).
152
CHAPTER
2
TRIGONOMETRY
By connecting these key points with a smooth curve and extending the curve
in both directions over the interval [ 7T, 47T], you obtain the graph shown in
Figure 2.43. Try using a graphing utility to check this result.
FIGURE 2.43
Amplitude and Period of Sine and Cosine Curves
DISCOVERY
U~e a
to graph
r
a ,in .\, where a
0.5. I. 2. and 3. (
a lie\\ in \\ hiel; -6.3 --+ } -+.) How doe, the I alue of a affect the graph '.' In the rest of this section, you will study the graphic effect of each of the
constants a, b, c, and d in equations of the forms
y = d
+
c)
a sin(bx
and
y = d
+
a cos(bx ­
A quick review of the transformations studied in Section 1.8 should help in this
investigation.
a sin x acts as a vertical stretch or vertical
The constant factor a in y
shrink of the basic sine curve, as shown in Example I. (If Ia I > I, the basic
sine curve is stretched, and if •a 1 < I, the basic sine curve is shrunk.) The
a sin x ranges between -a and a instead of
result is that the graph of y
between - 1 and 1. The absolute value of a is the amplitude of the function
y = a sin x.
DEFINITION OF AMPLITUDE OF SINE AND COSINE CURVES
The amplitude of y
and is given by
Amplitude
1
a I·
a sin x and y
=
a cos x is the largest value of y
SECTION 2.5
EXAMPLE:2
153
GRAPHS OF SINE AND COSINE FUNCTIONS
\erticai Shrink;ng <end Slretching
Sketch the graphs of y = ~ cos x and y = 3 cos x.
Because the amplitude of y = ~ cos x is
minimum value is -!. For one cycle, 0
The amplitude of y
(0,3),
(~,o),
4, the maximum value is ~ and the
:$
x
:$
277', the key points are
3 cos x is 3, and the key points are
(77',
3),
c;,o).
and
(277',3).
The graphs of these two functions are shown in Figure 2.44. Try using a
graphing utility to check this result.
y
3 cos x
Amplitude Determines Vertical Stretch or Shrink
FIGURE 2.44
Reflection in the
FIGURE 2.45
,X-
Axis
"",.,,,,
I;"D
~~-:-;
You know from Section 1.8 that the graph of y
-f(x) is a reHection (in
the x-axis) of the graph of y = f(x). For instance, the graph of y = 3 cos x
is a reflection of the graph of y = 3 cos x, as shown in Figure 2.45.
Because y = a sin x completes one cycle from x
0 to X
277', it
follows that y
a sin bx completes one cycle from bx = 0 to bx
277'. This
implies that y = a sin bx completes one cycle from x = 0 to x = 277'/b.
154
CHAPTER 2
TRIGONOMETR~
SCOVERY
10
draw th"
" !i.5. i. "nd :: ; L~e a
, c·,:.ngi;.' in \\hi;.;h 0
X s; 6.3
,,,,J ~:: Y s; 2.1 Hov., does the
•.. 'Lie:: of h affect the graph)
PERIOD OF SINE AND COSINE FUNCTIONS
Let b be a positive real number. The period of y
y = a cos bx is 271'/b.
a sin bx and
Note that if 0 < b < I, the period of y = a sin bx is greater than 271' and
represents a horizontal stretching of the graph of y
a sin x. Similarly, if
b > I, the period of y
a sin bx is less than 271' and represents a horizontal
shrinking of the graph of y = a sin x.
If b is negative, use the identities sin( - x) = sin x and cos( - x) =
cos x to rewrite the function.
EXA M PLE:3
Horizontal Stretching and Shrinking
Sketch the graph of
y
.
X
SIn ­
2'
SOLUTION
The amplitude is 1. Moreover, because b
L the period is 271' /(4) 471'. By
dividing the period-interval [0,471'] into four equal parts with the values 71',271',
and 371', you obtain the following key points on the graph.
(0,0),
(71', I),
(271', 0),
(371', -I),
and
(471',0)
The graph is shown in Figure 2.46. Use a graphing utility to check this result.
y
Y =\
SInX
I
- t
······r
~y_.sm.
~2" ....... " ............ "...' .
~
--rr
X
.
,
/
............ .. ;r.
".
17
217
....... ... .
_ _ _~
'(
141T
... . I J
Period: 41T
FIGURE 2.46
•••••••
•
111
SECTION 2.S
EXAMPLE 4
GRAPHS OF SINE AND COSINE FUNCTIONS
155
Horizontal Stretching ,,:'d '';
sin 3x.
Sketch the graph of y
SOLUTION
The amplitude is I, and because b = 3, the period is
y=
27T
27T
b
3
Dividing the period-interval [a, 27T /3] into four equal parts, you obtain the
following key points on the graph.
The graph is shown in Figure 2.47. Use a graphing utility to check this result.
FIGURE 2.47
EXAMPLE 5
l'sing a Graphing Uiiit)
Use a graphing utility to graph y = 2 cos O.4x and y
2 cos 4x on the same
screen. Determine the period of each function from its graph.
SOLUTION
3
'\
r~
OH!~~~~4Y~~~~44~~~18
I
\
27T
v
II
3
i
y = 2 cos O.4x I
FIGURE 2.48
The standard viewing rectangle is not appropriate for graphing most sine and
cosine functions. You should choose a viewing rectangle that accommodates
the amplitudes of the functions as well as any stretching or shrinking that may
occur. For these graphs, a y-scale ranging between - 3 and 3 will accommo­
date the amplitude of 2. Figure 2.48 shows the two graphs and an appropriate
2 cos
viewing rectangle. From the graph. you can estimate the period of y
OAx to be about 16 and the period of )' = 2 cos 4x to be about 1.5. Alge­
braically, the period of y
2 cos O.4x is
0.4
i
57T
15.71,
and the period of y
27T
7T
4
2
1.57.
2 cos 4x is
156
CHAPTER:2
TRIGONOMETRY
Translations of Sine and Cosine Curves
The constant c in the general equations
y
-= a sin(bx - c)
and
y
a cos(bx
c)
creates a horizontal shift of the basic sine and cosine curves. The graph of
y
a sin(bx
c) completes one cycle from bx - c
0 to bx - c = 27T. By
solving for x in the inequality 0 :::; bx - c :::; 27T, you can find the interval for
one cycle to be
c
b
-'-­
C
27T
- +­
b
b'
This implies that the period of y = a sin(bx - c) is 27T/b, and the graph of
y = a sin bx is shifted by an amount c lb. The number c /b is the phase shift.
~---~------~-----.-.-
..
,
~-~.--
1Il
,Jnt\\ the
\\ here
(L
a
hid:
6.:;"nu--2::;Y
!-l(I\\
Joe,
lh~
\alllc: of
::.i
affect
GRAPHS OF THE SINE AND COSINE FUNCTIONS
The graphs of y
a sin(bx
c) and y
lowing characteristics. (Assume b > 0.)
a cos(bx
c) have the fol­
Amplitude = ia I
Period = 27T /b
The left and right endpoints corresponding to a one-cycle interval of the
graphs can be determined by solving the equations bx - c = 0 and
bx - c
27T.
SECTION 2.5
157
GRAPHS OF SINE AND COSINE FUNCTIONS
Horizontal Shift
EXAMPLE 6
Describe the graph of
y
= ~ sin (x
- ~).
SOLUTION
Because a = ~ and b
the inequality
o
17
-:s
3
17
X -
-
:s
3
<:
x
-
J, the amplitude is ~ and the period is 217. By solving
217 717
3 '
you can see that the interval [17/3, 717/3] corresponds to one cycle of the
graph. Dividing this interval into four equal parts produces the following key
points.
I)
1117
6 ,- 2'
C;,o)
and
....... .. The graph is shown in Figure 2.49. FIGURE 2.49
The final type of transformation is the vertical shift caused by the constant
d in the equations
y
=
d
+ a sin(bx - c) and y
d
+ a cos(bx
c).
The shift is d units upward for d > 0 and downward for d < O. In other
words, the graph oscillates about the horizontal line y = d instead of the
x-axis.
7
EXAMPLE
Vertical Shift
Describe the graph of y
=2+
3 sin 2x.
SOLUTION
The amplitude is 3 and the period is 17. The key points over the interval [0, 17]
are
··2
(0,2),
Period:
FIGURE 2.50
11'
(~,5). (~'2).
C;,-l).
The graph is shown in Figure 2.50. and
(17,2).
....... .. 158
CHAPTER
2
TRIGONOMETRY
~~:'.~:.7'::.~
Finding an Equation for a Given Graph
Find the amplitude, period, and phase shift for the sine function whose graph
is shown in Figure 2.51. Write an equation for this graph.
SOLUTION
The amplitude for this sine curve is 2. The period is 21T. and there is a right
phase shift of 1T /2. Thus, you can write the following equation .
A Shifted Sine Curve
.\'
2 Sin( x -
~)
FIGURE 2.51
.... '" .'" ..
Try finding a cosine function with the same graph.
A SINE SHOW If you have a programmable calculator, try running the following "sine-show
program." This program simultaneously draws the unit circle and the correspond·
ing points on the sine curve. After the circle and sine curve are drawn, you can
connect points on the unit circle with their corresponding points on the sine
curve by repeatedly pressing ENTER. After the program is complete, the screen
should look like that shown in the figure. (This program is for the TI-81. Pro­
gram steps for other calculators are given in Appendix 8.)
:Rad
:ClrDraw
:Line(-1.25,
:DispGraph
.19,-:.25,
:O~N
:Param
:Simul
;-2.2S-Xmin
; 7T/2-X::,ax
:3-Xscl
. 9~Ymi:;
:1.:"9-Ymax
::~Yscl
:O-Tmin
: 6 . 3-T::.ax
:. 5-Tstep
:"-1.2 J.ccs T"-X
:"51::1
T"-Y i
: "sin T"-Y
,
: Lbl 1
:N.L1-N
: N7T/6. S-T
: 1.25+cos T-A
:sin ':'-8
: Line(A, B, C, 3)
:Pat;se
: f N=12
:Goto 2
:Goto 1
:Lbl 2
: Fc;nctiorl
:Seqt;e:1ce
:End
.19)
SECTION 2.5
WARM-UP
GRAPHS OF SINE AND COSINE FUNCTIONS
159
The following warm-up exercises involve skills that were covered in earlier sections_ You
will use these skills in the exercise set for this section.
In Exercises I and 2, simplify the expression.
27T
1.
2.
27T
In Exercise, 3-6, solve for x.
7T
3. 2..
5.
3
+ 6"
37TX
7T
4. 2..
0
0
6.
27T
3
37TX
+ 6"
2"
In Exercises 7-10. evaluate the trigonometric function without using a calculator.
7. sin
"
2
8. sin 7T
"
10. cos2
9. cos 0
SECTION 2.5 . EXERCISES
...........•.......................................................•..........................•.........................
In Exercises 1-10. determine the period and amplitude of the
function. Then describe the viewing rectangle for Exercises
1-6.
1.
y
3. \'
2 sin 2x
3
x
- cos
2
2
2. \
4. \'
5. \'
I
~
.
SlO ITr
3 cos 3x
2 sin
3
7. ,\
3 sin lOx
8.
5
.\
- cos­
2
4
1/
9. r
:1 sin 47T.\
Tr\"
10. \
3
cm,­
10
6.
y
5
- cos
')
7iX
2
160
CHAPTER 2
TRIGONOMETRY
In Exercises II 16. describe the relationship between the
graphs of f and g,
In Exercises 29-42. sketch the graph of the function. (Include
two full periods. and use a graphing utility to verify your result.)
11. fed
29. \' = -2 sin
ger)
13. fix)
!i'(x)
15.
g(x)
sin x
sinCr
12. fix) = cos x
g(r) = cos(x -'- 17)
17)
31. y = cos
14. j(x) = sin 3x
g(rl = sin(-3xl
cos 2"
-cos 2x
sin x
2 -'- sin x
16. fix)
g(r)
= cos 4x = _"! ... cos
35.
.r
217X
="J -
-
37. r
In Exercises 17-24. sketch the graphs of the two functions on
the same coordinate plane, (Include two full periods. and use a graphing utility to verify your resu)L)
17. fix)
g(rl
18. fix)
g(xl
19. fix)
g(xl
20. fix)
g(x)
2
22. g(x)
23. fix)
g(x)
24. I(x)
g(x)
217X
sin3
(\x - "4rr)
39. v= 3 cos(x +
)
41. \'
cos
I
3 cos 4x
\'
32.
\'
34.
\'
3
- sin
2
6
36. \ .
2 cos x
38.
-2 SIIl(,
.
3
I ,
\'
4
42. v
60171'
l7X
10 cos
40.
17)
l7X
17)
COS('
4
m
I
5 cos
3
x
In Exercises 43-54. use a graphing utility to graph the function,
(Include two full periods,)
.3
cos x cos x 2 cos 2,
-cos 4x 21.
g(x)
Sin
2 sin x 4 sin x sin x
sin
217X
-sin3
33. y
4x 30.
6.1
sin 3
x
2
sin
2
4 sin l7X
4 sin l7X
2 cos x
2 cos (x
-cos x
-cos(x
x
::
.3
43. y
= 3 cos(x
45.
= -::
\'
3
17) - .3
+
44.
cos (x- - -17)
\2
4/
47. y
=
49. \'
= cos( 2m
46. r
-2 sin(4x + 71')
-
~)
\'
+
4
cosC, +
~1
+
4/ 3 cos(6x
~
"
C
48.
\'
-4 sin ~x -
50.
\'
3 cos
52.
\'
5 sin( 17
17) 54.
\'
17)
'r3r)
(rrx
2" + 2'17)
\
51. y '" -01 sin ( -17X + 171
, )0
/
53. \' = 5 cos( 17 - 2x) + 2
2x)
4
-
2
+ 10
sin 120m
17)
In Exercises 55-58. use the graph of the trigonometric function
to find all real numbers x in the interval (- 217. 271'J that give the
In Exercises 25-28. sketch the graphs of I and g on the same
coordinate axes and show that ((xl = g(rl for all x, (Include
two full periods. and use a graphing utility to verify your result,)
26. fix) = sin x
sin x
25.
~)
g(x) g(x) = -cos(, .:.,
cos x
27. g(x)
f
sin\ x
\
7r\
-I! specified function value,
Function
55. sin x
-~
56. cos x
-)
57. co, x 28. fix)
g(x)
cos x
-cosc,
17) FIlIlC/lrll1
58. sin "
2
\/3
2
Value
SECTION 2.5
In Exercises 59 and 60. find a and d for the function j(x) =
a cos x -+- d so that the graph of j matches the
GRAPHS OF SINE AND COSINE FUNCTIONS
161
69. Piallo TUlling When tuning a piano, a technician strikes a
tuning fork for the A above middle C and sets up wave
motion that can be approximated by
v
= 0.00 I sin 880m.
where t is the time in seconds.
(a) What is the period p of this function,)
(b) The frequency j is
by j = 1/p. What is the fre­
quency of this note')
(c) Use a graphing utility to graph this function.
In Exercises 61-64, find G, b. and c so that the graph of the
function matches the graph in the
61. v
= a sin{bx
c)
62. y
70. Blood Pressure
P
=
a cos(hx
c)
64. v
=
100 - 20 cos
5m
3
approximates the blood pressure P in millimeters of mer­
cury at time I in seconds for a person at rest.
(a) Find the period of the function.
(b) Find the number of heartbeats per minute.
(c) Use a graphing utility to graph the pressure function.
- 1.5
63. y
The function
a sin(hx - c)
= a
sin(bx
c)
Sales [n Exercises 71 and 72. use a graphing utility to graph
the sales function over I year where S is sales in thousands of
units and t is the time in months. with t
1 corresponding to
January. Use the graph to determine the month of maximum
sales and the month of minimum sales.
2e' and
5 cos x. Approximate any points of intersection of
the graphs in the interval [-7T, 7T].
65. Use a graphing utility to graph the functionsj(x)
71. S
22.3 - 3.4 cos
g (xj
66. Use a graphing utility to determine the smallest illteger
value of a such that the graphs of j(x) = 2 In x and
g (x) = a cos x intersect more than once.
67. Respiratory Cvcle For a person at rest, the velocity v (in
liters per second) of air flow during a respiratory cycle is
72. S = 74.50
'iT1
6
'iTt
43.75 sin -
6
In Exercises 73-76. describe the relationship between the graphs
of the functions j and g.
73. 74.
75. 76.
7Tt
\' = 0.85 sin-.
3
where t is the time in seconds. (Inhalation occurs when
I' > O. and exhalation occurs when v < 0.)
(a) Find the time for one full respiratory (bl Find the number of cycles per minute. (el Use a graphing utility to graph the velocity function. 68. Respirator>: Cvcle After exercising for a few minutes. a
person has a respiratory cycle for which the velocity of air
flow is approximated by
\. ~
O~'2.57
7Tt
1.75 sin - .
2
Use this model to repeat Exercise 67. -2 -2
162
CHAPTER 2
TRIGONOMETRY
2.6
OTHER TRIGONOMETRIC GRAPHS
•............................................................•..........................
Graphs of Tangent and Cotangent Functions I Graphs of the Reciprocal
Functions I graphs of Combinations of Trigonometric Functions I
Combinations of Algebraic and Trigonometric Functions I
Damped Trigonometric Graphs
Graphs of Tangent and Cotangent Functions
Recall from Section 2.2 that the tangent functions is odd. That is, tan( - x)
-tan x. Consequently, the graph of y = tan x is symmetric with respect to the
origin. From the identity tan x
sin x/cos x, you know that the tangent is
undefined when cos x
O. Two such values are x
1T/2 = ± 1.5708.
I
7T
x
-­
tan x
undef.
2
1.57
-1255.8
:
1.5
-14.1
-·1
o
-1.56
0
I
1
1.5
1.57
1.56
14.1
1255.8
I~
.........
Period: 1T
Domain: all
1T
x ;/=
"2
±
n1T
Range: (-00, 00)
Venical asymptotes: x =
1T
"2 ± mT
_­
As indicated in the table, tan x increases without bound as x approaches 1T/2
from the left, and decreases without bound as x approaches 1T/2 from the
right. Thus. the graph of y
tan x has vertical asymptotes at x = 1T/2 and
1T/2. as shown in Figure 2.52. Moreover. because the period of the tangent
function is 1T, vertical asymptotes also occur when x
± n1T. The do­
main of the tangent function is the set of all real numbers other than x =
1T/2 ± n1T. and the range is the set of all real numbers.
Sketching the graph of a function with the form Y
a tan(bx - c) is
similar to sketching the graph of y = a sin(hx
c) in that you locate key
points which identify the intercepts and asymptotes. Two consecutive asymp­
totes can be found by solving the equations
hx - c =
FIGURE 2.52
I
undef.
2
and
bx
1T
C =-. 2
The midpoint between two consecutive asymptotes is an x-intercept of the
graph. After plotting the asymptotes and the x-intercept, plot a few additional
points between the two asymptotes and sketch one cycle. Finally. sketch one
or two additional cycles to the left and right.
REMARK The period of the function r
(I lan(iJx c) j, the distance between two
••••••
consecutive asymptotes. The amplitude of a tangent function is not defined.
-
SECTION 2.6
OTHER TRIGONOMETRIC GRAPHS
163
Sketching the Graph of a Tangent Function
EXAMPLE 1
x
tan -,
2
Sketch the graph of y
SOLUTION
By solving the inequality
-- <
I
<­
7T
222
i---- l
I
X
7T
I
I
1
I
I
I
I
_7>-,,+-¥-t---H-I--+-++-;HH......)(·
3
I
1
I
L
I
I
'1 I
-
.......
_
[an 2.
I",
. . . ....
y
-­
X
1
1 -3
.
you can see that two consecutive asymptotes occur at x = 7T and x = 7T.
Between these two asymptotes, plot a few points including the x-intercept, as
shown in the table. Three cycles of the graph are shown in Figure 2.53. Use
a graphing utility to confirm this result.
=
-
X
.. .
x
tan 2
7r
7r
0
2
2
0
1
1
......... ... FIGURE 2.53
EXAMPLE
Sketching the Graph of a Tangent Function
2
-3 tan lx.
Sketch the graph of y
SOLUTION
By solving the inequality
7T
7T
--<lx<­
2
2
7T
7T
-<x<­
4
4
you can see that two consecutive asymptotes occur at x = - 7T/4 and x = 7T/4.
Between these two asymptotes, plot a few points as shown in the table, and
complete one cycle. Four cycles of the graph are shown in Figure 2.54. Use
a graphing utility to confirm this result.
x
FIGURE 2.54
-3 tan 2x
7r
7r
-8
0
-
3
0
-3
8
............. ..
164
CHAPTER
2
TRIGONOMETRY
Period: 1T
Domain: all x
Range: (-00, 00)
By comparing the graphs
graph of
*' n1T
Vertical asymptotes: x
n1T
y
y
=a
In
Examples I and 2, you can see that the
tan(bx + c)
is increasing between consecutive vertical asymptotes if a > 0 and decreasing
between consecutive vertical asymptotes if a < O. In other words, the graph
for a < 0 is a reflection of the graph for a > 0,
The graph of the cotangent function is similar to the graph of the tangent
function. It also has a period of 11'. However, from the identity
y
.
FIGURE 2.55
=
cot x
cos x
sin x
=-­
you can see that the cotangent function has vertical asymptotes at x = n11'
(because sin x is zero at these x-values). The graph of the cotangent function
is shown in Figure 2.55.
Technofo91J Note _ _ _ _ _ __
You can use the tangent function on
your graphing utility to obtain the
graph of the cotangent function. For
example. to graph the function \. =
:: cot (xI3) from Ex<:mple 3. let
\'1
::/tanlxI3).
If you select the viewing rectangle
-9 :;; x :;; 18 and -6 :;; Y :;; 6, you
should obtain a graph similar to that
of Figure 2,56,
EXAMPLE
3
Sketching the Graph of a Cotangent Function
x
Sketch the graph of y
2 cot
'3'
SOLUTION
To locate two consecutive vertical asymptotes of the graph, you can solve the
following inequality.
x
0<-<11'
3
0< x < 311' Then, between these two asymptotes, plot the points shown in the following
table, and complete one cycle of the graph. (Note that the period is 311', the
distance between consecutive asymptotes.) Three cycles of the graph are
shown in Figure 2.56.
311'
X
2 cot
FIGURE 2.56
i
x
3
37r
2
4
911'
4
-
I
2
0
i
-2
" '" '" " " " " " "
SECTION 2.6
OTHER TRIGONOMETRIC GRAPHS
165
Graphs of the Reciprocal Functions
fJ
SCOVERY
Use a graphing utility to graph
the functions}" = sin x and
)'2 = csc.t "" l/sin.t on ttie
same viewing rectangle. How are
the graphs related? What happens
to the graph of the cosecant
function as x approaches the
zeros of the sine function?
Similarly, graphy, = cos.t and
Jz = sec x = I/cos x on the
same viewing rectangle. How are
these functions related?
Y
The graphs of the two remaining trigonometric functions can be obtained from
the graphs of the sine and cosine functions using the reciprocal identities
esc x
sin x
and
I
sec x = - - .
cos x
For instance, at a given value for x, the y-coordinate for sec x is the reciprocal
of the y-coordinate for cos x. Of course, when cos x = 0, the reciprocal does
not exist. Near such values for x, the behavior of the secant function is similar
to that of the tangent function. In other words, the graphs of tan x
(sin x)/(cos x) and sec x = l/(cos x) have vertical asymptotes at x
(?T/2) + n?T, because the cosine is zero at these x-values. Similarly,
cot x
(cos x)/(sin x) and csc x = I/{sin x) have vertical asymptotes where
0, that is, at x = n?T.
sin x
To sketch the graph of a secant or cosecant function, we suggest that you
first make a sketch of its reciprocal function. For instance, to sketch the graph
of y
csc x, first sketch the graph of y = sin x. Then take reciprocals of the
y-coordinates to obtain points on the graph of y = csc x. You can use this
procedure to obtain the graphs shown in Figure 2.57.
Cosecant: local minimum Period: 217
Domain: all x
1'117
Range: all y not in (-1, 1)
Vertical asymptotes: x = 1'117
Symmetry: origin
Period: 217
.
17
Domam: all x*-"2 + 1'117
Range: all y not in (-I, 1)
FIGURE 2.57
Symmetry: y-axis
*'
Cosecant:
local maximum
FIGURE 2.58
Vertical asymptotes: x =
17
"2 + n17
In comparing the graphs of the secant and cosecant functions with those
of the sine and cosine functions, note that the "hills" and "valleys" are inter­
changed. For example, a hill (or maximum point) on the sine curve corre­
sponds to a valley (a local minimum) on the cosecant curve. Similarly, a valley
(or minimum point) on the sine curve corresponds to a hill (a local maximum)
on the cosecant curve, as shown in Figure 2.58.
EXAMPLE 4
Graphing a Cosecant Function
Use a graphing utility
y
2 Sin(x
to
+~)
graph
and
\'
=
2 csc(x
+~).
166
CHAPTER Z
TRIGONOMETRY
SOLUTION
The two graphs are shown in Figure 2.59. Note how the "hills" and "valleys"
of each graph are related. For the function y
2 sin[x + (7T/4)], the ampli­
tude is 2 and the period is 27T. One cycle of the sine function corresponds to
the interval from x
-7T/4 to x
77T/4, Because the sine function is zero
at the endpoints of this interval, the corresponding cosecant function
y = 2 csc(x
+
~)
has vertical asymptotes at x = -7T/4. x
FIGURE
= 37T/4. and
77T/4.
....... .. 2.59
In Figure 2.60, we summarize the graphs, domains. ranges, and periods of
the six basic trigonometric functions.
Y
6
Domain: all Teals
Range: [-I, 1]
Period: 21T
y
Domain: all Teals
Range: [-I. 1]
Period: 21T
4
-+--J-.--f----:f--f---E-+_x
*'
-¥
:U
I
:
-2'
I
"
I
-3
Penod: 21T
4
I
,~
I
:u:
W
Ra~ge: (-00,
,
2
'(\W¥':
:
:
I :=cscx=~1
I
+ mr
-I] and [I, 00)
Domain: all x*'2
y
. 3:
:
I
' :
1
217'
7T
'-Ji~"'I!
-2:
:
:
-3
;
,
I
I
\
• x
1
I
ly=secx=~1
Graphs of the Six Trigonometric Functions
FIGURE 2.60
*'
1T
Domain: all x
n1T
Range: (-00. -1] and [I, 00)
Period: 21T
Y
y
Domain: all x
n1T
Range: (-<>0, 00)
Period: 1T
\
x
SECTION 2.6
OTHER TRIGONOMETRIC GRAPHS
167
Graphs of Combinations of Trigonometric Functions
Sums. differences. products. and quotients of periodic functions are also peri­
odic. The period of the combined function is the least common multiple of the
periods of the component functions. This period is important for determining
an appropriate viewing rectangle for a graphing utility.
Finding the Period and Relative Extrema of
a Function
EXAMPLE 5 Graph y = sin x - cos 2x. Find the period and the relative minimums and
maximums of this function.
SOLUTION
The period of sin x is 217 and the period of cos 2x is 17. Thus. the period of the
given function is 217, because 217 is the least common multiple of 217 and 17.
This conclusion is further reinforced by graphing the function, as shown in
Figure 2.61. From the graph, it appears that the function has two relative
maximums and two relative minimums in each complete cycle. For instance.
between 0 and 217, the graph appears to have relative maximums when x
17/2 = 1.57 and when x = 317/2 = 4.71. Using the zoom and trace features
of the graphing utility. you can find that the relative minimums occur when
x = 3.40 and when x
6.03.
=
FIGURE
2.61
Relative maximums:
(1.57,2.00)
(4.71.0.00)
Relative minimums:
(3.40,
(6.03, -l.J3)
1.13)
....... ..
Finding the Period and Range of a Function
EXAMPLE 6
Graph y = 2 sin 6x
sin 4x. Find the period and range of this function.
SOLUTION
_ 3 14 ~-+-,-.,..f-M--+-'-+--f---.Jl-'-+--\-;f--'-+---I3.14
The period of 2 sin 6x is 17/3 and the period of sin 4x is 17/2. Thus. the period
of the gi ven function is 17, because 17 is the least common multiple of 17/3 and
17/2. This conclusion is further reinforced by graphing the function. as shown
in Figure 2.62. In the interval from 0 to 17, the maximum y-value occurs when
x = 0.29 and y
2.89. The minimum value in this interval occurs when
x = 2.86 and y = -2.89. Thus, the range of the function is approximated by
=
FIGURE
2.62
-2.89
:S
Y
:S
2.89. Range
....... .. 168
CHAPTER 2
TRIGONOMETRY
Combinations of Algebraic and Trigonometric Functions
Functions that are combinations of algebraic and trigonometric functions are
not, in general. periodic.
EXAMPLE 7
The Graph of a Nonperiodic Function
Graph y
x + cos x. Find the domain and range of the function. Approxi­
mate any zeros of the graph.
SOLUTION
The graph of the function is shown in Figure 2.63. Notice that even though the
function is not periodic, it does have a pattern that repeats. Notice that the
x. Both the domain and
graph of y = x + cos x oscillates about the line y
range of the function are the set of all real numbers. Using the zoom feature,
x + cos x is approximately x = -0.739.
you can find that the zero of y
- 12.57 \'------'---'--.L-+t---'--'---'---'\ 12.57
FIGURE
EXAMPLE 8
Graph y
....... .. 2.63
A :Function Involving Absolute Value
Isin x I, and
find the domain and range of the function.
SOLUTION
FIGURE
2.64
The domain of the function is the set of all real numbers. The graph of the
function is shown in Figure 2.64. Notice that the minimum value of the
function is 0 and the maximum value is 1. Thus, the range of the function is
....... ..
given by 0 :s y :s 1.
SECTION 2.6
OTHER TRIGONOMETRIC GRAPHS
169
Damped Trigonometric Graphs
A product of two functions can be graphed using properties of the individual
functions involved. For instance, consider the function
J(x) = x sin x
as the product of the functions y
x and y
sin x. Using properties of
absolute value and the fact that Isin x I ::;; J. you obtain 0 ::;; Ix II sin x I ::;; Ix I.
Consequently,
I x I ::;;
x sin x ::;;
I x I'
which means that the graph of J(x)
and y
x. Furthermore, since
J(x)
J(x)
FIGURE 2.65
=
x sin x =::::x
x sin x = 0
at
at
x
x sin x lies between the lines y
1T
+
2
-x
n7r
x = n7r, the graph of J touches the line y
- x or the line \ = x at x = (1T /2) + WTr
and has x-intercepts at x
n1T. The graph of J. together with y
x and
y
- x, is shown in Figure 2.65.
In the functionJ(x) = x sin x, the factor x is called the damping factor.
By changing the damping factor, you can change the graph significantly. For
example, look in Figure 2.66 at the graphs of
.v = -x sin x
and
y
e- x sin 3x.
FIGURE 2.66
170
CHAPTER
2
TRIGONOMETRY
Damped Cosine Wave
EXAMPLE 9
Graph f(x) = T
,r;:
cos x.
SOLUTION
A graph of f{x) = 2 -x12 cos x is shown in Figure 2.67. To analyze this fuction
further, consider f(x) as the product of the two functions
y
=
2-,1"
and
cos x,
y
each of which has the set of real numbers as its domain. For any real number
x, you know that 2 - ,/2
0 and cos x I :S I. Therefore, I 2 - xj2 I I cos x I :S
2-,12, which means that
cos x
:S
2
Furthermore, since
f(x) =
rx/2
cos x
=2-
at
x
= n1T
and
o
1T
at
X=
the graph off touches the curve y =
r
and has intercepts at x
(1T/2) + fl1T.
FIGURE 2.67
2 + fl1T,
,/2
or the curve y ... ,. .. ... Use a graphing utility to graph the following functions for varying values of a, b,
e, and d.
GRAPHING UTILITIES y
y
y
d
d
d
+ a sin(bx + e)
+ a tan(bx + e)
+ a sec(bx + e)
y
y
y
=
d
d
d
+
+
+
a cos(bx
a cot(bx
a csc(bx
+ e) + c) + e) In a paper or a discussion, summarize the effects of the constants a. b, e, and d
in these graphs.
SECTION 2.6
WARM-UP 171
OTHER TRIGONOMETRIC GRAPHS
ni~foilow4ngwarni-up exercises iravolVe ski,Dstliat)'f~co~ered in. earliersediODS' YO\!,
will:~ tii~skms in the exercise setforthii'Sei:tl~~'. -.. - ; : - : . . ' c :
In Exerdsesl-4.
1. f(x)
•
findtllex~values in the interVal (0; 2~] for~iCb f(x)
sin x
,-
2. /{x) = cos x
'<
3. f{x}
is -l.O, or
L
= sin2x
,
In Exercises·S':'S, graph' the function.
5. Y
j xl
7. y = sin TrX
= cos 2x
8. y
In Exercises 9 and 10, evaluate f(x) when x = 0, Tr/6, Tr/4, Tr/3, and Tr/2.
9. f(x)
2.6 .
SECTION
x cos x
10. f{x)
x + sin x
EXERCISES
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• e
In Exercises 1-8, match the trigonometric function with its
graph and describe the viewing rectangle. [The graphs are la­
beled (a), (b), (cl, (d), eel, (f), (g), and (h).]
1. Y
sec 2x
'l
2. r "" tan 3x
3. \' = tan
x
-
2
4. Y
2 csc
,
•••••••••••••••••••••••
~
••••••••••••••••••••••••
j
j(
x
2
5. \'
cot 7TX
6. Y = ~ sec 7TX
-sec x
7. Y
8. y = - 2 esc 2 =
I
(b)
(a)
. ., ,
,-"--~----
r-~~~~--~~
II )1
t.
.
,
-,I
f
,,
,,
'n'
1
1
llli
,I ,
.
,
172
CHAPTER
Z
TRIGONOMETRY
In Exercises 9-30, sketch the graph of the function. (Include
two full periods, and use a graphing utility to verify your result.)
t
9. y
tan x
11. y = tan 2x
- ~ sec x
]5. y = -sec 7TX
13. Y
17. y = sec
x
19. Y
csc 2
21. y
23. Y
7TX -
cot
x
x
3
7TX
.
24. Y
7TX
2
=
~ tan x
~ cSC(X +~)
28. y = sec( 7T - x)
30. Y
2cot(X
+~)
In Exercises 31-40, use a graphing utility to graph the function.
(Include two full periods.)
x
tan 3
31. Y
32. Y
33. y = - 2 sec 4x
35. \' == tan(x
39.
~)
~ cot ( x - ~)
37. Y
v
= 2 sec(2x - 7T)
34. -" = sec
36. Y
38. Y
49. v
2
51. Y
53. .\'
=4
-
2 . x
2:
SID
2 cos
50. y
-3 + cos x
52. \' =5
.l sin 27TX
54.) = 2 + tan 7TX
7TX
I ..;.. csc x
7TX
-csc(4x - 7T)
~ sec( r.; + ~)
40. Y = 0, I tan(
r; -'- ~)
In Exercises 55-60, use a graphing utility to graph the func­
tion. Determine the period of the function and approximate
any relative minimums and maximums of the function through
one period.
55. y
sin x 1- cos x
57. I(x)
2 sin x + sin 2x
x
59. g(x) = cos x - cos '2
56. \'
cos x - cos 2x
58. I(x) = 2 sin x + cos 2x
I
x
60. g(x)
sin x - - SIn -
2
2
Function Value
41. tan x
1
42. cot x
-\13
43. sec x
44. esc x
-2
V2
In Exercises 45 and 46. use the graph of the function to deter·
mine whether the function is even, odd, or neither.
45./(x) = sec x
In Exercises 49-54, sketch the graph of the function. (Use a
graphing utility to verify your result.)
-tan 2x
In Exercises 41-44, use the graph of the trigonometric function
to find all real numbers x in the interval [-27T. 27T] that give the
specified function value.
Flmcrion 48. Consider the functions I(x) = tan(7Tx/2) and g(x)
1sec(7Tx/2) over the interval (- L I J.
(a) Use a graphing utility to graph I and g in the same
viewing rectangle,
(b) Approximate the interval where I < g.
26. -" = sec(x + 7T)
4
27. Y = CSc(7T - x)
29. Y
2
22. \' = 3 cot­
2
tan -
esc ­
20. Y
= ~ sec 2x
25. Y
10. \'
~ tan x
12. y = 3 tan 7TX
14. \'
t sec x 16. y
2 sec 4x 18. \' = -2 sec 4x
47. Consider the functions/(x) = 2 sin x and
= ~ esc x
over the interval (0. 7Tl.
(a) Use a graphing utility to graph I and g in the same
viewing rectangle.
(b) Approximate the interval where I > g,
(c) Describe the behavior of each of the functions as x
approaches 7T. How is the behavior of g related to the
behavior of I as x approaches 7T~
46. I(x} = tan x
In Exercises 61-64, use a graphing utility to graph the function
through three periods.
61. hex)
sin x + ~ sin 5x
62. hex} = cos x
~ cos 2x 63.),
- 3 + cos x + :; sin 2x 64. .\' = sin
TTX .,-
sin ­
7TX
2
SECTION 2.6
In Exercises 65-72, use a graphing utility (0 graph the given
function and the algebraic component of the function in the
same viewing rectangle. Notice that the graph of the function
oscillates about the graph of the algebraic component.
x + sin x
65. Y
67. f(x)
2 cos x
66. y
tx -
. 1Tl 69. get)
71. y
6S. f(x)
1+ sm
x2
=-
8
2x - sin x
1TX + sin -
72. y
2
1TS
- - sin
2
4
S
=
70. h(s)
2
2
x2
4 -
16
=
2- x / 4 cos 1TX
= e-
r2
!2
sin x
Function 92. f(x)
(2
Interval
sin t =
[0. 41T ]
~ sin 3x + ~ sin 5x
93. f(x) = sin x
94. f(x)
[0, 21T]
sin x ~2 - ~(cos
1TX
1T
+ I cos 31TX)
9
[0, 1T]
[0. 2]
+ 4 cos 1TX
In Exercises 73-76, use a graphing utility to graph the function
and the damping factor of the function in the same viewing
rectangle. Describe the behavior of the function as x increases
without bound.
73. f(x)
75. g(x)
In Exercises 9\-94, use a graphing utility to graph the function
over the specified interval.
91. f(t) =
+ cos x
x
173
OTHER TRIGONOMETRIC GRAPHS
74. f(t)
e-' cos t
76. h(l)
2-,2/ 4 sin I
95. Distance A plane flying at an altitude of 6 miles over
level ground will pass directly over a radar antenna (see
figure). Let d be the ground distance from the antenna to
the point directly under the plane and let x be the angle
of elevation to the plane from the antenna. Write d as a
function of x, and graph the function over the interval
0< x < 1T.
In Exercises 77 -80, use a graphing utility to graph the function
and the equations }' = x and y
- x in the same viewing
rectangle. Describe the behavior of the given function as x ap­
proaches zero.
77. f(x)
79. g(x)
= x cos x
Ixl sin x
I x sin x I
7S. f(x)
SO. g(x)
I x i cos x
FIGURE FOR 95
In Exercises 81 -86, use a graphing utility to graph the function.
Describe the behavior of the function as x approaches zero.
81. y
6
x > 0 82. y
- + cos x,
x
sin 2x,
x > 0
I
cos x
84. f(x) = - - ­
sin x 83. g(x)
4
x
= +
x
I
85. f(x) == sin x
86. hex)
x sin ­
x
96. Television Coverage A television camera is on a review­
ing platform 100 feet from the street on which a parade
will be passing from left to right (see figure). Express the
distance d from the camera to a particular unit in the
parade as a function of the angle x, and graph the func­
tion over the interval -1T/2 < x < 1T/2. (Consider x as
negative when a unit in the parade approaches from the
left. )
In Exercises 87-90, use a graphing utility to graph the func­
tions f and g in the same viewing rectangle. Use the graphs to
determine the relationship between the functions.
87. f(x) = sin x + cos( x +
8S. f(x)
89. f(x)
= sin
_
2
90. f(x) - cos
2
~),
== 1(I
cos( x +
sin x
x,
g(x)
1TX
2'
~),
I
g(x) = 0
g(x)
2 sin x
cos 2x)
()
I(
)
g x =="2 I + cos 1TX
TJ(
I
Camera
FIGURE FOR 96
174
CHAPTER 2
TRIGONOMETRY
97. Sales The projected monthly sales S (in thousands of
units) of a seasonal product is modeled by
S = 74
71'[
3t ..,- 40 sin­
6'
where t is the time in months, with f = I corresponding
to January. Graph this sales function over I year.
98. Sales
The projected monthly sales S (in thousands of
units) of a seasonal product is modeled by
101. Harmonic MOlion An object weighing W pounds is sus­
pended from the ceiling by a steel spring (see figure). The
weight is pulled downward (positive direction) from its
equilibrium position and released. The resulting motion of
the weight is described by the function
y = ~e-f/4 cos 41,
I
> 0,
where y is the distance in feet and I is the time in seconds.
Graph the function.
71'1
S
25 + 21 + 20 sin 6'
where I is the time in months, with I
I corresponding
to January. Graph this sales function over I year.
99. Predator-Prey Problem Suppose the population of a
certain predator at time I (in months) in a given region is
estimated to be
P = 10,000
Equilibrium
1
2m + 3000 sin 24' and the population of its primary food source (its prey) is
estimated to be
p
2m
15.000 + 5000 cos - .
.
24
FIGURE FOR 101
Graph both of these functions in the same viewing rectan­
gle, and explain the oscillations in the size of each popu­
lation.
100. Normal Temperatures The normal monthly high tem­
perature for Erie, Pennsylvania, is approximated by
H(I)
54.33
20.38 cos
71'1
6
71'1
- 15.69 sin - ,
6
and the normal monthly low temperature is approxi­
mated by
L(I) = 39.36 - 15.70 cos
71't
71'1
6
14.16 sin
6'
where t is the time in months, with 1
I corresponding
to January. (Source: NOAA) Use a graphing utility to
graph the functions over a period of I year, and use the
graphs to answer the following questions.
(a) During what part of the year is the difference between
the normal high and low temperatures greatest? When
is it smallest?
(b) The Sun is farthest north in the sky around June 21,
but the graph shows the warmest temperatures at a
later date. Approximate the lag time of the tempera­
tures relative to the position of the Sun.
SECTION 2.7
2.7
INVERSE TRIGONOMETRIC FUNCTIONS
175
INVERSE TRIGONOMETRIC FUNCTIONS
Inverse Sine Function I Other Inverse Trigonometric Functions I
Compositions of Trigonometric and Inverse Trigonometric Functions
Inverse Sine Function
Recall that for a function to have an inverse. it must be one-to-one. From
Figure 2.68 it is clear that y = sin x is not one-to-one because different values
of x yield the same y-value. If, however, you restrict the domain to the interval
-7T/2 :s x :s 7T/2 (corresponding to the darker portion of the graph in Fig­
ure 2.68), the following properties hold.
\. On the interval
l 7T/2, 7T/2], the function
2. On the interval
7T/2, 7T/2], Y
-I :s sin x
I.
3. On the interval [-7T/2, 7T/2], Y
y = sin x is increasing.
sin x takes on its full range of values,
sin x is a one-to-one function.
x:s 7T/2, y = sin x has a unique
Thus, on the restricted domain -7T/2
inverse called the inverse sine function. It is denoted by
y = arcsin x
y = sin-I x.
or
y
---_~r.~----~----~f------+----~~-----~. x
Sin x is one-to-one on this interval.
FIGURE 2.68
The notation sin -I x is consistent with the inverse function notation I-I (x),
which is used in Section 1.10. The arcsin x notation (read as "the arcsine of x ")
comes from the association of a central angle with its subtended arc length on
a unit circle. Thus, arcsin x means the angle (or arc) whose sine is x. Both
notations, arcsin x and sin -I x, are commonly used in mathematics, so remem­
ber that sin -I x denotes the inverse sine function rather than 1/sin x.
The values of arcsin x lie in the interval
7T/2, 7T/2]' The graph of y =
arcsin x is shown in Example 2. Try producing this graph with a graphing
utility. Note that the domain of y = arcsin x is -I :s x :s 1.
176
CHAPTER
2
TRIGONOMETRY
.. - - .
.
~
DEFINITfONOFINVER5.ESiNE·FUNCTION
When evaluating the in­
verse sine function, it helps to remem­
ber the phrase "the arcsine of x is the
angle (or number) whose sine is x."
REMARK
The inverse sine function is defined by
y
= arcsin
sin y
if and only if
x
x,
where
I 5 x 5 1 and
7T/2 5 Y 5 7T/2. The domain of \'
arcsin x is [ I, l], and the range is [-rr/2, 7T/2l.
As with trigonometric functions, much of the work with inverse trigono­
metric functions can be done by exact calculations rather than by calculator
approximations. Exact calculations help to increase your understanding of the
inverse functions by relating them to the triangle definitions of the trigonomet­
ric functions.
EXAMPLE
1
Evaluating the Inverse Sine Function
Find the values of the following (if possible).
Try verifying the answers
in Example I with your calculator.
What happens when you try to calcu­
late sin -I 2?
REMARK
A. arcsin (
-~)
•
B. SIn
~I
r­
V3
2
C. sin 12
SOLUTION
arcsin( - ~) implies that
A. By definition, y
.
I
2'
.
Because sin(
arcSin(
sin -"
5
\" 5
.
rr
2
-.
- 4, you can conclude that \'
rr/6)
-~)
B. Bydefinition,y
rr
2
for
Sin \" = - -
-rr/6 and
:
= sin-tV'3/2) implies that
V3
for
2 '
rr
2
rr 2
y5-.
.
V3/2, you can conclude that y
Because sin(rr/3)
7T/3 and
rr
3
c. I t is not possible to eval uate v
sin ~ I x when x
2, because there is no
angIe whose sine is 2. Remember that the domain of the inverse sine
function is [ 1, I].
• ....... .
SEC T I ON 2.7
INVERSE TRIGONOMETRIC FUNCTIONS
177
From Section 1.10, you know that graphs of Inverse functions are
x.
reflections of each other in the line y
EXAMPLE 2
Sketching the Graph of the Arcsine Function
Sketch a graph of v
arcsin x bv hand.
SOLUTION
By definition, the equations
y = arcsin x
sin y
and
x
are equivalent for 7T/2 ~ Y ~
Hence, their graphs are the same. By
assigning values to y in the second equation. you can construct the following
table of values.
!
0
V
x
sin v
-\
2
2
0
7T
7T
7T
6
4
2
v2
2
2
The resulting graph of y
arcsin x is shown in Figure 2.69. Note that it is the
reflection (in the line y = x) of the darker part of Figure 2.68. Be sure you see
that Figure 2.69 shows the entire graph of the inverse sine function. Remember
arcsin x is the closed interval [ 7T/2. 7T/2]' You can
that the range of y
verify this graph with a graphing utility for the function y == arcsin x .
FIGURE
2.69
.........
178
CHAPTER
2
TRIGONOMETRY
Other Inverse Trigonometric F'unctions
.y
The cosine function is decreasing on the interval 0 :5 x :5 1T, as shown in
Figure 2.70. Consequently, on this interval the cosine has an inverse function,
which is called the inverse cosine function and is denoted by
y == arccos x
Cos x is one-to-one on this interval.
or
y = cos - I x.
Similarly, you can define an inverse tangent function by restricting the
domain of y == tan x to the interval
1T /2, 1T /2). The following list summa­
rizes the definitions of the three most common inverse trigonometric functions.
The remaining three are discussed in the exercise set.
FIGURE 2.70 Domain
Function y
arcsin x if and only if sin y = x
y
arccos x if and only if cos y == x
y == arctan x if and only if tan y == x
1T
1T
-:5v<­
-}:5x:5
2
,- 2
O:5y:51T
1T
1T
- 2 <v<­
.
2
x<x<x
The graphs of these three inverse trigonometric functions are shown in
Figure 2.71.
Domain: [-I, IJ 1T
Range:
2
!!.]2 Domain: [-1. I J
Range: [0, 1TJ
Domain: (-00, 00)
\'
y
~~+----+----:if--'--+---+---
-2
-i11
FIGURE 2.71
-----------------
x
179
INVERSE TRIGONOMETRIC FUNCTIONS
SECTION 2.7
Evaluating Inverse Trigonometric Functions
EXAMPLE 3 Find the exact values of the following.
v2
A. arccos 2
B. arccos(-I)
C. arctan
°
SOLUTION
v2/2 and
A. Because cos(17'/4) =
17'/4 lies in [0,17'], it follows that
v2 17'
arccos - - = -.
2
4
B. Because cos 17' = -1 and 17' lies in [0, 17'], it follows that
arccos( - 1) C. Because tan
° ° °
arctan 0
REMARK In Example 4, had you set
the calculator to the degree mode, the
display would have been in degrees
rather than radians. This convention is
peculiar to calculators. By definition,
the values of inverse trigonometric
functions are always in radians.
17'.
=
and
lies in
17' 12. 17'/2), it follows that
....... . .
= O.
In Example 3, you were able to find the exact values of the given inverse
trigonometric functions without a calculator. In the next example, a calculator
is necessary to approximate the function values.
Tecbno(oepJ Note __.._ _ _ __
Most graphing utilities do not have
keys for evaluating the inverse
cotangent function, the inverse secant
function. or the inverse cosecant
function. Is it still possible to
calculate arcsec 3.4? If you let
x = arcsec 3.4, then sec x
3.4
and cos x = I/sec x
1/3.4.
Hence. using the inverse cosine
function key. x = 1.272. Similarly, to
evaluate the other inverse
trigonometric functions, use the
following identities.
arccsc x = arcsin ­
arccot x = ;
+
x
arctan (-x)
EXAMPLE 4 Using a Calculator to Evaluate Inverse Trigonometric
Functions
Use a calculator to approximate the values (if possible).
A. arctan ( -8.45)
B. arcsin 0.2447
C. arccos 2
SOLUTION
Function
A. arctan( -8.45)
B. arcsin 0.2447
C. arccos 2
Rounded 10
3 Decimal Places
Mode
Radian
Radian
Radian
1.45300 1005
0.2472102741
ERROR
-1.453
0.247
Note that the error in part (C) occurs because the domain of the inverse cosine
.. ...... .
function is [- I, I].
180
CHAPTER
z
TRIGONOMETRY
Compositions of Trigonometric and Inverse Trigonometric
Functions
Recall from Section 1.10 that inverse functions possess the properties
/(F1(x))
x
I(/(X))
and
=
x.
The inverse trigonometric versions of these properties are as follows.
REMARK Keep in mind that these in­
verse properties do not apply for arbi­
trary values of x and y. For instance.
arcSin(Sin
3;) =
arcsin(-l)
INVERSE PROPERTIES
I s x s I and -n'/2 s y s n/2. then If
sin(arcsin x)
7T
37T
2
2
- - if:.
In other words. the property
arcsin(sin y) = y is not valid for
values of y outside the interval
[- 7T /2. 7T /2].
If - I s x S I and 0
cos( arccos x) = x
C OV. E R Y
(a) Use a graphing utility to
graph y =arcsin(sin x). What
are the domairiand range of
the function? Explain why
arcsin(sin 4) dOes not equal 4.
(b) Use a graphing utility to
graph y
sin(arcsinx). What
are the domain and range of
the function? Explain why
sin(arcsin 4 lis not defined.
arcsin(sin y)
y. y S n, then and
arccos( cos y) = v. n/2 < y < n/2, then If
tan(arctan x)
EXAMPLE 5
f) IS
and
x
x
arctan(tan y)
and
y. Lsing Inverse Properties
If possible, find the exact values.
A. tan[arctan( -5)]
B.
. (. 5n)
arCSin
c.
SIn-:3
cos(cos-1n)
SOLUTION
A. Because - 5 lies in the domain of arctan x, the inverse property applies,
and you have tan[arctan(-5)] = -5.
B. In this case,
/3 does not lie within the range of the arcsine function,
7T /2 S x S 7T /2. However, 57T /3 is coterminal with
5n
which does lie in the range of the arcsine function, and you have
arcsin( sin 5;)
= arcsin [ sin( -~) ]
3
c. T he
Try verifying these results
with your calculator.
REMARK
expression cos( cos -I 7T) is not defined, because cos -I 7T is not
defined. Remember that the domain of the inverse cosine function is
[-1, lJ.
..
.......
SECTION 2.7
181
INVERSE TRIGONOMETRIC FUNCTIONS
Example 6 shows how to use right triangles to find exact values of
functions of inverse functions. Example 7 shows how to use triangles to
convert a trigonometric expression into an algebraic one. This conversion
technique is used frequently in calculus.
Evaluating Functions of Inverse Trigonometric
Functions
EXAMPLE 6 Find the exact values of the following.
=Vs
~)
A. tan( arccos
SOLUTION
2
= arccos ~, then cos u = ~. Because cos u is positive, u is a
first-quadrant angle. You can sketch and label angle u as shown in Fig­
ure 2.72. Consequently.
A. If you let u
FIGURE 2.72
tan(
arccos~)
= tan u
opp
Vs
adj
2
B. If you let u = arcsin( - ~), then sin u ==
~. Because sin u is negative, u
is a fourth-quadrant angle. You can sketch and label u as shown in
Figure 2.73. Consequently,
cos[ arcsin(
FIGURE 2.73
D] =
cos u
4
5
= adj
hyp
............ .... 7 Some Problems from Calculus
EXA M PLE
Write each of the following as an algebraic expression in x.
A. sin(arccos 3x),
n
l
,
B. cot(arccos 3x),
SOLUTION
./1
v
- (3.xi
\u
3x
FIGURE 2.74
If you Jet u
cos u
=
arccos 3x, then cos u
Because
3x
hyp'
you can sketch a right triangle with acute angle u, as shown in Figure 2.74.
From this triangle, you can convert each expression to algebraic form.
A. sin(arccos 3x)
REMARK In Example 7, a similar ar­
gument can be made for x-values ly­
ing in the interval [-1/3,0]. Why do
we restrict x < ~ in part (B)')
= 3x.
sin u =
B. cot(arccos 3x} = cot u
hyp
=­
adj
opp
=
VI -
9x 2 ,
3x o
I
­
X <
-3
1
3
O~x<-
.............. ....
182
CHAPTER 2
.......................... INVERSE
FUNCTIONS
TRIGONOMETRY
You have studied inverses for several types of functions. Match each of the func­
tions in the left column with its inverse function in the right column.
l. f(x)
(a)
(b)
(c)
(d)
(e)
x
2. f(x)
O:5x 3. f(x) 4. f(x)
5. f(x)
In x
6. f(x)
sin x, 7. f(x)
cos x,
8. f(x)
tan x. eX 1T
1T
2
2
-:5x:5
0
x:5
1T
1T
1T
2
2
-:5x:5
FI(x) = arcsin x
FI(x) = In x
FI(x) = Vx
FI(x) = arctan x
F I (x) = arccos x
(f) FI(x)
=
Vx
(g) FI(x)
:=
eX
(h) FI(x) =x
Provide reasons for your answers.
WARM-UP The following warm-up exer~ises involVe skills that were covered in earlier sectiou.s. You
will use these. skills in the exercise set for this section.
In Exercises 1-4, evaluate the trigonometric function without using a calculator.
1. Sin(
~)
2. cos
3. tan (
-~)
4. sin
1T
7T
4
In Exercises 5 and 6, find a real number x in the interval [- 7T /2, 7T/2] that has the same sine
value as the given value.
5. sin 27T
6
. 57T
• sJn
6
In Exercises 7 and 8, find a real number x in the interval [0, 7T] that has the same cosine value
as the given value.
7. cos 37T
8. cos(
~)
In Exercises 9 and 10, find a real number x in the interval (-7T/2. 7T/2) that has the same
tangent value as the given value.
9. tan 4rr
10. tan
37T
4
SECTION 2.7
183
INVERSE TRIGONOMETRIC FUNCTIONS
SECTION 2.7 . EXERCISES
.•.....•.......•......•.•............•...........................•......................................................
In Exercises 1-16. evaluate the expression without using a
calculator.
1. arcsin ~
3. arccos ~
2. arcsin 0
4. arccos 0 In Exercises 37-44, find the exact value of the expression
without using a calculator. (Hint: Make a sketch of a right
triangle, as illustrated in Example 6.)
37. sin(arctan ~)
38. sec(arcsin ~)
39. cos(arctan 2)
40. sin(arccos
r­
5. arctan
V3
3
v:) 7. arccos ( 9. arctan ( 11. arccos (
6. arctan ( I ) ~)
8. arcsin (
yJ)
10. arctan( yJ)
D
12. arcsin
V2
2
yJ
2
14. arctan ( 15. arctan 0
16. arccos I
13. arcsin
~) 41. cos(arcsin
43. sec[arctan( -
17.
19.
21.
23.
25.
27.
arccos 0.28
18,
20.
22.
24.
26.
28.
arcsin( -0.75)
arctan( - 2)
arcsin 0.31
arccos(-O.4l)
arctan 0.92
arcsin 0.45 arccos( -0.8)
arctan 15 arccos 0.26 arcsin( -0.125) arctan 2.8 In Exercises 45 and 46, use a graphing utility to graph I and g
in the same viewing rectangle to verify that the two are equal.
Identify any asymptotes of the graphs.
45./(x) = sin (arctan 2x),
29. I(x) = tan x,
30./(x)
sin x,
~),
2x
g(x) == ---;===
g(x) = - - x - ­
In Exercises 47-56, write an algebraic expression that is equiv­
alent to the given expression. (Hint: Sketch a right triangle, as
demonstrated in Example 7.)
47. cot(arctan x)
49. cos(arcsin 2x)
48. sin(arctan xl
50. sec (arctan 3x)
51. sin (arccos xl
52. cot ( arctan; )
53. tan ( arccos
In Exercises 29 and 30, use a graphing utility to graph/, g, and
y == x in the same viewing rectangle to verify geometrically that
g is the inverse of f (Be sure to properly restrict the domain of f)
42. csc[arctan( - fi))
44. tan[arcsin( -~)J
t)J
46·/(x) == tan(arccos
In Exercises 17-28, use a calculator to approximate the given
value. (Round your result to two decimal places.)
~)
55. csc (arccos
I'
~)
54. sec [arcsin(x - 1))
~)
-r-)h'
X
56. cos ( arcsin
In Exercises 57-60, fill in the blanks.
g (x) == arctan x g(x) == arcsin x
9
57. arctan x
~
In Exercises 31-36, use the properties of inverse trigonometric
functions to evaluate the expression.
58. arcsin
31. sin(arcsin 0.3)
33. cos[arccos( -0.1)]
32. tan(arctan 25)
34. sin[arcsin( -0.2)J
59. arccos 35. arcsin(sin 37T)
36. arccos (cos
7;)
60. arccos
).
arcsin (
6
x =I' 0
).
== arccos( 3
x - 2
(
2 == arctan
arcsin (
),
05x
6
)
Ix - 21
2
184
CHAPTER 2
TRIGONOMETRY
In Exercises 61-64. graph the function.
61. fLr)
arcsin!x - I)
62. fLI I
= -::; + arctan \
63. fCd 64. f(r)
68. Pholograph1' A television camera at ground level is
filming the lift-off of the space shuttle at a point 2000 feet
from the launch pad (see figure). If (j is the angle of eleva­
tion to the shuttle and s is the height of the shuttle in feet.
write (j as a function of s. Find (j when (a) s = 1000 and
(b) s = 4000.
arctan 2x
x
arccos ­
it
4
In Exercises 65 and 66. write the given function in terms of the
sine function by using the identity
,
A cos wI + B sin wI
Sin(wl + arctan-).
A,
65. f(1I
3 cos 21 + 3 sin 21
66. fUi =
4
7TI
I
s
B'
Verify your result by using a graphing utility to graph both
forms of the function
cos
;
3 sin 7TI
1-1------ 2000ft - - - - - 1
FIGURE FOR 68
67. PllOf()graphr
A photographer is taking a picture of a
4-foot-square painting hung in an art gallery. The camera
lens is I foot below the lower edge of the painting (see
figure). The angle {3 subtended by the camera lens x feet
from the painting is given by
{3 = arctan
4x
+ 5
x > O.
(a) Use a graphing utility to graph (3 as a function of x.
(b) Move the cursor along the graph to approximate the
distance from the picture when {3 is maximum.
(c) Identify any asymptote of the graph and discuss its
meaning in the context of the problem.
FIGURE FOR 67 69. Docking a Baal A boat is pulled in by means of a winch
located on a dock 12 feet above the deck of the boat (see
figure). If (j is the angle of elevation from the boat to the
winch and s is the length of the rope from the winch to the
boat, write () as a function of s. Find (j when (a) s
48 feet
and (b) s
24 feet.
FIGURE FOR 69
SECTION 2.7
70. Area
INVERSE TRIGONOMETRIC FUNCTIONS
185
In calculus, it is shown that the area of the region
bounded by the graphs of y
0, r
I/(x' + I L x
(I,
and x
b is given by
73. Define the inverse cosecant function by restricting the do­
main of the cosecant to the intervals [- To /2. 0) and
(0. 7T /2). and sketch its graph.
Area = arctan b - arctan a
74. Use the results of Exercises 71-73 to evaluate (hc follo\'.·
ing without using a calculator.
(see figure), Find the area for the following values of a
and b,
(a) a = 0, b
(b) a
-l.b=1
(e) a = 0, b = 3
(d) a =
I.b=3
(a) arcsecY2
(b) arc sec I
(c) arceot -
(d) arc sec 2
In Exercises 75-79. verify the identity with a calculator. then
prove the identi ty.
75. arcsin( -
-arcsin x
x)
76. arctan( - x) = -arctan x
77. arccos( xl =
78. arctan x + arctan
79. arcsin x
FIGURE FOR 70
+
72. Define the inverse secant function by restricting the domain
of the secant to the intervals [0, 1T /2) and (1T /2. 1TJ, and
sketch its graph.
1T
2. x
x
0
To
arccos x
')
80. The Chebyshev polynomial of degree
formula T,,(x)
71. Define the inverse cotangent function by restricting the
domain of the cotangent to the interval (0. 1T). and sketch
its graph,
arccos x
1T
11
11
cos(n arccos xl for
is defined by the
1:0; x
I and
I. 2. 3..
(a) Show that T,,(x)
(b) Show that T1(xj
(C) Find
=
I.
x.
the quadratic polynomial T,Ln.
(d) Graph T,(x) and 4x'
3x on the same viewing
rectangle. What do you observe')
186
CHAPTER 2
TRIGONOMETRY
2.8
ApPLICATIONS OF TRIGONOMETRY
Applications Involving Right Triangles
Harmonic Motion
/
Trigonometry and Bearings
/
Applications Involving Right Triangles
In keeping with our twofold perspective of trigonometry, this section includes
both right triangle applications and applications that emphasize the periodic
nature of the trigonometric functions.
In this section, the three angles of a right triangle are denoted by the letters
A, B, and C (where C is the right angle), and the lengths of the sides opposite
these angles are denoted by the letters a, b, and c (where c is the hypotenuse).
B
&J
,
a
0
A)4.2
b =' 19.4
FIGURE 2.75
I
EXAMPLE 1 Solving a Right Triangle, Given One Acute Angle and
One Side
Solve the right triangle having A = 34.20 and b
ure 2.75.
= 19.4,
as shown in Fig­
C
SOLUTION
Because C = 90°, it follows that A
+
B = 90° and
To solve for a, use the fact that
tan A opp = ~
adj
b
->
a
= b tan A. Thus,
a
=
19.4 tan 34.2° "'" 13.18.
Similarly. to solve for c, use the fact that
cos A
adj
hyp
b
c -- c
b
cos A
Thus,
c
=
19.4
= 23.46.
cos 34.2° ....... .. SEC TI 0 N 2.8
187
ApPLICATIONS OF TRIGONOMETRY
Finding a Side of a Right Triangle
EXAMPLE 2
B
A safety regulation states that the maximum angle of elevation for a rescue
ladder is n°. If a fire department's longest ladder is 110 feet what is the
maximum safe rescue height?
/
r---+---'---'-i
SOLUTION
c=llOft
A sketch is shown in Figure 2.76. From the equation
a
sin A = -.
C
it follows that
a
= c sin A =
1 10(sin 72°) = 104.6 feet.
.........
FIGURE 2.76
EXAMPLE
3
Finding a Side of a Right Triangle
At a point 200 feet from the base of a building, the angle of elevation to the
bottom of a smokestack is 35°, whereas the angle of elevation to the lOp is 53°,
as shown in Figure 2.77. Find the height s of the smokestack alone.
SOLUTION
T
Note from Figure 2.77 that this problem involves two right triangles. In the
smaller right triangle, use the fact that tan 35°
a/200 to conclude that the
height of the building is
s
r
a = 200 tan 35°.
Now. from the larger right triangle, use the equation
a
1
tan 53°
-"-'---:,,.,...~;..jJ
a
+s
=-­
200
to conclude that a
+s
s = 200 tan 53°
FIGURE 2.77
Observer Horizontal
'.,...:...;=:..:==..,.
Object
,: Angle of
Angle of
depression
'-. elevation
Observer
Object
Horizontal
FIGURE 2.78
200 tan 53°. Hence. the height of the smokestack is
a
200 tan 53° - 200 tan 35°
125.4 feet.
.... , ....
Examples 2 and 3 used the term angle of elevation to represent the angle
from the horizontal upward to an object. For objects that lie below the horizon­
tal. it is common to use the term angle of depression, as shown in Figure 2.78.
In Examples I through 3. you found the lengths of the sides of a right
triangle. given an acute angle and the length of one of the sides. You can also
find the angles of a right triangle given only the lengths of two sides, as
demonstrated in Example 4.
188
CHAPTER 2
TRIGONOMETRY
EXAMPLE 4
Finding an Acute Angle of a Right Triangle
A swimming pool is 20 meters long and 12 meters wide. The bottom of the
pool is slanted so that the water depth is 1.3 meters at the shallow end and
4 meters at the deep end. as shown in Figure 2.79. Find the angle of depression
of the bottom of the pool.
FIGURE 2.79
REMARK
Note that the width of the
pool, 12 meters, is irrelevant to the
problem.
SOLUTION
Using the tangent function, you can see that
tan A = opp = 2.7 = 0.135. adj
20 Thus, the angle of depression is given by
A
= arctan 0.135
= 0.13419 radians
= 7.69°.
........ . Trigonometry and Bearings
In surveying and navigation, directions are generally given in terms of bear­
in~s. A bearing measures the acute angle a path or line of sight makes with a
fixed north-south line. For instance, in Figure 2.BO(a), the bearing is S 35° E,
meaning 35 degrees east a/south. Similarly, the bearings in Figure 2.BO(b) and
(c) are N BO° W and N 45° E, respectively.
N
N
i
W·-=""""'I---E
W ----,"'---..... E
S
s
s
(a) S 35° E
FIGURE 2.80
0
(b) N80 W
(el N 45° E
SECTION 2.8
EXAMPLE 5
ApPLICATIONS OF TRIGONOMETRY
189
Finding Directions in Terms of Bearings
A ship leaves port at noon and heads due west at 20 knots (nautical miles per
hour). At 2 P.M., to avoid a stonn, the ship changes course to N 54° W, as
shown in Figure 2.81. Find the ship's bearing and distance from the port of
departure at 3 P.M.
nm
= 2(20 nm)
_I A
FIGURE 2.81
SOLUTION
Because the ship travels west at 20 knots for two hours, the length of AB is 40.
Similarly, BD is 20. In triangle BCD, you have B
90°
54° = 36°. The
two sides of this triangle are detennined as follows.
sin B =
b
~
20
20 sin 36"
and
d
cos B = ­
• 20
d = 20 cos 36°
Now, in triangle ACD. you can detennine angle A as follows.
b
20 sin 36°
tan A = - - =
= 0.2092494
d + 40
20 cos 36° + 40
A = arctan 0.2092494 "" 0.2062732 radians = 11.82°
The angle with the north-south line is 90°
bearing of the ship is
N 78.18° W.
Bearing
Finally, from triangle ACD, you have sin A
b
11.82° = 78.18°. Therefore, the
b Ie, which yields
20 sin 36°
c=--=
sin A
sin(l1.82)
c = 57.4 nautical miles.
Distance from pOri
• • • • • • II • •
til
190
CHAPTER
2
TRIGONOMETRY
Harmonic Motion
The periodic nature of the trigonometric functions is useful for describing the
motion of a point on an object that vibrates, oscillates. rotates. or is moved by
wave motion.
For example. consider a ball that is bophing up and down on the end of
a spring, as shown in Figure 2.82. Suppose that 10 centimeters is the maximum
distance the ball moves vertically upward or downward from its equilibrium
(at rest) position. Suppose further that the time it takes for the ball to move
from its maximum displacement above zero to its maximum displacement
below zero and back again is t = 4 seconds. Assuming the ideal conditions of
perfect elasticity and no friction or air resistance, the ball would continue to
move up and down in a uniform and regular manner.
From this spring you can conclude that the period (time for one complete
cycle) of the motion is
Period = 4 seconds
and that its amplitude (maximum displacement from equilibrium) is
Amplitude = 10 centimeters.
Motion of this nature can be described by a sine or cosine function. and is
called simple harmonic motion.
10 em
10 em
10 em
Oem
o em
o em
-10 em
-10 em
-10 em
Maximum negative
displacement
Equilibrium
Maximum positive
displacement
Simple Harmonic Motion
FIGURE 2.82
SECTION 2.8
ApPLICATIONS OF TRIGONOMETRY
th~t mov~s'o[;iddordi~~te
191
.. A· point
HiieisiIl:siillpJeJiarlllonic motion
ifitSdistancedf~omthe.origt{~tjiIne
byeith~r
.
.
-:
'-, - "', .-',.
.. ',- .,,' -.' :..i$ , .
. .... .
'
.'
,-
-~.
-
".
t given
-
"
:',
where aand·:w. ~ert!alnuinbers stich thatw >0. J11emotion has amplitudela j, periOd 2 'IT! w, and rr~c:jtiency'w /2 'IT. EXAMPLE
Simple Harmonic Motion
6
Write the equation for the simple harmonic motion of the ball described in
Figure 2.82, where the period is 4 seconds. What is the frequency of this
motion?
SOLUTION
Because the spring is at equilibrium (d
d
= 0)
when
t
= 0,
use the equation
a sin wt.
Moreover, because the maximum displacement from zero is 10 and the period
is 4, you have
= Ia I =
Amplitude
2 'IT
Period = -
w
= 4
10
->
w
'IT
2
Consequently, the equation of motion is
'IT
d
10 sin
"2('
Note that the choice of a = 10 or a =
10 depends on whether the ball
initially moves up or down. The frequency is given by
Frequency
w
'IT /2
1
= - - = - cycle per second.
2 'IT
2 'IT
4
...... ... 192
CHAPTER
2
TRIGONOMETRY
FIGURE 2.83
One illustration of the relation between sine waves and harmonic motion
is seen in the wave motion resulting from dropping a stone into a calm pool
of water. The waves move outward in roughly the shape of sine (or cosine)
waves, as shown in Figure 2.83. As an example, suppose you are fishing and
your fishing bob is attached so that it does not move horizontally. As the waves
move outward from the dropped stone, your fishing bob will move up and
down in simple harmonic motion, as sho'wn in Figure 2.84.
y
1
I
1
Simple Harmonic Motion
EXAMPLE 7
Given the equation for simple harmonic motion,
37T
d = 6 cos-t
4 '
A fishing bob moves in a vertical
direction as waves move to the right
FIGURE 2.84
find the following.
A. The maximum displacement
B. The frequency
c. The value of d when t = 4
D. The least positive value of t for which d
=
0
SOLUTION
The given equation has the form d = a cos wI, with a = 6 and w
=
37T /4.
A. The maximum displacement (from the point of equilibrium) is given by
the amplitude. Thus, the maximum displacement is 6. w
37T/4
3
I
. f'
B. Frequency = 27T = ---z;:- = "8 cyc e per umt 0 time
C. d
=
6cos[3;(4)]
6 cos 37T = 6(-1) = -6
D. To find the least positive value of t for which d = 0, solve the equation
d
=
37T
6 cos - [
4
TechnofolJlJ Note
You can use your graphing utility to
solve part D as follows. Graph the
function YI = 6 cose:,) on the
viewing rectangle 0 :5 X :5 6,
-7 :5 Y :5 7, and observe that the
first x -intercept is approximately
x = 0.7. Using the zoom and trace
features, you will see that x "" 0.667.
0
to obtain
37T
4
7T 37T 57T
2' 2 ' 2
-1=- ­
10
2
t = -3' 2, ­3 ' ....
Thus, the least positive value of t is [ = ~.
....... .. SECTION 2.8
APPLICATIONS OF TRIGONOMETRY
193
Many other physical phenomena can be characterized by wave motion.
These include electromagnetic waves such as radio waves, television waves,
and microwaves. Radio waves transmit sound in two different ways. For an
AM station, the amplitude of the wave is modified to carry sound (AM stands
for amplitude modulation). See Figure 2.85(a). An FM radio signal has its
frequency modified in order to carry sound, hence the term frequency modu­
lation. See Figure 2.85(b).
(a) AM: Amplitude modulation
(b) FM: Frequency modulation
FIGURE 2.85
Radio Waves
Suppose you are teaching a class in trigonometry. Write two "right triangle prob­
l~ms" that you think would be reasonable to ask your students to solve. (Assume
that your students have 5 minutes to solve each problem.)
You
BE THE
INSTRUCTOR
194
CHAPTER 2
TRIGONOMETRY
covered in earlier sections. You
amplitude arid period of the function.
8. f(~) =
.:
10. g (x)
! sin
1TX
X
= 0.2 cot '4
SECTION 2.8 . EXERCISES
...................•..................•...•.•...•..•.....•........•..................•....................•............•
In Exercises I -10. solve the right triangle shown in the figure.
(Round your result to two decimal places.)
In Exercises II and 12. find the altitude of the isosceles triangle
shown in the figure. (Round your result to two decimal places.)
B
c
a
c
A
b
FIGURE FOR 1-10
1. A
=
b == 10
20°.
3. B = 71°.
b
5. A
12° 15'.
7. a ==
6,
9. b = 16,
b=
C
24
c = 430.5
10
52
(j
2. B == 54°, c == 15
4. A = 8Ao, a = 40.5
6. B
= 65°
8. a
25,
12'.
10. b = 1.32.
C
a
=
b
FIGURE FOR 11 AND 12
14.2
== 35
9.45
C
f
11. () = 52°.
b = 4 inches
12. ()
b
18°,
10 meters
SECTION 2.8
13. Length oj a Shadow If the sun is 30° above the horizon,
find the length of a shadow cast by a silo that is 70 feet high
(see figure).
FIGURE FOR 13
ApPLICATIONS OF" TRIGONOMETRY
195
16. Height The length of a shadow of a tree is 125 feet when
the angle of elevation of the sun is 33° (see figure). Approx­
imate the height h of the tree.
FIGURE FOR 16
14. Length oj a Shadow The sun is 20° above the horizon.
Find the length of a shadow cast by a building that is 600
feet high (see figure).
17. Angle oj Elevation An amateur radio operator erects a
75-foot vertical tower for his antenna (see figure). Find the
angle of elevation to the top of the tower at a point on level
ground 50 feet from the base.
/~
1
600 ft
FIGU RE FOR 14
15. Height A ladder of length 16 feet leans against the side of
a house (see figure). Find the height h of the top of the
ladder if the angle of elevation of the ladder is 74°.
FIGURE FOR 15
FIGURE FOR 17
18. Angle oj Elevation The height of an outdoor basketball
backboard is 124 feet, and the backboard casts a shadow
17 ~ feet long (see figure). Find the angle of elevation of the
sun.
FIGURE FOR 18
196
CHAPTER
2
TRIGONOMETRY
19. Angle ofDepression
A spacecraft is traveling in a circular
orbit 100 miles above the surface of the earth (see figure).
Find the angle of depression from the spacecraft to the
horizon. Assume that the radius of the earth is 4000 miles.
23. Height From a point 50 feet in front of a church. the
angles of elevation to the base of the steeple and to the top
of the steeple are 35° and 47° 40', respectively (see figure).
Find the height of the steeple.
Orbit
FIGURE FOR 23
Angle of
depression
FIGURE FOR
19
24. Height
From a point 100 feet in front of a public library,
the angles of elevation to the base of the flagpole and to the
top of the pole are 28° and 39° 45', respectively. The
flagpole is mounted on the front of the library's roof (see
figure). Find the height of the pole.
20. Angle of Depression
Find the angle of depression from
the top of a lighthouse 250 feet above water level to the
water line of a ship 2 miles offshore.
21. Airplane Ascent
When an airplane leaves the runway (see
figl'.re), its angle of climb is 18° and its speed is 275 feet per
second. Find the altitude of the plane after 1 minute.
lOOft----l
FIGURE FOR 24
25. Navigation An airplane flying at 550 miles per hour has
a bearing of N 52° E. After flying for 1.5 hours, how far
FIGURE FOR 21
22. Mountain Descent A sign on the roadway at the top of a
mountain indicates that for the next 4 miles the grade is
12S (see figure). Find the change in elevation for a car
descending the mountain.
north and how far east has the plane traveled from its point
of departure?
26. Navigation A ship leaves port at noon and has a bearing
of S 27° W. If the ship is sailing at 20 knots, how many
nautical miles south and how many nautical miles west has
the ship traveled by 6:00 PM?
27. Navigation
A ship is 45 miles east and 30 miles south of
port. If the captain wants to sail directly to port, what
bearing should be taken?
28. Navigation
FIGURE FOR 22
A plane is 120 miles north and 85 miles east
of an airport. If the pilot wants to fly directly to the airport,
what bearing should be taken?
SECTION 2.8
29. Surveying A surveyor wishes to find the distance across a
swamp (see figure). The bearing from A to B is N 32° W.
The surveyor walks 50 yards from A, and at point C the
bearing to B is N 68° W. (a) Find the bearing from A to C.
(b) Find the distance from A to B.
APPLICATIONS OF TRIGONOMETRY
197
32. Distance Between Towns
A passenger in an airplane
flying at 30,000 feet sees two towns directly to the left of
the airplane. The angles of depression to the towns are 28°
and 55° (see figure). How far apart are the towns')
30,000 ft
I
1
FIGURE FOR 32
FIGURE FOR 29
33. Altitude ofa Plane
30. Location of a Fire Two fire towers are 20 miles apart,
tower A being due west of tower B. A fire is spotted from
the towers, and the bearings from A and Bare N 76° E
and N 56° W (see figure). Find the distance d of the fire
from the line segment A.B. [Hint: Use the fact that d =
20/(cot 14° + cot 34°).]
d
FIGURE FOR 30
A plane is observed approaching your
home, and you assume it is traveling at 550 miles per hour.
If the angle of elevation of the plane is 16° at one time, and
I minute later the angle is 57°, approximate the altitude of
the plane.
34. Height of a Mountain In traveling across flat land, you
notice a mountain directly in front of you. The angle of
elevation (to the peak) is 3.5°. After you drive 13 miles
closer to the mountain, the angle of elevation is 9°. Approx­
imate the height of the mountain.
35. Length A regular pentagon is inscribed in a circle
of radius 25 inches. Find the length of the sides of the
pentagon.
36. Length A regular hexagon is inscribed in a circle of ra­
dius 25 inches. Find the length of the sides of the hexagon.
37. Wrench Size Use the figure to find the distance J across
the flat sides of the hexagonal nut as a function of r.
31. Distance Between Ships An observer in a lighthouse 300
feet above sea level spots two ships directly offshore. The
angles of depression to the ships are 4° and 6.5" (see figure).
How far apart are the ships?
r-·······_-x
FIGURE FOR 31
'1
FIGURE FOR 37
198
CHAPTER 2
TRIGONOMETRY
38. Bolt Circle
The figure shows a circular sheet 25 cm in
diameter containing 12 equally spaced bolt holes. Deter­
mine the straight-line distance between the centers of adja­
cent bolt holes.
----------------------- l'
40.
6
____________ 1
!
~
6
____~~__~~____~1
f--9~",,__4
1....-·--------36 -----------l·1
Harmonic Motion
In Exercises 41-44, for the simple har­
monic motion described by the given trigonometric function,
find (a) the maximum displacement, (b) the frequency, and (c)
the least possible value of ( for which d == O.
FIGURE FOR 38
Trusses
In Exercises 39 and 40, find the lengths of all the
pieces of the truss.
41. d = 4 cos 8m
43. d = k sin 120"t
42. d
44. d
~ cos 20"t
= t!:. sin 792m
45. Tuning Fork A point on the end of a tuning fork moves
in simple harmonic motion described by d = a sin wt, Find
w given that the tuning fork for middle C has a frequency
of 264 vibrations per second.
46. Wave Motion A buoy oscillates in simple harmonic mo­
tion as waves go past. At a given time it is noted that the
buoy moves a total of 3.5 feet from its low point to its high
point, and that it returns to its high point every 10 seconds,
Write an equation that describes the motion of the buoy if,
at t = 0, it is at its high point.
CHAPTER 2
CHAPTER
111T 1. -
21T
2.­
3. -110°
4. -.405°
9
In Exercises 5-8, convert the angle measurement to decimal
form. (Round your result to two decimal places.)
5. 135 0 16' 45"
7. 5° 22' 53"
6. -234° 50"
8. 280 0 8' 50"
In Exercises 9-12, convert the angle measurement to D° M' 5"
form.
9. 135.2r
ll. -85.15°
199
2 . REVIEW EXERCISES
In Exercises 1-4, sketch the angle in standard position, and list
one positive and one negative coterminal angle.
4
REVIEW EXERCISES
0
10. 25.1
12. -327.85°
In Exercises 31-34. find the remaining five trigonometric func­
tions of satisfying the given conditions. (Him: Sketch a right
triangle.)
e
e ~. tan e < 0
e=
sin e > 0
sin e = ~. cos e < 0
cos e = - ~. sin e > 0
31. sec
32. tan
33.
34.
In Exercises 35-40. evaluate the trigonometric function with­
out using a calculator.
1T rr
35. tan 3
36. sec
51T 37. sin 3
/5rr)
38. cot~ ­
6 J
40. esc 270°
39. cos 495°
4
In Exercises 13-16, convert the angle measurement from radi­
ans to degrees. (Round your result to two decimal places.)
51T 31T
13. -
14.
15. -3.5
16. 1.75
7
5
In Exercises 41 -44. use a calculator to evaluate the trigonomet­
ric function. (Round your result to two decimal places.)
41. tan 33°
42. csc lOse
121T
43. s e e -
44.
5
In Exercises 17-20, convert the angle measurement from de­
grees to radians. (Round your result to four decimal places.)
17. 480° 19. -33 0 45'
18. -16.5°
20. 84° 15'
Sin( ~)
e
e
In Exercises 21-24, find the reference angle for the given angle.
21. 252 0
23. 61T
5
22. 640°
17rr
24.­
3
In Exercises 25-30, find the six trigonometric functions of the
(in standard position) whose terminal side passes
angle
through the point.
e
25. (12. 16)
27.(-7.2)
29. (-4. -6)
26. (x.4x)
28. (4. -8)
30. (~.~)
e
In Exercises 45-48. find two values of
in degrees
~
< 360°) and in radians (0 ~ < 21T) without using a
calculator.
(0°
45. cos
e=
47. esc
(J
2
-2
46. sec
48. tan
e is undefined
V3
(J =
3
e
In Exercises 49-52, find two values of
in degrees
(0 0
< 360°) and in radians (0 ~ < 2rr) by using a cal­
culator.
e
e
sec e =
49. sin
51.
e
0.8387
1.0353
50. cot
52. esc
e=
e=
1.5399
11.4737
200
C HAP T E R 2
TR IGONOMETRY
In Exercises 53-56, write an algebraic expression for the given
expression.
53. sec[ arcsin(x - I)]
54. tan ( arccos
85. Alritude of a Triangle
the figure.
~)
/
2
55. Sin(arccos 4 x x 2 )
10m
56. csc(arcsin lOx)
/\50
In Exercises 57 -70. sketch a graph of the function through two
full periods. (Use a graphing utility to verify your resulL)
57. Y
3 cos
58. Y = -2 sin
27TX
59. fix) = 5 sin
~
60. fix)
5
1
7TX
- - cos-
61. fix)
4
4
66. h(r)
67. f(r)
csc( 3r -
,
!!.)'
2
tower is 225 feet. and it casts a shadow of length 105 feet
(see figure), Find the angle of elevation of the Sun.
-1)
3 cos(r + 7T)
=
sec(r -
~)
68. fir) = 3 csc(2r +
!!.1
4/
7TO
"8
I lOcos ( 120m
70. E(r)
-7T3)
In Exercises 7 J -80, use a graphing utility to graph the func­
tion.
x
73. Y =
.
- - sm x
4
71. fix)
x
"7"
3
cos 7T X
75. h(O) = 0 sin 7TO
77. y = arcsin
86. Angle of Elevarion The height of a radio transmission
4
_
7T4)
tan (,t
FIGURE FOR 85
7TX
64. g(r)
65. h(r)
8 cos(
0
62. fix) = -tan­
63. g(r) = ~ sin(r - 7T)
69. f(O) = cot
7TX
2x
Find the altitude of the triangle m
x
2
7T
79. fix) = - + arctan x
2
72. g(x) = 3 ( sin
74. Y
76. fit)
78. y
4 -
x
4
TTX
"3
+ cos
+
I)
7TX
2.5e·'/4 sin 2m
2 arccos x
80. fix) = arccos(x .. 7T)
In Exercises 8 J -84, use a graphing utility to graph the func­
tion. Use the graph to determine if the function is periodic. If
the function is periodic, find any relative maximum or mini­
mum points through one period.
81. fix)
e""'
83. g(x) = 2 sin x cos2 x
82. g(x)
84. h(x)
FIGURE FOR 86
87. Shuttle Height An observer 2.5 miles from the launch
pad of a space shuttle measures the angle of elevation to
the base of the vehicle to be 28° soon after lift- off (see
figure). How high is the shuttle at that instant? (Assume the
shuttle is still moving vertically.)
= sin e'
4 sin 2 x cos 2 x
T
J
r--__________
h.
I
~l
//~
/ / / 2.5 mi 1~
FIGURE FOR 87
.
Observer
------~-
CHAPTER 2
From city A to city B. a plane flies 650 miles at
a bearing of N 48° E, From city B to city C, the plane flies
810 miles at a bearing of S 65° E. Find the distance from
A to C and the bearing from A to C.
88. Distance
A train travels 2.5 miles on a straight
track with a grade of 10 10' (see figure). What is the
vertical rise of the train in that distance?
89. Railroad Grade
FIGURE FOR 89
90. Distance Between Towns A passenger in an airplane
flying at 35,000 feet sees two towns directly to the left of
the airplane. The angles of depression to the towns are 32°
and 76° (see figure). How far apart are the towns?
Town 2
FtGURE FOR 90
REVIEW EXERCISES
201
91. Using calculus. it can be shown that the sine and cosine
functions can be approximated by the polynomials
x 3 Xl
X7
sin X = x - - +­
3!
5!
7!
and
cos x = 1
x 2 x·
-+­
2!
4!
x6
6"
where x is in radians,
(a) Use a graphing utility to graph the sine function and its
polynomial approximation in the same viewing rec­
tangle.
(b) Use a graphing utility to graph the cosine function and
its polynomial approximation in the same viewing
rectangle,
(c) Study the patterns in the polynomial approximations
of the sine and cosine functions and guess the next
term in each, Then repeat parts (a) and (b). Do you
think your guesses were correct? How did the accuracy
of the approximations change when additional terms
were added?
202
CHAPTER
Z
TRIGONOMETRY
CHAPTERS 1 AND 2 . CUMULATIVE TEST
......•...............................•........•........................................................................
Take this test as you would take a test in class. After you are
done. check your work against the answers in the back of the
book.
In Exercises I and 2, solve the equation.
1. I -
4(x
- 3)
=~
2. 3x 2 +
X -
10 = 0
In Exercises 3-6, sketch the graph of the equation without the
aid of a graphing utility.
3. x
5.y
3y + 12 = 0
~
4. y
6.}'
=
x 2 4x + I
3-ix-2i
7. A line passes through the point (4, 9) with slope m ~.
Find the coordinates of three additional points on the line.
(There are many correct answers.)
8. Find an equation of the line passing through the points
(- L I) and (3, 8).
x/ex - 2) at the
9. Evaluate (ifpossihle) the functionf(x)
specified values of the independent variable.
(a) f(6)
(b) f(2)
(c) f(2r)
(d) f(s + 2)
10. Express the area A of an equilateral triangle as a function
of the length s of its sides.
11. Express the angle 41T/9 in degree measure and sketch the
angle in standard position.
12. Express the angle 1200 in radian measure as a multiple of
1T and sketch the angle in standard position.
13. The terminal side of an angle (j in standard position passes
through the point (12. 5). Evaluate the six trigonometric
functions of the angle.
14. If cos r
~ where 1Tj2 < r < 'IT, find sin t and tan t.
15. Use a calculator to approximate two values of
(j(0° ~ (j < 360°) such that sec (} = 2.125. Round your
answer to two decimal places.
16. Sketch a graph of each of the functions through two
periods.
(a) y =
3 sin 2x
(b) f(x) = 2 cos(x
1TX
(c) g(x) = tan 2
(d) her) = sec
.:)
2/
t
17. Write a sentence describing the relationship between the
graphs of the functions f(x) = sin x and g.
(a) g(x)
10 + sin x
(c) g(x) = sin( x +
~)
(b) g(x) = sin
(d) g(x)
'lTX
:2
-sin x
]8. Consider the function fix)
sin 3x
2 cos x.
(a) Use a graphing utility to graph the function.
(b) Approximate (accurate to one decimal place) the zero
of the function in the interval [0, 3 J.
(c) Approximate (accurate to one decimal place) the max­
imum value of the function in the interval [0. 3J.
19. Evaluate the expression without the aid of a calculator.
4
V3
(a) arcsin
(b) arctan
20. Write an expression that is equivalent to sin(arccos 2x).
21. From a point on the ground 600 feet from the foot of a cliff,
the angle of elevation to the top of the cliff is 320 30'. How
high is the cliff?