Trigonometric Functions - Birmingham Public Schools

Trigonometric Functions 4.1 Radian and Degree Measure

4.2 Trigonometric Functions: The Unit Circle

4,3 Right Triangle Trigonometry

44 Trigonometric Functions of Any Angle

4.5 Graphs of Sine and Cosine Functions

4.6 Graphs of Other Trigonometric Functions

4.7 Inverse Trigonometric Functions

4,8 Applications and Models

The Big Picture In this chapter you will learn how to

▪ describe an angle and convert between degree and radian measures.

a identify a unit circle and its relationship to real numbers.

▪ evaluate trigonometric functions of any angle. use fundamental trigonometric identities.

a sketch graphs of trigonometric functions. a evaluate inverse trigonometric functions. al evaluate the composition of trigonometric

functions. la use trigonometric functions to model and

solve real-life problems.

The Mauna Loa Observatory in Hawaii conducts research to understand the global carbon cycle. It is

located far from pollution sources that would affect the gases being measured. (Source: NOAA/Climate

Monitoring and Diagnostic Laboratory)

, Important Vocabulary As you encounter each new vocabulary term in this chapter, add the term and its definition to your notebook glossary.

• trigonometry (p. 284) • angle (p. 284) • initial side (p. 284) • terminal side (p. 284) • vertex (p. 284) • positive angles (p. 284) • negative angles (p. 284) • coterminal angles (p. 284) • central angle (p. 285) • radian (p. 285) • complementary angles (p. 287) • supplementary angles (p. 287)

• degree (p. 287) • linear speed (p. 289) • angular speed (p. 289) • unit circle (p. 295) • sine (pp. 296, 303) • cosecant (pp. 296, 303) • cosine (pp. 296, 303) • secant (pp. 296, 303) • tangent (pp. 296, 303) • cotangent (pp. 296, 303) • periodic (p. 298) • period (pp. 298, 326)

• hypotenuse (p. 303) • opposite side (p. 303) • adjacent side (p. 303) • reference angle (p. 316) • sine curve (p. 323) • amplitude (p. 325) • phase shift (p. 327) • damping factor (p. 339) • inverse sine function (p. 345) • inverse cosine function (p. 347) • inverse tangent function (p. 347)

illightional Resources Text-specific additional resources are available to help you do well in this course. See page xvi for details.

283

Figure 4.3

Positive angle (counterclockwise)

Negative angle (clockwise)

284 Chapter 4 • Trigonometric Functions

.1 Julian...adIlewea Angles 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 and was used in the development of astronomy, navigation, and surveying. With the development of calculus and the physical sciences in the 17th century, 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 and vibrations, including sound waves, light rays, planetary orbits, vibrating strings, pendulums, and orbits of atomic particles. The approach in this text incor-porates both perspectives, starting with angles and their measure.

(a) (b) Figure 4.1

An angle is determined by rotating a ray (half-line) about its endpoint. The starting position of the ray is the initial side of the angle, and the position after rotation is the terminal side, as shown in Figure 4.1(a). The endpoint of the ray is the vertex of the angle. This perception of an angle fits a coordinate system in which the ori-gin is the vertex and the initial side coincides with the positive x-axis. Such an angle is in standard position, as shown in Figure 4.1(b). Positive angles are generated by counterclockwise rotation, and negative angles by clockwise rotation, as shown in Figure 4.2. Angles are labeled with Greek letters a (alpha), s (beta), and 0 (theta), as well as uppercase letters A, B, and C. In Figure 4.3, note that angles a and s have the same initial and terminal sides. Such angles are coterminal.

gadii

te.401

You Should Learn:--'. • How to describe angles • How to use radian measure • How to use degree measure • How to use angles to model

and solve real-life problems

You Should Learn It: You can use radian measures of angles to model and solve real. life problems. For instance, in Exercise 93 on page 293 you are asked to determine measures of angles through which figure skaters jump.

Del

One in le

Becau full re Theref 7712 r over Si dian That for s a

Becau obtain

$1 These

6

Figure 4.2

C Figure 4

6 /radians \r

4 radians r 5 radians

Figure 4.5

1 2 IT ;volution = IT radians - re

2

1 27T IT ;volution = —

4 = —

2 radians - re

4

1 2- 7r -6

revolution = — = — radians 6 3

the radian measure of an angle of one full revolution is 27r, you can 3 le following. radians

Because obtain t

4.1 • Radian and Degree Measure 285

11110r4

Learn: angles

ian measure ree measure les to model ife problems

Measure

e way to measure angles is in radians. This type of measure is especially use-

ilculus. To define a radian, you can use a central angle of a circle, one in co vertex is the center of the circle, as shown in Figure 4.4.

pose

nition of a Radian

One radian is the measure of a central angle that intercepts an arc s equal in length to the radius r of the circle. See Figure 4.4.

adia

.earn If

measures of solve real-

nstance, in e 293 yo:i are measures of ch figurt,

Arc Length = radius when 0 = 1 radian.

Because the circumference of a circle is 27rr, it follows that a central angle of one Figure 4.4

full revolution (counterclockwise) corresponds to an arc length of s = 27rr.

Therefore, 27r radians corresponds to 360°, 7r radians corresponds to 180°, and

ir/2 radians corresponds to 90°. Moreover, because 27r ----, 6.28, there are just

Over .-

siir radius lengths in a full circle, as shown in Figure 4.5. In general, the ra-

mta,

eass/urr measure a central angle 0 is obtained by dividing the arc length s by r.

0, where 0 is Measured in radians. Because the units of measure 2 radians ,-- r 1 radian Thdiaant

----.... for s and r are the same, this ratio is unitless—it is simply a real number.

r \ r

These and other common angles are shown in Figure 4.6.

ADM 4.6

STUDY TIP

One revolution around a circle of radius r corresponds to an angle of 27r radians because

s 27rr - = = 27r radians. r


Recall that the four quadrants in a coordinate system are numbered I, II, III, and IV. Figure 4.7 shows which angles between 0 and 27 lie in each of the four quad-rants. Note that angles between 0 and 7/2 are acute and that angles between 7r/2 and ir are obtuse.

e =

Quadrant II Ir —2 < 0 < ir

9 - 0

Quadrant ifi Quadrant IV

7r<0<-2 37c —37r <6< 2

3ir u = — 2

Figure 4.7

Two angles are coterminal if they have the same initial and terminal sides. For instance, the angles 0 and 27 are coterminal, as are the angles 7r/6 and 137r/6. You can find an angle that is coterminal to a given angle 0 by adding or subtract-ing 27r (one revolution), as demonstrated in Example 1. A given angle 0 has infi-nitely many coterminal angles. For instance, 0 = 7r/6 is coterminal with

—6

+ 2n7r, where n is an integer.

=

Quadrant I

STUDY TI

The phrase "the terminal o lies in a quadrant" is of abbreviated by simply sa3 lies in a quadrant." The te sides of the "quadrant an 7r/2, 7, and 37/2 do not within quadrants.

pair

ide of en ing ..0

n flal les" 0, lie

C01710ffieflillfy,

Figure 4.9

Two poSi

t heir sum other) if th

EXAMPL

If possible.

Solution a. The con

2

The sup

b. Becaus only po,

The Interactive CD-ROM and Inte versions of this text show every e with its solution; clicking on the Try button brings up similar problems. Guided Examples and Integrated Examples show step-by-step solutions to additional examples. Integrated Examplec are related to several concepts in the section.

Degree A second \ (1°) is Nu tex. To me a circle, as sponds to

Because 2 are related

360°

From the 1

EXAMPLE 1 Sketching and Finding Coterminal Angles

a. For the positive angle 137r/6, subtract 27 to obtain a coterminal angle

13 6

— 27r = See Figure 4.8(a).

b. For the positive angle 37r/4, subtract 27 to obtain a coterminal angle

57r —4

— zir = — —4 •

See Figure 4.8(b).

c. For the negative angle — 27r/3, add 27r to obtain a coterminal angle

— 27r + 27r = 7r —3 3 •

See Figure 4.8(c).

37r 2

(a)

Figure 4.8 (b) (e)

7 VI, rminal side of t" is often 'ply saying ‘, 6 ' The terminal -ant angles" 0, do not lie

M and Internet NI/ every exampi. ig on the Try Ii 7 problems. ntegrated -step solutions tegrated Examr ,ncepts in the

complimentary angles

Figure 4.9

b. Becau only p

;e 47r/5 is greater than 7r/2, it has no complement. (Remember to use mitive angles for complements.) The supplement is

IT

Degree

47r IT

5 5 •

Measure

Supplementary angles


Two °sitive angles a and 13 are complementary (complements of each other) if P

is 7r/2. Two positive angles are supplementary (supplements of each their °In if

their sum is IT. See Figure 4.9. other/

r

ExAmPLE 2 Complementary and Supplementary Angles

If possible, find the complement and the supplement of (a) 27r/5 and (b) 47r15.

i Solution I a. The complement of 27715 is

27r 57r 47r 7T _ = =

5 10 10 10

pplement of 277-/5 is

= 27r 37r _ • 5 5

A second way to measure angles is in terms of degrees. A measure of one degree (1°) is equivalent to a rotation of 1/360 of a complete revolution about the ver-tex. To immure angles, it is convenient to mark degrees on the circumference of a circle, as shown in Figure 4.10. So, a full revolution (counterclockwise) corre-sponds to 360°, a half revolution to 180°, a quarter revolution to 90°, and so on.

Because 2ir radians corresponds to one complete revolution, degrees and radians are related by the equations

--= 27r rad and 180° = 71- rad.

latter equation, you obtain

IT 180°

180 rad and 1 rad =

IT

d to the conversion rules at the top of the next page.

IT

2

The su

IT

90° = ,1(360°)

120° 60°= —61 (360°) 135°A e

l 0= *(360°)

150° 30° = i (360°) 180° 0°

225° W 315° 3

360°

30° 210°

240° 2700 300° Figure 4.10

x

360°

From the

1°

Which lea

45°

90° 180°


Conversions Between Degrees and Radians ir rad

1. To convert degrees to radians, multiply degrees by 180° •

180° 2. To convert radians to degrees, multiply radians by

7T rad

To apply these two conversion rules, use the,basic relationship 7)- rad =180°. (See Figure 4.11.)

360°

APPli The ra. circle. length

where

EXA

A circl tral an

SOW/ To use

STUDY Tli) Figure 4.11

When no units of angle measure are specified, radian measure is implied. For instance, if you write (3 = IT or 0 = 2, you should mean 0 = radians or 0 = 2 radians.

EXAMPLE 3 Converting from Degrees to Radians

Express each angle in radian measure.

a. 135° b. —270°

Solution

a. 135° = (135 deg)( — 37r rad 7T rad

180 deg) 4

EXAMPLE 4 Converting from Radians to Degrees

Express each angle in degrees.

7T a. --- rad b. 2 rad

Solution

a. —IT 5 ra 2

d = ( 7T

) .06

)(180 deg) 90. Multiply by

2 A r sad 7T

b. 2 rad = (2 r4(180 deg ) 360

rad 7T = 114.59°

71- rad ) _ 37r rad

b. —270° = (-270 deg)

180 deg 2

Multiply by

Multiply by

Multiply by80

.

With calculators it is convenient to use decimal degrees to denote fractional parts of degrees. Historically, however, fractional parts of degrees were expressed in minutes and seconds, using the prime (') and double prime (") notations, respectively. That is,

1 = one minute =

1" = one second = 36loo(11

Consequently, an angle of 64 degrees, 32 minutes, and 47 seconds was represented by

= 64° 32 '47". Many calculators have spe-

cial keys for converting angles in degrees, minutes, and seconds (D° M' S") into decimal degree form, and vice versa.

Then, u,

Note tha radian m

The for a particl

Cons id If s is tir ticle is

Lin

Moreov s, the m!

length s given by

s 7.0

where 0 is measured in radians.

I .

APPlications measure formula 0 = s/r can be used to measure arc length along a

The radian snecifically, for a circle of radius r, a central angle 0 intercepts an arc of

circle; r

Length of circular arc

Length of circular arc

Substitute for r and O.

_ 167r — 3

Simplify.

Linear and Angular Speed


EXAMPLE 5 Finding Arc Length

A circle has a radius of 4 inches. Find the length of the arc intercepted by a cen-

tral angle of 240°, as shown in Figure 4.12.

Solution

To use the formula s = rO, first convert 240° to radian measure.

n- rad

deg) (180 deg) 2400 = (240 Convert from degrees to radians.

'

47T -= —

3 radians

Then, using a radius of r = 4 inches, you can find the arc length to be

16.76 inches. Use a calculator.

Note that the units for re are determined by the units for r because 0 is given in radian measure and therefore has no units.

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.

Simplify.

A computer simulation to accompany this concept appears in the Interactive CD-ROM and Internet versions of this text.

Figure 4.12

r TV

is convenient rees to denote legrees. rer, fractional !.re expressed mds, using louble prime ;ctively. That

rte =

rid = 361613(i0

ngle of 64 s, and 47 iented by

ors have spe rting angles ;, and seconds cimal degree ;a.

). • 44

er a particle moving at a constant speed along a circular arc of radius r. le length of the arc traveled in time t, then the linear speed of the par-

arc length s 'ear speed =

time t.

Moreover, if 0 is the angle (in radian measure) corresponding to the arc length S, the angular speed of the particle is

central angle 0 gular speed = time t •

An

Consid If s is t tide is

Li

in Exercise one-half rac

1.


--

EXAMPLE 6 Finding Linear Speed

The second hand of a clock is 10.2 centimeters long, as shown in Figure 4.13. Find the linear speed of the tip of this second hand.

Solution In one revolution, the arc length traveled is

s = 27rr

= 271(10.2)

= 20.47r centimeters.

The time required for the second hand to travel this distance is

t = 60 seconds = 1 minute.

So, the linear speed of the tip of the second hand is

G!-

10.2 cm

Figure 4.13

Linear speed = —t

20.47r centimeters

60 seconds

1.068 cm/sec.

EXAMPLE 7 Finding Angular and Linear Speed

A 10 inch radius lawn roller makes 1.2 revolutions per second, as shown in Figure 4.14.

a. Find the angular speed of the roller in radians per second.

b. Find the speed of the tractor that is pulling the roller.

Solution a. Because each revolution generates 27r radians, it follows that the roller turns

(1.2)(270 = 2.47r radians per second. In other words, the angular speed is 10 in. Figure 4.14

Angular speed = —t

2.47r radians 1 second

= 2.47r rad/sec.

b. The linear speed is

Linear speed = —st

r0 = — t

10(2.470 inches — 75.4 in./sec.

1 second

In Exercise the angle Ii :

5.(a) —5

77T 6. (a) —

7. (a) —

8. (a) —1

9. (a) 3.5

10. (a) 5.63

In Exercis position.

3 77-

11. (a) — 4

12. (a) —

In Exercise in radian m the given a I he interact/1',

'leP-bY-step lAercises. The (mided

10.2 cm

3 2.

in Exercises 1-4, estimate the angle to the nearest 15. (a)

one.half radian.

1.

iv 2

iv 0 =

12 7r --•""Is- 0

3rc 2

(b)

IT 117r (b) -

[2 9

(b) -2

(b) 2.25

3 (b) -2.25

7. (a) - -

8. (a) -1

9. (a) 3.5

10.(a) 5.6

2


IExercises

All11111110111111111111111111111111111111

(b)

4.

In Exercises 5-10, determine the quadrant in which

the angle lies. (The angle is given in radian measure.) 117r

77r - 17. (a) 4

5. (a) -5

(b) -5

77r 777-

*6. 117r 18. (a) -8

(a) -4-

(b) 4

In Exercises 11-14, sketch the angle in standard position.

16. (a)

27r (b) -

15

8 7T (b) -35

In Exercises 19-22, find (if possible) the complement and supplement of the angle.

.57r (b)

71, 77r

4 (b)

117 13.(a) r (b) 97r

14.(a) 4 6

In Exercises 15-18, determine two coterminal angles 01 radian measure (one positive and one negative) for the given angle. rile Interactive CD-ROM and Internet versions of this text contain

Exercises. They solutions to all odd-numbered Section and Review

G also provide Tutorial Exercises, which link to Sided Examples

for additional help.

In Exercises 23-26, estimate the number of degrees in the angle.

23.

26.

19.(a)

20. (a) -1-1-1-2

21. (a) 1

22. (a) 3

(b) -3 25.

24.

In Exercises 47-54, convert the angle measure degrees to radians. Round your answer to three deei

In Ex mal places. 75.

47. 115° 4

50

8.

.

49. -216.35°

82.7

46:520

51. 642° 52. 0.54°

53. -0.78° 54. 395°

77. 32 In Exercises 55-58, express the angle in de measure. (Do not use a calculator.)

31T 7T 55. (a) -

2 (b) --

6

7 71" IT56. (a) - (b) -

12 9

7 77" 13 7T 57. (a) - (b) -

3 60

157r 281r

In Exe central arc of 1

79. 15 58. (a) (b) 6 15

80. 22

In Exercises 59-66, convert the angle measure from 81. 14.

radians to degrees. Round your answer to three dui. 82. mal places.

80

7r 60. -81T

257r 61.

8 59. -f

13

62. 6.57r 63. -4.2Tr 64. 4.8

65. -2 66. -0.48

In Exercises 67-70, use the angle-conversion capabil ities of a graphing utility to convert the angle measure to decimal degree form. Round your answer to three decimal places if necessary.

39. (a) 24° (b) 126° 67. (a) 64° 45' (b) - 124° 30'

40. (a) 87° (b) 167° 68. (a) 275° 10' (b) 9° 12'

41. (a) 79° (b) 150° 69. (a) 85° 18 ' 30" (b) - 408°16'

42. (a) 130° (b) 170° 70. (a) - 125° 36" (b) 330° 25"

In Exe. circle o

. 1

83. 14

84. 9 fe

85. 6 m

86. 40 c

Distance the dista sphere o same me

292 Chapter 4 Trigonometric Functions

In Exercises 27-30, determine the quadrant in which the angle lies.

27. (a) 1500

(b) 282°

28. (a) 7.9°

(b) 257.5°

29. (a) - 132° 50' (b) -336° 30'

30. (a) -260.25° (b) -2.4°

In Exercises 31-34, sketch the angle in standard position.

31. (a) 60° (b) 1500

32. (a) -270° (b) -120°

33. (a) 405° (b) 780°

34. (a) -450° (b) -570°

In Exercises 35-38, determine two coterminal angles in degree measure (one positive and one negative) for the given angle.

35. (a) (b)

\36. (a)

0 = 114°

----

37. (a) 300°

38. (a) -445°

(b) 230°

(b) -740°

(b)

In Exercises 39-42, find (if possible) the complement and supplement of the angle.

Prr"

87. Miai

Erie 88. Johai

Jerus

•1

89. Dffe sPher

In Exercises 43-46, express the angle in radian mea-sure as a multiple of 77. (Do not use a calculator.)

43. (a) 30° (b) 150°

44. (a) 315° (b) 120°

45. (a) -20° (b) -240°

46. (a) - 270° (b) 144°

In Exercises 71-74, use the angle-conversion capabil

ities of a graphing utility to convert the angle measur(

to D° M ' S " form.

71. (a) 280.60

72. (a) -345.12°

73. (a) 4.5

74. (a) -0.355

(b) -115.8°

(b) 310.75°

(b) -3.58

(b) 0.7865

Radius r

83. 14 inches

84. 9 feet

85. 6 meters


measure fr

T to three deet.

Exercises 75-78, find the angle in radians.

76.

version capabil-e angle measure answer to three

) — 124° 30'

) 9° 12'

) —408° 16'25'

) 330° 25"

version capabil.

e angle measure

) —115.8°

) 310.75°

) —3.58

) 0.7865

In Exercises 79-82, find the radian measure of the

central angle of a circle of radius r that intercepts an

arc of length s.

Radius r Arc Length s

79. 15 inches 8 inches

80. 22 feet 10 feet

81. 14.5 centimeters 35 centimeters

82. 80 kilometers 160 kilometers

In Exercises 83-86, find the length of the arc on a circle of radius r intercepted by a central angle 0.

Central Angle 0

180°

60°

—21T radians 3

37r . 86.40 centimeters —

4 radians

Distance Between Cities In Exercises 87 and 88, find the distance between the cities. Assume that earth is a sphere of radius 4000 miles and the cities are on the same meridian (one city is due north of the other).

City Latitude

87.Miami 25° 46 ' 37"N

Erie 42° 7 ' 15"N 88. Johannesburg, South Africa 26° 10' S

Jerusalem, Israel 31° 47 'N

89.Difference in Latitudes Assuming that earth is a Sphere of radius 6378 kilometers, what is the differ-

ence in latitude of two cities, one of which is 600 kilometers due north of the other?

90. Difference in Latitudes Assuming that earth is a sphere of radius 6378 kilometers, what is the differ-ence in latitude of two cities, one of which is 800 kilometers due north of the other?

91. Instrumentation The pointer on a voltmeter is 6 centimeters in length. Find the angle through which the pointer rotates when it moves 2.5 centimeters on the scale.

92. Electric Hoist An electric hoist is used to lift a piece of equipment 1 foot. The diameter of the drum on the hoist is 10 inches. Find the number of degrees through which the drum must rotate.

10 in.

1 ft

93. Figure Skating The number of revolutions made by a figure skater for each type of axel jump is given. Determine the measure of the angle generated as the skater performs each jump. Give the answer in both degrees and radians.

(a) single axel:

(b) double axel:

(c) triple axel:

94. Linear Speed of an Earth Satellite An earth satellite in circular orbit 1250 kilometers high makes one complete revolution every 90 minutes. What is its linear speed? Use 6400 kilometers for the radius of the earth.

83.7°

— 46.52°

0.54°

395°

ngle in dem. e

_ 77

6

7F

9

137r

60

287r

15

e measure from er to three deci.

257r 61.

8

64. 4.8

6 cm

78.

101. The angles of a triangle can have radian n ieasur4 2 /7- Ir , IT

102. Writing If the radius of a circle is increasing

how is the length of the intercepted arc ch angingl the magnitude of a central angle is held c onstant,

Explain your reasoning.

and

103. Writing Write a short paragraph defining and giv. ing examples of coterminal angles.

104. Geometry Show that the area of a circular s of radius r with central angle 0 is A = vst 0 is measured in radians.

Area of a Circular Sector In Exercises 105 and use the result of Exercise 104 to find the area of sector.

105. 106. 12 ft

5 ft

107. Graphical Reasoning The formulas for the a[edd

of a circular sector and arc length are A = r I 2e an s = rO, respectively. (r is the radius and 0 is the angle measured in radians.)

(a) If 0 = 0.8, express the area and arc 11 functions of r. What is the domain of ea tion? Graph the functions using a grapl-ity. Use the graphs to determine which changes more rapidly for changes in r> 1. Explain.

(b) If r = 10 centimeters, express the area and arc

length as functions of 0. What is the domain ol

each function? Use a graphing utility to graph and identify the functions.

Review

ength as ch fund-Ling util-function r when

110. f(x) = 2 - x5 111. f(x) = -(x + 3)5

In Exercises 112-115, graph the exponential functO

112. f(x) = 6x

114. f(x) = 6- x

113. f(x) = 6x - 2

115. f(x) = 6x +1

The Unit Cil The two historic introducing the is based on the I

Consider th

2 ± y2

known in Figi

(0, 1)

(- 1. 0)

(0, --I)

Figure 4.15

Imagine that numbers corre corresponding

Figure 4.16

As the real flume responds to a corresponds to fereace of 27r,

gen.eral, eac P"iti°1-1 ) WhOS

F '°r111111a s = r arc intercepted


95. Angular Speed A car is moving at a rate of 40 miles per hour, and the diameter of its wheels is 2.5 feet.

(a) Find the rotational speed of the wheels in revo-lutions per minute.

(b) Find the angular speed of the wheels in radians per minute.

96. Circular Saw Speed The circular blade on a saw has a diameter of 7.5 inches and rotates at 2400 rev-olutions per minute.

(a) Find the angular speed in radians per second.

(b) Find the linear speed of the saw teeth (in feet per second) as they contact the wood being cut.

7.5

97. Floppy Disk The radius of the magnetic disk in a 3.5-inch diskette is 1.68 inches. Find the linear speed of a point on the circumference of the disk if it is rotating at a speed of 360 revolutions per minute.

98. Speed of a Bicycle The radii of the sprocket assemblies and the wheel of a bicycle are 4 inches, 2 inches, and 14 inches, respectively. If the cyclist is pedaling at a rate of 1 revolution per second, find the speed of the bicycle in (a) feet per second and (b) miles per hour.

4 in. 2 in.

Synthesis

True or False? In Exercises 99-101, determine whether the statement is true or false. Justify your answer.

99. A degree is a larger unit of measure than a radian.

100. An angle that measures -1260° lies in Quadrant III.

In Exercises 108-111, sketch the graph of y X5 and

the specified transformations.

108. f(x) = (x - 2)5 109. f(x) = x5 - 4

You Should Learn:

• How to identify a unit circle and its relationship to real numbers

• How to evaluate trigonometric functions using the unit circle

• How to use the domain and period to evaluate sine and cosine functions

• How to use a calculator to evaluate trigonometric functions

You Should Learn It:

Trigonometric functions are used to model the movement of an oscillating weight. For instance, in Exercise 59 on page 301, the displacement from equilibrium of an oscillating weight suspended by a spring is modeled as a func-tion of time.

Ei)

4.2 • Trigonometric Functions: The Unit Circle 295

adian measures

; increasing aud ; held constant

arc chatiging,;

efining and gtv,

circular sector = 1 ,2 p

1 2' ‘-'5 Where

S 105 ard 106, the area of the

12 ft

6x —2

The Unit Circle lie two historical perspectives of trigonometry incorporate different methods for

,oducing the trigonometric functions. Our first introduction to these functions

is based on the unit circle. Consider the unit circle given by

x2 + y2 -= 1 Unit circle

as shown in Figure 4.15.

(0, 1)

(0,—1

Figure 4.15

Imagine that the real number line is wrapped around this circle, with positive numbers corresponding to a counterclockwise wrapping and negative numbers corresponding to a clockwise wrapping, as shown in Figure 4.16.

Figure 4.16

As the real number line is wrapped around the unit circle, each real number t cor-responds to a point (x, y) on the circle. For example, the real number 0 corresponds to the point (1, 0). Moreover, because the unit circle has a circum-ference of 2, the real number 277- also corresponds to the point (1, 0).

In general, each real number t also corresponds to a central angle 0 (in standard Position) whose radian measure is t. With this interpretation of t, the arc length formula s -= r0 (with r = 1) indicates that the real number t is the length of the 45e intercepted by the angle 0, given in radians.

las for the area .e A = -}r 20 and us and 0 is the

d arc length as in of each func-a graphing util-which function

ages in r when

he area and arc s the domain of utility to graph

I of y = x5 and

5_ 4

-(x + 3)5

ntial function.

A computer animation of this concept appears in the Interactive CD-ROM and Internet versions of this text.


The Trigonometric Functions

From the preceding discussion, it follows that the coordinates x and y are two functions of the real variable t. You can use these coordinates to define the six trigonometric functions of t.

sine cosecant cosine secant tangent cotangent

These six functions are normally abbreviated sin, csc, cos, sec, tan, and cot, respectively.

Definitions of Trigonometric Functions

Let t be a real number and let (x, y) be the point on the unit circle correspond-ing to t.

sin t = y

cos t = X

1 csc t = —

y,

1 sec t = —

x,

y 0

tan t = — x * 0 cot t = y * 0

Note that the functions in the second column are the reciprocals of the corre-sponding functions in the first column.

In the definitions of the trigonometric functions, note that the tangent and secant are not defined when x = 0. For instance, because t = 7r/2 corresponds to (x, = (0, 1), it follows that tall(r/2) and sec(7r/2) are undefined. Similarly, the cotangent and cosecant are not defined when y = 0. For instance, because t = 0 corresponds to (x, = (1, 0), cot 0 and csc 0 are undefined.

In Figure 4.17, the unit circle has been divided into eight equal arcs, corre-sponding to t-values of

rrr0 — —371- 7r —57T —37 —77T and 27r.

4 ' 2' 4 4 2 4

Similarly, in Figure 4.18, the unit circle has been divided into twelve equal arcs, corresponding to t-values of

71" 7T 77- 27r 577- 77r 47r 377- 57r 117r ,

6' 3' 2' 3 6 "if and

CI' L 7T.

6 3 2 3 6

Using the (x, y) coordinates in Figures 4.17 and 4.18, you can easily evaluate the trigonometric functions for common t-values. This procedure is demonstrated in Examples 1 and 2.

ExA:1F EvaloLne

a. t = -6

SO/Utia For each t Then use

. t= IT,,

Si

co

tan

b. t = 57r

Figure 4.17

Figure 4.18

A computer animation of this conc

appears in the Interactive CD-RC Internet versions of this text.

(if 2 (0, 1) /

sin

cog

tan

C. t = 0 c

sin

cos

tan

c

sin

cos

Lin

' .4) (0, 1) /

2 -1)

EV1 2 ' 2

I) 2

Fl

In

sin IT = y = 0 1

CSC IT = - is undefined.

cos IT = X = 1 1

sec IT = = 1 x

. of this concept 'ive CD-ROM and is text.

y 0 x. tan 12= = = 0 AT - 1

Cot '7r= y- is undefined.


•\ri 2

EXAMPLE 1 Evaluating Trigonometric Functions

valuate the six trigonometric functions at each real number.

IT 57r b. t = —4 c. t = 0 d. t = 77- . t = -6-

Solution For Fo ea t-value, each ue, begin by finding the correspdnding point (x, y) on the unit circle. Then use the definitions of trigonometric functions listed on page 296.

a. t , 77/6 corresponds to the point (x, y) = (-1372, 1/2).

. IT 1 csc -

7r 1 = - = 2 sin -6 = y = -2

6 y

IT -,./ IT 1 2 2-, cos - = X = —

6 2 sec - = - = — =

ir y 1/2 1 IT X tan

6 x -,12 --` cot =

; =

b. t --- 57/4 corresponds to the point (x, y) = (- -..//2, -

57T 1 2 y

,-2

sin —4 = y = - —2 csc -h--4 = - = = - 2

57r 57r 1 2 cos —4 = x = - - sec-h-- = -

x = --- = --,/,

57r y --.,//2 57r X tan — = 1

4= =

x - -4,/2 cot —

4 = ; = 1

c. t = 0 corresponds to the point (x, y)>= (1, 0).

1 csc 0 = - is undefined.

1 sec 0 = - = 1

x

y 0 x tan 0 = = = 0 cot 0 =. - undefined.

d. t= 77- corresponds to the point (x, = (- 1, 0).

sin 0 = y = 0

cos 0 = x = 1


Evaluate the six trigonometric functions at t = -

Solution Moving clockwise around the unit circle, it follows that t = - 7r/3 corresponds to the point (x, y) = (1/2, -

csc( - -7r) = -- 3

2

sec(_-) = 2 3

= _/ cot( - )

_ - 2

_ 1

- 2

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 Figure 4.19. Because r = 1, it follows that sin t = y and cos t = x. Moreover, because (x, y) is on the unit circle, you know that - 1 y 1 and - 1 x 1. So, the values of sine and cosine also range between - 1 and 1.

- 1 5_ y 1 and

- 1 < x < 1

- 1 5_ sin t 1 - 1 cos t 1

_

With your graphing utility in radian and parametric mod enter X1T = COS T and YlT * IT and use the following settings. Tmin = 0, Tmax = 6.3, Tstep = 0.1 Xmin = -1.5, Xmax = 5, Xscl = 1 Ymin = -1, Ymax = 1, YL

scG1 r=apjh the entered egliations

and describe the graph. 2. Use the trace feature to move

the cursor around the graph, What do the t-values represent? What do the x-and y-values represent? What are the least and gre est values for x and y9

ExAM

Beca

sin

b. Beca

C 0 S

Recall fi j(-t)

The co

co

The sin

sin tan

Adding 27r to each value of t in the interval [0, 2] completes a second revolution around the unit circle, as shown in Figure 4.20. The values of sin(t + 27r) and cos(t + 27r) 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 + 27rn) = sin t

and

cos(t + 2n) = cos t

for any integer n and real number t. Functions that behave in such a repetitive (or cyclic) manner are called periodic.

Definition of a Periodic Function

A function f is periodic if there exists a positive real number c such that

At ± = f (0

for all t in the domain of f. The smallest number c for which f is periodic is called the period of f

-1 v <

-1 < x < Figure 4.19

22 2

3n ,- 3n t=-+LIL 4 4

t = tr, 3r,

5n 5n , = -4 '-4 + 27r,

27.1. 71E + '

t = + 2r ... 2 ' 2 ' 2 Figure 4.20

Evalual

When ev calculatd

Most functions tive recir csc(71-78),

csc - 8

and enter

M

EXAM p

Functic R. sin 27r/ b. cot 1.5

Th ; cosine and secant functions are even.

— = cos t sec( —t) = sec t

, cosecant, tangent, and cotangent functions are odd.

cos(

Th ; sine

and

STUDY TIP

From the definition of periodic function it follows that the sine and cosine functions are peri-odic and have a period of 27r. The other four trigonometric functions are also periodic, and more will be said about that in Section 4.6.

STUDY TIP

When evaluating trigonometric functions with a calculator, remember to enclose all frac-tional angle measures in paren-theses. For instance, if you want to evaluate sin 0 for 0 = 7r/6, you should enter

SIN 6 0 ENTER

rather than (SIN) E) E) 6 ENTER

The first set of keystrokes yields the correct value of 0.5. The second set yields the incorrect value of 0.

Calculator Keystrokes Display

0.8660254

0.0709148

Mode

Radian

Radian

(SIN (112 30 TAN) 15 1=1

ENTER

ENTER


g utility in etric modes,

YlT ing settings, = 6.3,

ax = 1.5

= 1,

red equations Le graph. eature to rno nd the graph, values A do the x-Tresent? !,ast and great-

and y?

EC

pAPLE 3 Using the Period to Evaluate the Sine and Cosine

ecause 137r = + —7r, you have

6 6

137r 1 = sin( 27r + —1r) = sin —

7r = —. ;in 6 6 6 2

71T IT ecause — —

2 = — 47r + —

2, you have

777" IT ;OS(— —2 ) = cos( — 47r + = cos = 0.

2

en and Odd Trigonometric Functions

Recall from Section 1.2 that a function"' is even if f(- 0 = f(0, and it is odd if —f(t).

:de sin( —t) = — sin t

tan( —t) = — tan t c s c ( — = — csc t

cot(— = — cot t

x \

t= 0, zit, •

luating Trigonometric Functions with a Calculator When evaluating a trigonometric function with a calculator, you need to set the calculator to the desired mode of measurement (degrees or radians).

Most calculators do not have keys for the cosecant, secant, and cotangent functions. To evaluate these functions, you can use the x-1 key with their respec-tive reciprocal functions sine, cosine, and tangent. For example, to evaluate csc( r/8), use the fact that

IT 1 8 sin(7r/8)

niter the following keystroke sequence in radian mode.

0 7rE800 x-1 ENTER Display 2.6131259

KAMM 4 Using a Calculator

Eva

vnction

fl 21r/3

A 1.5

ir 5. t = -4

77r 7. t = -6

47r 3 9. t = -

3/r 11. t = -

2

6. t = -3

57r 8. t =

10. t = -57T 3

12. t = 7r

In Exercises 13-22, evaluate (if possible) the sine, cosine, and tangent of the real number.

13. t = -7r IT

4 14. t = -

3

37. sin t =

(a) sin(-

(b) csc(-0

39. cos(- = (a) cos t

(b) sec(-0

41. sin t =

(a) sinerr - 0

(b) sin(t +

38. sin(-0 =

(a) sin t

(b) csc t

40. cos t =

(a) cos(-0

(b) sec(-

42. cos t

(a) cos(7r - t)

(b) cos(t + 7r)


•

In Exercises 1-4, determine the exact values of the six trigonometric functions of the angle O.

1. 2.

3. 4.

In Exercises 5-12, find the point (x, y) on the unit cir-cle that corresponds to the real number t.

3 71" 21. t = - -

2 22. t = -27

In Exercises 23-28, evaluate (if possible) lit e si \ trigonometric functions of the real number.

3 7T 7T 23. t = -

4 24. t = 5

6

7/ 3 71 25. t = -

2 26.

7r 3 7T 27. t = - -3 28. t = - -

2

In Exercises 29-36, evaluate thectrigonometric hole. 2.50

2.75 29. sin 57r 30. cos 57r

3.00 31. cos--87r 97r

3 32.

sin-h-

3.25 33. cos(- 37) 34. sin (- 370

35. sin(--97r) 36. cos( - -87r) 4 3 3.75

3.50

4.00 In Exercises 37-42, use the value of the trigonometric function to evaluate the indicated functions.

Estimation and a strail trigonornett

53. (a) sin !

54. (a) sin (

Estimation and a straigl equation, wl

55. (a) sin t

56. (a) sin t

2.25

tion using its period as an aid.

In Exercises 43-52, use a calculator to evaluate tk

expression. Round to four decimal places.

7T 7T 7T 43. sin - ,-, 44. tan g

15. t = - -6 16. t = - -71- 4

45. csc 1.3 46. cot 1 17. t = 18. t = 47. cos(- 1.7) 48. cos(- 2.5)

4 3 49. csc 0.8 50. sec 1.8

117r S IT 19. t =

6 20. t = 51. sec 22.8 52. sin(- 0.9) 3

Asible) the six mber.

IT

6

IT

2

3 71-

mometri fun.

Estimation In Exercises 55 and 56, use the figure

and a straightedge to approximate the solution of the equation, where 0 t < 2/r.

55.(a) sin t = 0.25

56.(a) sin t = -0.75

A

1.75 - 1.50 2.00 1.25

0.75

0.50

0.25 ,6.25

/ I 1.2

6.00

5.75 3.75

4.00 5.50 4.25 5.25

4.50 4.75 5.00

(b) cos t = -0.25

(b) cos t = 0.75

1. 2.25

2.50

1.00

2.75

3.00

3.25

3.50

-1 ft 4

--.1111r ir:stiination In Exercises 53 and 54, use the figure :nct a straightedge to approximate the value of the trigonometric function.

(b) cos 2 53, (a) sin 5

54. (a) sin 0.75 (b) cos 2.5

le trigonometric tctions.

3 -t) = 8

in t

;se t 3 =

OS(-

.ec(- 0 4

- 5

:os(77- -

.os(t + 7r)

to evaluate th aces.

-2.5)

.8

-0.9)

FIGURE FOR 53-56

57. Electric Circuits The initial current and charge in the electrical circuit shown in the figure are zero. When 100 volts is applied to the circuit, the current is

I = 5e-21 sin t

if the resistance, inductance, and capacitance are 80 ohms, 20 henrys, and 0.01 farads, respectively. Approximate the current t = 0.7 seconds after the voltage is applied.


58. Electrical Circuits Approximate the current in the electrical circuit in Exercise 57 at t = 1.4 seconds after the voltage is applied.

59. Harmonic Motion The displacement from equilib-rium of an oscillating weight suspended by a spring is

y(t) = -1

cos 6t 4

where y is the displacement in feet and t is the time in seconds (see figure). Find the displacement when (a) t = 0, (b) t = and (c) t =

-Oft Oft

ft 4 4

Equilibrium Maximum negative Maximum positive displacement displacement

Simple Harmonic Motion

60. Harmonic Motion The displacement from equilib-rium of an oscillating weight suspended by a spring and subject to the damping effect of friction is

y(t) = -1 e t cos 6t 4

where y is the displacement in feet and t is the time in seconds. Find the displacement when (a) t = 0, (b) t = and (c) t =

61. Verify that cos 2t * 2 cos t by approximating cos 1.5 and 2 cos 0.75.

62. Verify that sin(ti + t2) * sin t1 + sin t2 by approxi-. mating sin 0.25, sin 0.75, and sin 1.

E(t)

4 ft


Synthesis

True or False? In Exercises 63 and 64, determine whether the statement is true or false. Justify your answer.

63. Because sin( -t) = -sin t, it can be said that the sine of a negative angle is a negative number.

64. tan a = tan(a - 67r)

65. Exploration Let (x1, y1) and (x2, y2) be points on the unit circle corresponding to t = t1 and t = ir - t1, respectively.

(a) Identify the symmetry of the points (x1, yi) and (x2, y2).

(b) Make a conjecture about any relationship between sin t1 and sin(7t - t1).

(c) Make a conjecture about any relationship between cos t1 and cos(7r - t1).

66. Exploration Let (x1, y1) and (x2, y2) be points on the unit circle corresponding to t = t1 and t = t1 + 7r, respectively.

(a) Identify the symmetry of the points (x1, y1) and (x2, y2).

(b) Make a conjecture about any relationship between sin t1 and sin(ti + 7r).

(c) Make a conjecture about any relationship between cos t1 and cos(ti + 7r).

67. Use the unit circle to verify that the cosine and secant functions are even and that the sine, cosecant, tangent, and cotangent functions are odd.

68. Think About It Because f(t) -= sin t is an odd function and g(t) = cos t is an even function, what can be said about the function h(t) = f(t)g(t)?

69. Think About It Because f(t) = sin t and g(t) = tan t are odd functions, what can be said about the function h(t) = f(t)g(t)?

Review

In Exercises 70-73, find the inverse of the one-to-one function f. Use a graphing utility to graph both l and f - IL in the same viewing window.

70. f(x) = (3x - 2)

71. f(x) = + 1

72. f(x) = - 4, x 2

73. f(x) = 2x 1'

x> -1 x +

In Exercises 74-77, sketch the graph of the ratioti function. Be sure to label all asymptotes. Use a grami41 ing utility to verify the graph.

74. f(x) = x

2x 3

75. f(x)

76. f(x) = x2 - 3x + 8 x - 2

x3 - 6x + x - 1 77' f(x) = 2x2 - 5x - 8

78. Average Cost The average cost (in dollars p

pound) of recycling a waste product x (in pounds)

C(x) x > 0.

Find the average cost of recycling x =-

= 450,000 + 5x

10,000 pounds, x = 100,000 pounds, and x 1,000,000 pounds. According to this model, what is the limit_ ing average cost as the number of pounds increasesi

79. Inflation If the inflation rate averages 4.5% over the next 10 years, the approximate cost C of goods or services t years from now is

C(t) = P(1.045)t

where P is the present cost. If the price of a AI! presently $69.95, estimate the price 10 years from now.

80. Population The population P of a city is P = 1 35,2000 318r, where t = 0 represents 1990 Use a graphing utility to graph the model and deter-mine when the population reaches 190,000. Verifi your answer algebraically.

81. Population The population p of a certain speciest years after it is introduced into a new habitat is giver by p(t) = 12001(1 + 3e-0 ).

(a) Determine the population size that was intro

duced into the habitat.

(b) Determine the population after 5 years.

(c) After how many years will the populatio 800?

x2 The Six 1

Our second tive. Conside

i n Figure 4.: hypotenuse,

e (the side

Side adj ace I

Figure 4.21

Using the le trigonometri

sine

In the follow such angles t

Let 0 be an. of the angle

sin ,=

csc 6 =

The abbrevi Sides of a ri

Opp t

hyp

Note that thO co-e-

sponclin!

4.3 • Right Triangle Trigonometry 303

of the L'ational s. Use a graph.

ng x = 10,000 = 1,000,000

hat is the hut. )unds increases?

ages 4.5% over )St C of goods or

(in dollars Per x (in pounds) is

5x + x — 6

?rice of a tire is !. 10 years from

of a city is epresents 1990. nodel and deter-190,000. Verify

certain species t / habitat is given

that was intro-

5 years.

e population be

The Six Trigonometric Functions

our second look at the trigonometric functions is from a right triangle perspec-

tive. Consider a right triangle, one of whose acute angles is labeled 0, as shown

in Figure 4.21. Relative to the angle 0, the three sides of the triangle are the

hypotenuse, the opposite side (the side opposite the angle 0), and the adjacent

side (the side adjacent to the angle 0). .

a)

fa. a)

-o

Side adjacent to 0

Figure 4.21

Using the lengths of these three sides, you can form six ratios that define the six trigonometric functions of the acute angle 0.

sine cosecant cosine secant tangent cotangent

In the following definitions it is important to see that 00< 0< 90° and that for such angles the value of each trigonometric function is positive.

Right Triangle Definitions of Trigonometric Functions Let 0 be an acute angle of a right triangle. Then the six trigonometric functions of the angle 0 are defined as follows.

sin 0 = opp opp hyp hyp adj

cos 0 = adj

tan 0 =

hyp hyp adj csc sec 0 = cot 0 = opp adj opp

The abb .eviations "opp", "adj", and "hyp" represent the lengths of the three sides of right triangle.

opp = the length of the side opposite 0 adj = the length of the side adjacent to 0 hyp the length of the hypotenuse

Note that the functions in the second row above are the reciprocals of the corresponding functions in the first row.

You Should Learn:

• How to evaluate trigonometric functions of acute angles

• How to use the fundamental trigonometric identities

• How to use a calculator to evaluate trigonometric func-tions

• How to use trigonometric functions to model and solve real-life problems


You can use trigonometric functions to model and solve real-life problems. For instance, Exercise 69 on page 312 shows you how trigonometric functions can be used to approximate the height of a buidling.


It

Georg Joachim Rhaeticas (1514-1576), the leading Teutonic mathematical astronomer of the 16th cemn was the first to define the trigonometric functions as ratio of the sides of a right triangle.

--

EXA M P

Use the sin 60°,

SoNtiOn Try using lengths o

opP

sin 6

and

cos 6

For 0 3

sin 30

and

cos 3

Because t trigonomet and 4.24.

sin 30°


Find the exact values of the six trigonometric functions of 0, as shown in Figure 4.22.

3

Figure 4.22

Solution By the Pythagorean Theorem, (hyp) 2 = (opp) 2 (adj) 2, it follows that

hyp = ./42 + 32 = -,/25 = 5.

So, the six trigonometric functions of 0 are

opp 4 sin 0= =

hyp 5

hyp 5 csc = =

opp 4

adj 3 cos =

hyp 5

hyp 5 sec O= =

adj 3

opp 4 tan 0 = =

adj 3

adj 3 cot 0 = =—.

opp 4


In Example 1, you were given the lengths of the sides of the right triangle, but not the angle O. A more common problem in trigonometry is finding the trigonomet-ric functions for a given acute angle 0. To do this, you can construct a right tri-angle having 0 as one of its angles.

opp 1 tan 45° ==—=l

adj 1

sin 60° =

Figure 4.23

In the box co mplemen definitions acute angle

sin(90°'

tan(90.

sec00

EXAMPLE 2 Evaluating Trigonometric Functions of 45°

Find the exact values of sin 45°, cos 45°, and tan 45°. sin 45°

Solution Construct a right triangle having 45° as one of its acute angles, as shown in Figure 4.23. Choose 1 as the length of the adjacent side. From geometry, you know that the other acute angle is also 45°. So, the triangle is isosceles and the length of the opposite side is also 1. Using the Pythagorean Theorem, you find the length of the hypotenuse to be

opp 1 adj 1 sin 45° = = = cos 45° = = =

hyp ,/2 2 hyp 2 2


EXAMPLE 3 Evaluating Trigonometric Functions of 300 and 600

Use the equilateral triangle shown in Figure 4.24 to find the exact values of

so 60° cos 60°, sin 30°, and cos 30°.

igyll/uisi iOni: the Pythagorean Theorem and the equilateral triangle to verify the

lengths of the sides given in Figure 4.24. For 0 = 60°, you have adj = 1,

opp and hyp = 2. Therefore,

sin 60° = h°P

ypP =

and adj 1

cos 60° = = _ hyp 2

For o 30°, adj = op p = 1, and hyp = 2. So,

• 30. = opp = 1

sin hyp 2

and

adj cos 30° =

hyp 2

Because the angles 30°, 45°, and 60° (7r/6, 7/4, and IT/3) occur frequently in trigonometry, you should learn to construct the triangles shown in Figures 4.23 and 4.24.

Sines, Cosines, and Tangents of Special Angles

In the box, note that sin 30° = = cos 60°. This occurs because 30° and 60° are complementary angles, and, in general, it can be shown from the right triangle definitions that cofunctions of complementary angles are equal. That is, if 0 is an acute angle, the following relationships are true.

/111

sin(90° — = cos 0

tan(90° — = cot 0 cot(90° — = tan 0

cos(90° — = sin 0

sec(90° — = csc 0 csc(90° — = sec 0

thaetict leading Inca' 16th ce ury, fine the ctions a atios ight tria,


sin 30°

sin 45°

sin 60°

Figure 4.24

7F 1 sin — = —

6 2 cos 30° = cos —

6 =

2

. IT -N/ 7T = sin — = cos 450 = cos — =

4 4 2 2

. 7T -• 7F 1 = sin --= 2 cos 60° = cos --=

3 3 2

7T tan 30° = tan =

6 3

'7T tan 45° = tan = 1

4

7T tan 60° = tan —

3 =

elect a number t and use our graphing utility to calculate (cos 02 + (sin 02. Repeat th experiment for other values and explain why the answet always the same. Is the resi, true in both radian and deg, modes?

EXA

Use tf.

a. cos

b. sec

of t is

It ee

Fundamental Trigonometric Identities

Reciprocal Identities

sin 0 = csc 0

1 csc 0 = .

sin 0

Quotient Identities

cos 0 = sec 0

1 sec 0 =

cos 0

1 tan 0 —

cot 0

cot 0 — tan 0

1 1

1

tan 0 — cos 0

• cos 0 cot 0 =

sin 0

sin 0

Pythagorean Identities

sin2 0 + c0s2 0 = 1

1 + tan2 0 = sec2 0

1 + cot 2 0 = csc2 0

Evalu To use degree in Sect

EXA

Use a

SONti Begin

5°

Then u

se

Appli Many triangl angle a are giv angle y

izontal depress

0.6

0.8


Trigonometric Identities In trigonometry, a great deal of time is spent studying relationships between trigonometric functions (identities).

Note that sin2 0 represents (sin 0)2 not sin(02); c0s2 0 represents (cos 0)2, not cos (02); and so on.

EXAMPLE 4 Applying Trigonometric Identities

Let 0 be an acute angle such that sin 0 = 0.6. Find the values of (a) cos 0 and (b) tan 0 using trigonometric identities.

Solution

a. To find the value of cos 0, use the Pythagorean identity

sin2 0 + c052 0 = 1.

So, you have

(0.6) 2 + c052 0 = 1

c0s2 0 = 1 — (0.6)2 = 0.64

cos 0 = = 0.8.

b. Now, knowing the sine and cosine of 0, you can find the tangent of 0 to be

sin 0 = —

0.6 = 0.75. tan 0 =

cos 0 0.8

Try using the definitions of cos 0 and tan 0, and the triangle shown in Figure 4.25, to check these results. Figure 4.25

111111m..-

and use YOUr

calculate Repeat this I [er values of t ie answe'• is s the res It and degree '4

V 0.6


EXAMPLE 5 Using Trigonometric Identities

Use trigonometric identities to transform one side of the equation into the other.

b. (sec 0 + tan 0)(sec 0 — tan 0) = 1 a, cos 0 sec 0 = 1

SoNtiOn

a. cos 0 sec 0 = 1

/ se6 0 = 1 \sec 0

1 = 1

Write original equation.

Reciprocal identity

Divide out common factor.

b. (sec 0 + tan 0)(sec 0 — tan 0) = 1 Write original equation.

sec2 0 — sea tan 0 + sec 0 tan 0 — tan2 0 = 1 Distributive Property

sec20 — tan20 = 1 Simplify.

1 = 1 Pythagorean identity

Evaluating Trigonometric Functions with a Calculator

To use a calculator to evaluate trigonometric functions of angles measured in degrees, first set the calculator to degree mode and then proceed as demonstrated

in Section 4.2.

EXAMPLE 6 Using a Calculator

Use a calculator to evaluate sec(5° 40' 12").

Solution Begin by converting to decimal form.

40 12 5° 40 ' 12" = 50 +

()° (3600)° = 5.67°

60)

Then use a calculator to evaluate sec 5.67°.

1 sec(5° 40 '12") = sec 5.67° = — 1.00492 cos 5.67°

STUDY TIP

Remember that throughout this text, it is assumed that angles are measured in radians unless noted otherwise. For example, sin 1 means the sine of 1 radian and sin 10 means the sine of one degree.

Applications Involving Right Triangles Many applications of trigonometry involve a process called solving right triangles. In this type of application, you are usually given one side of a right tri-angle and one of the acute angles and asked to find one of the other sides, or you are given two sides and asked to find one of the acute angles. In Example 7, the angle you are given is the angle of elevation, which denotes the angle from the hor-izontal upward to the object. In other applications you may be given the angle of depression, which denotes the angle from the horizontal downward to the object.

Angle of elevation

71.5°

x = 50 ft


EXAMPLE 7 Solving a Right Triangle

A surveyor is standing 50 feet from the base of a large tree, as shown in Figure 4.26. The surveyor measures the angle of elevation to the top of the tree as 71.5°. How tall is the tree?

In Exa

equatic

v011 nil

value of

can use

Figure 4.26

Algebraic Solution

From Figure 4.26, you can see that

opp

tan 71.5° =

adj

= Y

where x = 50 and y is the height of the tree. So, the height of the tree is

Numerical Solution

Use a graphing utility to create a table, as shown in Figure 4.27(a). The first col umn is the base of the triangle, x = 50. The second column is the height y, mid varies from 10 to 200. The quotient y/x is shown in the third column. Becaust tan 71.5 = y/x 2.9887, you can compare 2.9887 to the value of y/x in tht table. Because 2.9887 is between 2.8 and 3, the height y is between 140 and 150 Create a second table that shows the value of y/x when the values of y incr from 140 to 150, as shown in Figure 4.27(b). Because 2.9887 is between and 3, the height is between 149 and 150 feet. Repeating the process to cre table that shows values of y/x when the values of y increase from 149.1 to results in an estimate of the tree height of 149.4 feet.

EXAM

A skate

length. I

y = x tan 71.5°

50(2.98868)

149.4 feet.

Li L2 L3 50 100 2 50 110 2.2 50 120 2.4 50 130 2.6 50 140 NEM 50 150 3 50 160 3.2 L3(14)=2.8

(a)

Figure 4.27

Li L2 L3 0 0

0 0

0 0

0

Ifl tr, tn in in cn in

144 2.88 145 2.9 146 2.92 147 2.94 148 2.96 149 IIIIM 150 3

L3(10)=2.98

(b)

Figure 4.29

&Mull(' From the

rise

run

Which im

C 2

So, C

tan 0

Then,

You obtain

r EXAMPLE 8 Solving a Right Triangle

A person is 200 yards from a river. Rather than walk directly to the river, the per-son 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 4.28. ›,

o Solution 0 t-st From Figure 4.28, you can see that the sine of the angle 0 is

pp 200 = 1 sin = °- =

hyp 400 2.

Now, you should recognize that U = 30°.

Figure 4.28

(a). The first coi-e height y, Which column. Because Ale of y/x in the een 140 and 150, ues of y increase ' is between 2.98 rocess to create a :.om 149.1 to 150


example 8, you were able to recognize that 0 = 300 is the acute angle that

tis4e

5

the equation sin 0 = Suppose, however, that you were given the

equation sin 0 = 0.6 and asked to find the acute angle 0. Because

STUDY TIP

Calculators and graphing utilities have both degree and radian modes. As you progress through this chapter, be sure you use the correct mode.

—1

= 0.5000 and sin 45° = —1

0.7071 2

ou might guess that 0 lies somewhere between 30° and 45°. A more precise Yvalue of 0 can be found by using the inverse key on a calculator. To do this, you

can use the following keystroke sequence in degree mode.

.6 ENTER Display 36.8699

So. you can conclude that if sin 0 = 0.6, then 0 36.87°.


A skateboard ramp requires, a rise of one foot for each three feet of horizontal

length. In Figure 4.29, find the lengths of sides b and c and find the measure of 0.

Figure 4.29

Solution From the given specifications, you can write

rise 1 4 —run

= — = — 3

—ft

which implies that b = 12 feet. Using the Pythagorean Theorem, you can write

C2 = 42+122 Pythagorean Theorem

C2 = 160 Simplify.

C = 4,/173. Extract positive square root.

SO, c 4-117) 12.65 feet. To solve for 0, you can write

4 1 tan 0 = —12 =

Then, using the calculator keystrokes

ID 1 M 3

You obtain 0 18.43°.

ENTER

310 Chapter 4 0 Trigonometric Functions

CSC 0 -

sin 0

11' cos 0

tan 32.

5.

In Exercises 1-6, find the exact values of the six trigonometric functions of the angle 0 given in the fig-ure. (Use the Pythagorean Theorem to find the third side of the triangle.)

2.

4.

24

15

6. 6

18

12

In Exercises 7-10, find the exact values of the six trigonometric functions of the angle 0 for each of the triangles. Explain why the function values are the same.

7.

0

10.

In Exercises 11-18, sketch a right triangle corre-sponding to the trigonometric function of the acute

angle O. Use the Pythagorean Theorem to detern1 the third side and then find the other five trigon0niet. ric functions of 0.

11. sin 0 -= 12. cot 0 = 5

13. sec 0 = 4 14. cos 0 =

15. tan 0 = 3 16. csc 0 =

17. cot 0 = 18. sin 0 =

In Exercises 19-24, use the given function value and the trigonometric identities to find the indicated trigonometric functions.

1 19. sin 60° = , cos 60° = -2

(a) tan 600 (b) sin 300

(c) cos 30° (d) cot 60°

1 20. sin 30° = , tan 30° = - 3

(a) csc 30° (b) cot 600

(c) cos 300 (d) cot 30°

21. csc 0 = 3, sec 0 = 4

(b) cos 0 (a) sin 0

(d) sec(90° - (c) tan 0

tan 0 = 2-,/g 22. sec 0 = 5,

(b) cot 0 (a) cos 0

(c) cot(90° - (d) sin 0

23. cos a = -1

4

(b) sin a (a) sec a

(d) sin(90° - a) (c) cot a

24. tan 13 = 5

(a) cot p (b) cos p (d) csc p 13) (c) tan(90° -

In Exercises 25-32, use trigonometric identities

transform one side of the equation into the other.

25. tan 0 cot 0 = 1 26. csc 0 tan 0 sec 0

27. tan a cos a = sin a 28. cot a sin a = cos a

29. (1 + cos 0)(1 - cos 0) = sin2 0

Exercises

1.

3.

9.

8. Elk

tar

Iii Exercis(

don by ha'

33, (a) COS

tan

36.

((ba))

sin

(b) csc

In Exercisi function. F

391. (a) Sec 40. (a) tan

4. (a)

42. (a) co,.

43. (a) cot

44. (a) sec

45. (a) csc

46. (a) sec

In Exercis (0° < 0<9 a calculatc

47. (a) sir

48. (a)

49. (a)

CO

set

0. (a) tar

1. (a) Cs(

52. (a) co

(b) cos 0 = 0.0175

(b) cos 0 = 0.8746

(b) tan 0 = 0.4663

(b) cos 0 = 0.3746

53. (a) sin 0 = 0.8191

54. (a) cos 0 = 0.9848

55. (a) tan 0 = 1.1920

56. (a) sin 0 = 0.3746

cos 60° 34. (a) csc 30°

IT

-6 (b)

44.(a)

45.(a)

46.(a)

sec 1.25

csc 1

sec(_ 1) 2

sin 0 =

cos 0 = 2

sec 0 = 2

tan 0 =

csc 0 = 3

cot 0 = -3

(b) csc 0 = 2

(b) tan 0 = 1

(b) cot 0. = 1

(b) cos 0 =

(b) sin 0 = -2

(b) sec 0 =

20

61. Solve fory. 62. Solve for r.

25

75°

(b)

(b)

(b)

(b)

cos 65°

cot 71.5°

csc 48° 7'

sec 8° 50 ' 25"


30. (CSC A + cot Ccsc 0 - cot 0) = 1

Sill 0 cos 0 + - = csc 0 sec 0

31' cos 0 sin tan p + cot p

32. tan 13 m to deterimile we trigononiet

= ese2 p

35. (a) cot -7T 4

in Exercises 33-38, evaluate the trigonometric func-

tion bY hand.

33. (a)

7T (b) sin -

4 (b) cos 45°

57. Solve for y. 58. Solve for x.

7T 7T

37. (a) cos -6

38. (a) tan -3

(bf sec 600 (b) cot 30° 105

In Exercises 39-46, use a calculator to evaluate each function. Round your answers to four decimal places.

39.(a) sin 25°

40.(a) tan 18.5°

41.(a) sec 42° 12'

42.(a) cos 8° 50 ' 25"

IT 43.(a) cot -

16

In Exercises 47-52, find the value of 0 in degrees (0° < 0< 90°) and radians (0 < 0 < 7r/2) without using a calculator.

59. Solve forx.

38

60. Solve for r.

63. Height A 6-foot person walks from the base of a broadcasting tower directly toward the tip of the shadow cast by the tower. When the person is 132 feet from the tower and 3 feet from the tip of the tower's shadow, the person's shadow starts to appear beyond the tower's shadow.

(a) Draw a right triangle to represent the problem. Show the known quantities on the triangle and use a variable to indicate the unknown height of the tower.

(b) Write an equation involving the unknown.

(c) What is the height of the tower?

c identities to

the other.

= sec 0

1 a = cos a

- 0)

a) 47.(a)

48. (a)

49. (a)

50. (a)

51.(a)

52.(a)

5 _ 3 - 7

17 - 4

3

nction vain d the indica

In Exercises 53-56, find the value of 0 in degrees (00 < 0 < 900) and radians (0 < 0 < 7r/2) by using a calculator.

In Exercises 57-62, solve for x, y, or r, as indicated.

sm - 3

(b) csc 450

36. (a)

(b) tan -16

(b) cos 1.25

(b)

(b)

60 56

(x2, Y2),Ps%,

30°,' -

X

56 60

- A 1-• 4- 100 ft

w

0

cos

73.Geo'ne

Synthesis

True or F whether the answer.

74.sin 60° c

76.sin 450

77. cot 2 10°

78.Explora

(a) Use Rou

Il(b) Clas tions valu

(c) Fort funct funct


64. Length A 30-meter line is used to tether a helium-filled balloon. Because of a breeze, the line makes an angle of approximately 750 with the ground.

(a) Draw a right triangle to represent the problem. Show the known quantities on the triangle and use a variable to indicate the unknown height of the balloon.

(b) Use a trigonometric function to write an equa-tion involving the unknown.

(c) What is the height of the balloon?

65. 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 100 feet and sights to point C. From this sighting, it is determined that 0 = 58°. How wide is the river? Verify your result numerically using a graphing utility.

66. Distance From a 60-foot observation tower on the coast, a Coast Guard officer sights a boat in diffi-culty. The angle of depression of the boat is 4.5°. How far is the boat from the shoreline? Verify your result numerically using a graphing utility.

60 ft

67. Angle of Elevation A ramp 20 feet in length rises to a loading platform that is 31 feet off the ground.

(a) Draw a right triangle to represent the problem. Show the known quantities on the triangle and use a variable to indicate the unknown angle of elevation of the ramp.

(b) Use a trigonometric function to write an equa-tion involving the unknown.

(c) Find the angle of elevation of the algebraically.

(d) Use the table feature of a graphing uti approximate the angle of elevation numer

68. Duquesne Incline The track of the Dui Incline in Pittsburgh, Pennsylvania, is about 8 long, and the angle of elevation is 30°. The a speed of the cable cars is about 320 feet per r

(a) How high does the Duquesne Incline rise' (b) What is the vertical speed of the cable c

feet per minute)?

69. Jin Mao Building You are standing 75 from the base of the Jin Mao Building in Shi China. You estimate that the angle of elevatiot top of the building is 80°. What is the appro height of the building? Suppose one of your is at the top of the building. What is the d between you and your friend?

70. Machine Shop Calculations A steel plate has the form of one fourth of a circle with a radius of 60 cen. timeters. Two 2-centimeter holes are to be drilled the plate positioned as shown in the figure. Find the coordinates of the center of each hole.

71. Machine Shop Calculations A tapered shai diameter of 5 centimeters at the small end an centimeters long. If the taper is 3°, find the di d of the large end of the shaft.

rani

lily to ically

qUesne 00 feet Verae rlinute.

arS (j

Meteriflghj

to the ximate friends istance

t has a d is 15 ameter

72, Geome circle o construc perpend minal si actual m the poin to appr 20° angl


of the ralTID

wiling utility to

Lon numerically.

the Duquesne is about 800 feet

The average feet per minute. cline rise? ie cable cars on

ding 75 meters ing in Shanghai, f elevation to the the approximate a of your friends t is the distance

71 &and*, Use a compass to sketch a quarter of a

circle of radius 10 centimeters. Using a protractor,

construct an angle of 20° in standard position. Drop a

perpendicular from the point of intersection of the ter-minal 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

20° angle.

73.Geometry Repeat Exercise 72 using a 75° angle.

Synthesis


74.sin 60° csc 60° = 1 75. sec 30° = csc 60°

76. sin 45° + cos 45° = 1

77. cot2 10° — c5c2 10° = —1

78. Exploration

(a) Use a graphing utility to complete the table. Round your results to four decimal places.

0 0° 20° 40° 60° 80°

sin 0

cos 0

tan 0

(b) Classify each of the three trigonometric func-tions as increasing or decreasing for the table values.

(c) For the values in the table, verify that the tangent function is the quotient of the sine and cosine functions.

79. Exploration Use a graphing utility to complete the table and make a conjecture about the relationship between cos 0 and sin(90° — 0). What are the angles 0 and 90° — 0 called?

0 0° 20° 40° 60° 80°

cos 0

sin (90° — e)

80. Numerical Analysis A 3000-pound automobile is negotiating a circular interchange of radius 300 feet at a speed of s miles per hour. The relationship between the speed and the angle 0 (in degrees) at which the roadway should be banked so that no lat-eral frictional force is exerted on the tires is tan 0 = 0.672s2/3000.

(a) Use a graphing utility to complete the table.

10 20 30 40 50 60

0

(b) The speeds are incremented by 10 miles per hour in the table, but 0 does not increase by equal increments. Explain.

Review

In Exercises 81-84, sketch the graph of the equation and identify all x- and y-intercepts.

81. y = —x — 9 82. 2x + y = 10

83. —3x + 8y = 16 84. 12x — 7y = 22

In Exercises 85-88, determine the quadrant in which the angle lies.

85. 146° 86. — 290.8°

87. —310° 30' 88. — 122° 50'

eel plate has the radius of 60 cen-a to be drilled in figure. Find the

le.

iered shaft has a 111 end and is 15 ind the diameter

1011.


• How to evaluate trig(' lc functions of any angl

• How to use reference angles

metric

to evaluate trigonometric

Definitions of Trigonometric Functions of Any Angle

Let 0 be an angle in standard position with (x, y) a point on the terminal side of 0 and r = ..Jx2 +y2 0.

sin 0 = -r

tan 0 = -y, x 0

sec 0 = -x

-, x 0 0

cos 0 = ;

cot 0 = y 0

y 0

Given tan

Solution Note that gent is nel

tan 0

and the fac r

sin 0

• How to evaluate trigoncinchic functions of real numbers

You Should Learn It

You can use trigonometric li

ise

tions to model and solve real_life problems. For instance, Exercise 103 on page 321 shows you how trigonometric functions can be used to model the average daily temperature in a city.

Because r = ./x2 + y2 cannot be zero, it follows that the sine and cosine func-tions are defined for any real value of 0. However, if x = 0, the tangent and secant of 0 are undefined. For example, the tangent of 90° is undefined. Similarly, if y = 0, the cotangent and cosecant of 0 are undefined.


Let (-3, 4) be a point on the terminal side of 0. Find the sine, cosine, and tangent of 0.

(-3,4)

—3 —2 —1

Figure 4.30

Solution

Referring to Figure 4.30, you see that x = -3, y = 4, and

r= ± y2 = (_3)2 ± 42 = = 5.

y 4 x 3 y 4 So, you have sin 0 = - = -, cos 0 = - = - - and tan 0 = - =

' x 3.

sec 0.

EXAMPI

Evaluate ti and 37r/2.

Solution To begin, ( 4.32. For e

sin 0 =

• IT sin -2

Sill 77

. 37; sin

2

Introduction In Section 4.3, the definitions of trigonometric functions were restricted to acute angles. In this section, the definitions are extended to cover any angle. If 0 is an acute angle, the definitions here coincide with those given in the previous section.

7411111 r1.111 You Should Le

The s igns easily frOl

followS ti

Fi

( ge

\n: illb

i 1 A P

functions

0 < 0 < 2

• x > 0 y > 0

s, • x > 0

• y < 0

< 0 < 2

2

Figure 4.32

4.4 • Trigonometric Functions of Any Angle 315

Learn:—."474' .te trigonometric ay angle ference angles ;onometric

le trigonometric ;al numbers I

Learn It:

aometric func-1 solve real-life :ance, Exercise ;hows you how :lions can be average daily ity.

ns of the trigonometric functions in the four quadrants can be determined file

sig the definitions of the functions. For instance, because cos 0 = xlr, it

c3s1lY fr°11° that cos 0 is positive wherever x> 0, which is in Quadrants I and IV. mo

ws --- (Remember, r is always positive.) In a similar manner you can verify the results

wn in Figure 4.31.


--5

and cos 0 > 0, find sin and sec 0. sin 0: - Given • 4 cos 6: —

tan 0: +

SolUtiOn Note that 0 lies in Quadrant W because that is the only quadrant in which the tan-

gent is negative and the cosine is positive. Moreover, using

tan 0- x 4

trid the fact that y is negative in Quadrant IV, you can let y 16 25 = and you have

_ =

Y -5 sin 0 = Exact value

r

- 0.7809 Approximate value

Quadrant II sin 9: + cos 8: — tan 8: —

= - 5 and x = 4. So,

n —2

< <

x < 0 y > 0 s

x < 0 . y < 0 •

< 0it < 2

Quadrant IV sin 0: — cos 9: + tan 8: —

Quadrant I sin 61: + cos 0: + tan 0: +

Quadrant III

li• sec 0

---- 1.6008. Approximate value

r - - Exact value

x 4

Figure 4.31

EXAMPLE 3 Trigonometric Functions of Quadrant Angles

Evaluate the sine and cosine functions at the four quadrant angles 0, 7r/2, IT, and 37r/2.

Solution To begin, choose a point on the terminal side of each angle, as shown in Figure 4.32. For each of the four given points, r = 1, and you have the following. 2 (0, 1)

Sill -y

=-0

=O cos 0 = -r

= -1

= 1 x 1

r 1

= y = 1 = IT X =

0 = o COS = - —

r 1 2 r 1

y 0 x -1 cos IT = —

r r 1 1

37r x 0 cos —

2 = -

r = = 0

sin -7r 2

sin IT •

. 37r y — 1 sin — 2 r 1

(x, y) = (1,0)

(x, y) = (0,I)

(x, y) =(-1,0)

(x, y) = (0, — 1)

(0,-l)

0

Definition of Reference Angle

Let 0 be an angle in standard position. Its reference angle is the acute angle 0' formed by the terminal side of 0 and the horizontal axis.

I I I 11

I I I "1


Reference Angles

The values of the trigonometric functions of angles greater than 900 (or less than 0°) can be determined from their values at corresponding acute angles called ref-erence angles.

A computer simulation to aco , concept appears in the Interacii ROM and Internet versions of II

Trig on( To see I. c„nsider i nition. Y

sin

For the r

sin f

and

tan

So, it fo

true for cases, th

0 lies.

lam To find

1. Det 2. Dec

ate

By usin section, instance values o lowing angles a

Trigonom

0 (deg

0 (rad

sin

cos 0

ny

let

Figure 4.33 shows the reference angles for B in Quadrants II, III, and IV.

Quadrant II

Reference angle: 0'

Reference angle: 0'

Reference angle: 0'

Quadrant IV Quadrant

0' = rc— 0 (radians) 0' = 180° — 0 (degrees)

Figure 4.33

= 27r— 0 (radians) 0' = 360° — 0 (degrees)

0' = 0— rr (radians) 0' = 0— 180° (degrees)

EXAMPLE 4 Finding Reference Angles

Find the reference angle 0'.

a. 0 = 300° b. 0 = 2.3 c. 0 = —135°

Solution

a. Because 300° lies in Quadrant IV, the angle it makes with the x-axis is

0' = 360° — 300° = 60°. Degrees

b. Because 2.3 lies between 7112 1.5708 and ir 3.1416, it follows that it is in Quadrant II and its reference angle is

0' = 7r — 2.3 0.8416. Radians

First, determine that — 135° is coterminal with 225°, which lies in Quadrant III. So, the reference angle is

0' = 225° — 180° = 45°. Degrees

Figure 4.34 shows each angle 0 and its reference angle 0'.

C.

225° and —135° are coteriminal.

V

Figure 4.34

Figure 4.35

Library of Functions The trigonometric functions are important for modeling periodic behavior, such as business cycles, planetary orbits, pendulums, and light rays. These functions play a prominent role in calculus.

Consult the Library of Functions Summary inside the front cover for a description of the trigonometric functions.

STUDY TIP

Learning the table of values at the left is worth the effort. Doing so will increase both your efficiency and your confi-dence. Here is a pattern for the sine function that may help you remember the values.

0 0° 30° 45° 60° 90°

sin 0

2 2 2 2 2

Reverse the order to get cosine values of the same angles.

= 60°

2.3

' and —135° ;oterminal.

—135°


Trigonometric Functions of Real Numbers

see how a reference angle is used to evaluate a trigonometric function,

lconsider

the point (x, y) on the terminal side of 0 as shown in Figure 4.35. By def-

inition, you know that

sill = Y1: and tan 0 = Y.

F t“ he right triangle with acute angle 0' and sides of lengths and I y I, you have ot

•sin 0' oPP

= I YI

hyp r

to accompan, Interactive CI ions of this te,

oPP I YI tan 0 = =

adi Ix

So, it follows that sin 0 ancrsin 0' are equal, except possibly in sign. The same is

hue for tan 0 and tan 0' and for the other four trigonometric functions. In all

cases, the sign of the function value can be determined by the quadrant in which

0 lies.

toting Trigonometric Functions of Any Angle

To find the value of a trigonometric function of any angle 0:

1.Determine the function value for the associated reference angle 0'. 2. Depending on the quadrant in which 0 lies, prefix the appropri-

ate sign to the function value.

By using reference angles and the special angles discussed in the previous section, you can greatly extend the scope of exact trigonometric values. 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. For convenience, the fol-lowing table gives the exact values of the trigonometric functions of special angles and quadrant angles.

Trigonometric Values of Common Angles

0 (degrees) 0° 30° 45° 60° 900 180° 270°

0 (radians) 0 IT —6

IT 4

IT

3 IT

2 IT

3 IT

T

sin 0 _

0 1

2

i. 0 —1 2 2

c056 _

1

—2

0 —1 0 2 2

tan f) 0

1 -,,i3- Undef. 0 Undef. 3

(b) (a) Figure 4.36


EXAMPLE 5 Trigonometric Functions of Nonacute Angles

Evaluate the trigonometric functions.

117r a. cos b. tan( — 210°) c. csc

3 4

Solution a. Because 0 = 47r/3 lies in Quadrant III, the reference angle is

0' = (47r/3) — IT = 7r/3, as shown in Figure 4.36(a). Moreover, the cosine is negative in Quadrant III, so

47r „ IT cos = ) cos —

3

1 = 2'

b. Because —210° + 3600 = 1500, it follows that —210° is coterminal with the second-quadrant angle 150°. Therefore, the reference angle is

= 180° — 150° = 30°, as shown in Figure 4.36(b). Finally, because the tangent is negative in Quadrant II, you have

tan( — 210°) = (— ) tan 30°

= — 3

c. Because (117r/4) — 27r = 37r/4, it follows that 117r/4 is coterminal with the second-quadrant angle 3,7/4. Therefore, the reference angle is 0' = Ir — (37r/4) = 7r/4, as shown in Figure 4.36(c). Because the cosecant is positive in Quadrant II, you have

117r „ IT csc —

4 = +) csc —

4

1

sin(7/4)

=

The fundamental trigonometric identities listed in the previous section (for an acute angle 0) are also valid when 0 is any angle in the domain of the function.

(c)

EXAM

Jet L:0'; (b) tan v

500 0

a. Usin

Beca

b. Usin

tan

You can example.

EXAM

a. Use a

b. Use a

Solutioi

Funct

a. cot 41 sin(—

b. To so folio

Eqt

tan e

The di III (tdi


EXAMPLE 6 Using Trigonometric Identities

Let 0 be an angle in Quadrant such that sin =- I.. Find (a) cos 0 and

0)) tan o by using trigonometric identities.

SeihsT n the Pythagorean identity sin2 0 + c0s2 U = 1, you obtain

(1 2 + cos2 = 1

)

cos2 0 = 1 — —1

= 9 9

Because cos U < 0 in Quadrant II, you can use the negative root to obtain

2-12: cos 0 — — =

3 •

. Using the trigonometric identity tan 0 = sin 0/cos 0, you obtain

1/3 tan 0 =

1

2,/2

= — 4

You can use a calculator to evaluate trigonometric functions, as shown in the next example.

EXAMPLE 7 Using a Calculator

a. Use a calculator to evaluate cot 4100 and sin(— 7).

b. Use a calculator to solve tan 0 = 4.812, 0 27r.

Solution

Function Mode

a. cot 4100 Degree (1) TAN 410 M sin( — Radian Dl 1(- ))7 M SIN

Display

0.8390996

— 0.6569866

Graphing Calculator Keystrokes

b. To solve the equation tan 0 = 4.812, you can use the inverse tangent key, as follows.

Equation Mode Graphing Calculator Keystrokes Display

tan U =- 4.812 Radian 4.812 ENTER 1.365898912

The angle 0 1.366 lies in Quadrant I. A second value of U lies in Quadrant III (tangent is positive) and is

0 = ir + 1.366 4.507.

ThL Interactive CD-ROM and Internet

versions of this text offer a built-in graph-ing calculator, which can be used with the Examples, Explorations, and Exercises.

ENTER

ENTER

TAN-1

•

Set your graphing utility to degree mode and enter tan 90. What happens? Why? Now se your graphing utility to radian mode and enter tan(ir/2). Explain the graphing utility's answer.


AlLEJ(ercises In Exercises 1-4, determine the exact values of the six trigonometric functions of the angle O.

1. (a) y (b)

(4,3)

2. (a)

(b)

In Exercises 5-12, the given point is on the terminal side of an angle in standard position. Determine the exact values of the six trigonometric functions of the angle.

In Exercises 19-26, find the values of the six metric functions of O.

rig°11o.

Function Value Constraint

19. sin 0 = 0 lies in Quadrant

20. cos 0 = 0 lies in Quadrant

21. tan 0 = sin 0 < 0

22. csc 0 = 4 cot 0 < 0

23. sec 0 = - 2 0 0 71"

7T 37r 24. cot 0 is undefined. - < 0 -

2 - 2

25. sin 0 = 0 sec 0 = - 1

26. tan 0 is undefined. csc 0 = -1

In Exercises 27-30, the terminal side of 0 lies on the given line in the specified quadrant. Find the values of the six trigonometric functions of O.

Line Quadrant

27. y = -x II

28. y = III

29. y = 2x III

30. 4x + 3y = 0 IV

In Exercises 31-38, evaluate the trigonome tion of the quadrant angle.

31. sec 32. tan -7 33. 2

34. csc 7r. 35. sec 0 36.

ic futie.

7T Cot -

2

csc - 2

37. cot IT 38. csc - 5. (7, 24) 6. (8, 15) 7. (5, - 12) 2

8. (-24, 10) 9. (-4, 10) 10. (-5, - 6) In Exercises 39-50, find the reference angle 0'

11. (- 2, 9) 12. (6, -14)

sketch 0 and 0' in standard position.

In Exercises 13-18, state the quadrant in which 0 lies.

13. sin 0 < 0 and cos 0 < 0

14. sin 0> 0 and cos 0 > 0

15. sin 0> 0 and tan 0 < 0

16. sec 0> 0 and cot 0 < 0

17. cot 0 > 0 and cos 0 > 0

18. tan 0> 0 and csc 0 < 0

39. 0 -= 208° 40. 0 = 322°

41. 0= -245° 42. 0= - 145°

43. 0 = _2920 44. 0 = - 950

11 71" 17 7T 45. 0 = 46. 0 =

3 6

47. 0 = 3.5 48. 0 = 4.8

49. 0 = -3.68 50. 0 = - 1.72

gent 0

51. 21

53. - 55.

5 57. 3

59. -

11 61. 4

63. -

In Exe trigon decim

65. sin

67. tan

69. co

71. sec

73. sin

75. tan

77. csc

In Exe Give y radian calcpla

79. (a)

80. (a)

81. (a)

82. (a)

83. (a)

84. (a)

Exercises 51-64, evaluate the sine, cosine, and tan-

of the angle without using a calculator. gel'

51. 225° 53. -750°

55. -2400

51T 57. 3

59.

11'ff 61. 4

77r 63.

in Exercises 65-78, use a calculator to evaluate the

trigonometric function. Round you answer to four

decimal places.

the six trigs:100.

traint

uadrant II. uadrant III.

3'n-

2

1

1

52. 300°

54. _4950

56. -330°

7r 58. -4

7r 60. -

1 62.

07

3

2 64.

07

3

65. sin 100

67. tan 240°

69. cos(- 110°)

71. sec( - 280°)

73. sin 0.65

75. tan -7 9

77. csc 9

66. sec 225°

68. csc 3300

70. cot( - 220°)

72. sin( - 195°)

74. sin(- 0.65)

tan(9

1573- 78. co

s 14 )

76.

80. (a) cos 0 =


In Exercises 85-96, use a calculator to approximate two values of 0 (0° 0 < 360°) that satisfy the equa-tion. Round your answers to two decimal places.

cof 0 lies on the id the values of

nometric func-

7r 33. cot -

2

37 36. csc -

2

angle 0' and

22°

145°

95°

77r 5

8

1.72

In Exercises 79-84 find two solutions of the equation. Give your answers in both degrees (0° 0 360°) and radians (0 5_ 0 5 27r). Do not use a calculator.

79. (a) sin 0 = (b) sin 0 - - 21

2

2/ csc 0 = (b) cot 0 = - 1

3

csc 0 = - (b) csc 0 = 2

2-13- 1 sec 0 = (b) cos 0 =

2 3

cot 0 = - (b) sec 0 =

85. sin 0 = 0.8191 86. cos 0 = 0.8746

87. tan 0 = 0.6524 88. cot 0 = 0.7521

89. sec 0= - 1.2241 90. csc 0 = -1.0038

91. sin 0 = -0.4793 92. tan 0 = -2.1832

93. cos 0 = 0.9848 94. sin 0 = 0.0175

95. tan 0 = 1.192 96. cot 0 = 5.671

In Exercises 97-102, find the indicated trigonometric value in the specified quadrant.

Function Quadrant Trigonometric Value

97. sin 0 = - IV cos 0

98. cot 0 = -3 II sin 0

99. tan 0 = III sec 0

100. csc 0 = -2 IV cot 0

101. cos 0 = I sec 0

102. sec 0 = - III tan 0

103. Average Temperature The average daily temper-ature T (in degrees Fahrenheit) for a certain city is

T = 45 - 23 cos[ 21T365 (t 32)]

where t is the time in days, with t = 1 correspond-ing to January 1. Find the average daily tempera-tures on the following days.

(a) January 1

(b) July 4 (t = 185)

(c) October 18 (t = 291)

104. Sales A company that sells seasonal products forecasts monthly sales over a 2-year period to be

S = 23.1 + 0.442t + 4.3 sin -771 6

where S is measured in thousands of units and t is the time (in months), with t = 1 representing January 2000. Estimate sales for the following months.

(a) February 2000 (b) February 2001

(c) September 2000 (d) September 2001

(b) cos 0 = - 2

81. (a)

82. (a)

83. (a)

84. (a)

114. y = 3x+2

116. y = ln(x + 1

113. y = 2x-1

115. y = ln(x - 1)


105. Distance An airplane, flying at an altitude of 6 miles, is on a flight path that passes directly over an observer. If 0 is the angle of elevation from the observer to the plane, find the distance from the observer to the plane when (a) U -= 300, (b) 0 = 90°, and (c) 0 = 120°.

6 mi

Synthesis


106. sin 151° = sin 29°

11/r) 107. csc(- 7

6) = csc( 6

108. _cot() = cot( --4

109. Conjecture


0 00 20° 40° 60° 80°

sin 0

sin(180° - 0)

(b) Make a conjecture about the relationship between sin 0 and sin(180° - 0).

110. Conjecture


0 0 0.3 0.6 0.9 1.2 1.5

37T ) cos--- 0

- sin 0

(b) Make a conjecture about the relationship 3 7r

between cos (-2

- 0) and - sin 0.

111. Explain how reference angles are used to rind value of trigonometric functions of any angle 0

112. Writing Consider an angle in standard I s with r = 12 centimeters, as shown in the fi Write a short paragraph describing the chang the magnitudes of x, y, sin 0, cos 0, and tan 0 increases continually from 0° to 90'.

„

Review

In Exercises 113-116, sketch the graph the ft Identify any intercepts and asymptotes.

In Exercises 117-120, solve the equation. Round y answer to three decimal places.

4500 118.

+ e = 5(

4 2x

120. ln,/x + 10

In Exercises 121-124, sketch a right triangle corn, sponding to the trigonometric function of the acute angle O. Use the Pythagorean Theorem to determine the third side and then find the other five trigonomet• ric functions of O.

8 121. cot 0 = 122. sec 0 = 8

7 2 123. tan 0 = 124. cos 0 = 4

24

=1

117. 43 -x = 726

119. ln x = -6

12114ffigiu '01‘101 mmen,

I III V111"11:7T1

4.5 • Graphs of Sine and Cosine Functions 323

used to find ih : any angle 0, tan

.dard posnio,, rn Ifl the

figur:

g the change 1, and tan 0 as I°.

1 = 8

Basic Sine and Cosine Curves

this section you will study techniques for sketching the graphs of the sine and Di osine functions. The graph of the sine function is a sine curve. In Figure 4.37,

black portion of the graph represents one period of the function 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 func-

tion is shown in Figure 4.38. To produce these graphs with a graphing utility,

make sure you have set the graphing utility to radian mode.

Recall from Section 4.2 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, 1], and each function has a period of 277-. Do you see how this information

is consistent with the basic graphs given in Figures 4.37 and 4.38?

Range: < y

Figure 4.37

Range: —I 5.y 1

Figure 4.38

The table below lists key points on the graphs of y = sin x and y = cos x.

X 0

IT IT 71-

IT

—3-3 7r 37r

T ' 3IT T 27T

sin x o 1 7 2

,ii .,/ i '

,ii 0 — 1 0

2 2 2

cos x

- 1

2

_

0 1 2 2 2

di the function. S.

3x +2

ln(x + 1)

mi. Round your

)0 = 50 t

x + 10 =1

triangle corre-m of the acute m to determine lye trigonomet-

(x, y)

You Should Learn:

O How to sketch the graphs of basic sine and cosine functions How to use amplitude and period to help sketch the graphs of sine and cosine functions

o How to sketch translations of graphs of sine and cosine functions

• How to use sine and cosine functions to model real-life data


You can use sine and cosine functions to model real-life data. For instance, Exercise 83 on page 332 shows you how a sine function can be used to model the monthly precipitation for Seattle, Washington.

Note from Figures 4.37 and 4.38 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 function is odd whereas the Cosine function is even.


To sketch 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. See Figure 4.39

Arnplit

in the re'

a. b, • al

Intercept (Lt 1 2 '

Maximum Intercept Minimum Intercept Intercept Minimum Intercept Maximum

(0, 1) Yaximum = x (27r, 1) and

Y

A quick nves ti gat

The cons vertical s

and if la

y a sin value of

y = a sin

Defini

The am between

Am

37r _ 2'

(27r, 0) Full

period

Figure 4.39

EXAMPLE 1 Using Key Points to Sketch a Sine Curve

Sketch the graph of y = 2 sin x on the interval H 7T, 477-1.

Solution Note that

y = 2 sin x = 2(sin

indicates that the y-values for the key points will have twice the magnitude of the graph of y = sin x. Divide the period 2T- into four equal parts to get the key points for y = 2 sin x.

(0,0), (- 2), (i3O), — 2), 2

and (2i3O)

By connecting these key points with a smooth curve and extending the curve in both directions over the interval [— 7T, 4,7r], you obtain the graph shown in Figure 4.40. Use a graphing utility to confirm this graph. Be sure to set the graphing utility to radian mode.

(0, 0) Quarter Half period period

Period: 27r Three-quarter period

Quarter period

Period: 27r

an connect the points on the unit circle with their corresponding points on the sin curve by pressing ENTER Discuss the relationship that is illustrated.

the program SINE How given on our website, college.hmco.com, into your graphing utility. This program simultaneously draws the unit circle and the corresponding points on the sine curve, as shown below. After the circle and sine curve are drawn, you

Half period

Three-quarter period

Full period

Solution a. Becaus

minim get the

(0,

b. A simil Points i

On the sa 1

a. y — 2

EXAMP

0D,

The gra' utility t

Figure 4.40

Full period 411

xter

1 1111%

SINES 10

into your us program-ws the unit-;spondii .:urve, as r the cip drawn, you ints on r.le ir its on the s ENTER

aship that

.19

.19

e a graphing utility to graph = a sin x, where a = 0.5, 1, 2, d 3. How does the value of a

affect the shape of the graph? What happens to the graph if a

negative?

STUDY TIP

When using a graphing utility to graph trigonometric functions, pay special attention to the viewing window you use. For instance, try graphing y = [sin(104]/10 in the standard viewing window in radian mode. What do you observe? Use the zoom feature to find a viewing window that displays a good view of the graph.

y

--3 CO X

Figure 4.41


Amplitude and Period of Sine and Cosine Curves

b th, rest of this section you will study the graphic effect of each of the constants

c, and d in equations of the forms

+— d a sin(bx — -- I •

d + a cos(bx — c).

A qui review of the transformations studied in Section 1.3 should help in this ck investigation.

The constant factor a in y = a sin x acts as a scaling factor—a vertical stretch or

vertir•al shrink of the basic sine curve. If la > 1, the basic sine curve is stretched,

and if I a I <1, the basic sine curve is shrunk. The result is that the graph of

v , a sin x ranges between — a and a instead of between —1 and 1. The absolute

value of a is the amplitude of the function y = a sin x. The range of the function

_1, ..-.. a sin x is —a y ._. a.

finition of Amplitude of Sine and Cosine Curves

The amplitude of y = a sin x and y = a cos x represents half the distance between the maximum and minimum values of the function and is given by

Amplitude = 'al.

EXAMPLE 2 Scaling: Vertical Shrinking and Stretching

On the same coordinate axes, sketch the graph of each function. 1

= — cos x 2

Solution a. Because the amplitude of y = cos x is 1, the maximum value is and the

minimum value is — 1. Divide one cycle, 0 x 27r, into four equal parts to get the key points

(o, —2), (172, o),

1

b. A similar analysis shows that the amplitude of y = 3 cos x is 3, and the key points are

(0 3) (IT2 13)

The graphs of these two functions are shown in Figure 4.41. Use a graphing _I utility to confirm these graphs. I.

a. y b. y = 3 cos x

(,o), and (2

7r, —2). 1

(7r, —3), and (27r, 3).


You know from Section 1.3 that the graph of y = —f(x) is a reflection 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 4.42.

Because y = a sin x completes one cycle from x = 0 to x = 27r, it follows that y = a sin bx completes one cycle from x = 0 to x = 2i-/b.

Period of Sine and Cosine Functions Let b be a positive real number. The period of y = a sin bx and y = a cos bx is given by

27r Penod = —b .

Note that if 0< b <1, the period of y = a sin bx is greater than 27r and rep-resents a horizontal stretching of the graph of y = a sin x. Similarly, if b> 1, the period of y = a sin bx is less than 277- and represents a horizontal shrinking of the graph of y = a sin x. If b is negative, the identities sin( — x) = — sin x and cos( —x) = cos x are used to rewrite the function.

EXAMPLE 3 Scaling: Horizontal Stretching

x Sketch the graph of y = sin

Solution The amplitude is 1. Moreover, because b = the period is

27r 27r = = 47r.

Now, divide the period-interval [0, 47r] into four equal parts with the values IT,

27r, and 37r, to obtain the key points on the graph

(0,0), (7r, 1), (27r, 0), (37r, — 1), and (47r, 0).

The graph is shown in Figure 4.43 Use a graphing utility to confirm this graph.

MEMIIMMEMEMMINIMEIMMINIMME MIMMUMMEMEMEMMEOEMMEMMEM MMEMMINUMEMMEMEMMINIMMEMM 1111111111MAIMEMMEWMUMMEMERME UMW AMINIMMOMMMUMEMO MOO

MEMMOMMEINEMOMM MEM MENAMMEMERIV M inied111111111111 MEMMEEEE EEEMMEM MMEMMINIM PerqoffiMMIONIMMINI Figure 4.43

-4.71 Rh A

Figure 4.42

, 1,

Use a graphing utility t(_ grl y = sin bx, where b = 0.5

ph

and 2. How does the value of affect the graph?

This im

y a Si

TranS1

om

creates

Cpa r a si

The le ing the

The gr

charact

Graf

A

1

In general, to divide a period-interval into four equal parts, successively add "period/4," starting with the left endpoint of

the interval. For instance, for the

period-interval [— 7r/6, ir/21 of length 27/3, you would successively add

27r/3 _ IT

4 6

to get —17-16, 0, 7r/6, 7r/2 7r/2.

STUDY P4

, and

You see t atheg

duces th

EXAM

Analyze

Ake& The amp tions

(7r

se a graphing utility to

Use a viewing window in which —1.6 x 6.3 and —2 5_ y 2. How does the value of c affect the graph?

1 ---

•

il. y = sin(x + c)

here c = — 7r/ 4, 0, and 7714.

( T!1

le a period-!qual parts, period/4," ;ft endpoint of stance, for the 7r/6, 7r/2] of would

/6, 7r/3, and

Graphical Solution

Use a graphing utility set in radian mode to graph y = (1/2) sin(x — 7r/3), as shown in Figure 4.44. Use the minimum, maximum, and zero or root features of the graphing utility to approximate the key points (1.047, 0), (2.618, 0.5), (4.189, 0), (5.760, —0.5), and (7.330, 0).

7.85 1.57

= sin (x — 5-) 2

Zero X=1._0471976 Y=0

Figure 4.44


= —3 cos

ility to aph b = O. 1,

he value of b

Transiations of Sine and Cosine Curves

stant c in the general equations Tile con

y a sin(bx — and y = a cos(bx —

creates horizontal translations (shifts) of the basic sine and cosine curves.

,ornparing y = a sin bx with y = a sin(bx — c), you find that the graph of

a sin(bx c) completes one cycle from ,bx — c = 0 to bx — c = 277-. By

olving for x, you can find the interval for one cycle to be

Left endpoint Right endpoint

< X < b b b

Period

This implies that the period of y = a sin(bx — c) is 277-/b, and the graph of

y = a sin bx is shifted by an, amount c/b. The number c/b is the phase shift.

s of Sine and Cosine Functions

The graphs of y = a sin(bx — c) and y = a cos(bx — c) have the following characteristics. (Assume b > 0.)

Amplitude = la I Period = 27r/b

The left and right endpoints of a one-cycle interval can be determined by solv-ing the equations bx — c = 0 and bx — c = 27r.

-

EXAMPLE 4 Horizontal Translation

IT Analyze the graph of y = sin(x —

2 3

Algebraic Solution

The amplitude is and the period is 27r. By solving the equa-tions

IT and x — —

3 = 2ir

IT X = —

3

you see that the interval [7r/3, 77r/3] corresponds to one cycle of the graph. Dividing this interval into four equal parts pro-duces the following key points.

0), ( -5 1 T C ITT— 3 6 2 3

7 Ir x =

(117r 1)

6

Mathel

and .

in„ eleCtr

EXAMF

Throughc

The table

= —3 cos(274

—4

Figure 4.45


EXAMPLE 5 Horizontal Translation

Use a graphing utility to analyze the graph of y = —3 cos(2 7rx + 47r).

Solution The amplitude is 3 and the period is 27r/27r = 1. By solving the equations

27rx + 47r = 0 and 2.irx + 47r = 27r

27rx —41T 27rx = —21T

x = — 2 x = — 1

you see that the interval [-2, —1] corresponds to one cycle of the graph. Dividing this interval into four equal parts produces the key points

(-2, —3), (— 7/4, 0), (— 3/2, 3), (— 5/4, 0), and (-1, —3).

The graph is shown in Figure 4.45.

The final type of transformation is the vertical translation 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 d units downward for d < 0. In other words, the graph oscillates about the horizontal line y = d instead of about the x-axis.

EXAMPLE 6 Vertical Translation

Use a graphing utility to analyze the graph of y = 2 + 3 cos 2x.

Solution The amplitude is 3 and the period is IT. The key points over the interval [0, 7r] are

(0, 5), (77./4, 2), (7r/2, —1), (37r/4, 2), and (7r, 5).

The graph is shown in Figure 4.46. Compared with the graph off (x) = 3 cos 2x, the graph of y = 2 + 3 cos 2x is shifted upward two units.

A computer animation of this e appears in the Interactive CD-R Internet versions of this text.

iThe a graphing utility to graph y = d + sin x, where d = —2, 1, and 3. How does value of d affect the graph?

=2+3 cos 2x

y Wept

a. Use a

b. Find t

c. A boat in the •

S011itiO a. Begin

or cosi

The dit is twic

The co and mi

—1.57 which midnig becaus you ca

EXAMPLE 7 Finding an Equation for a Graph

Find the amplitude, period, and phase shift for the sine function whose graph is shown in Figure 4.47. Write an equation for this graph.

Solution The amplitude for this sine curve is 2. The period is 27r, and there is a right phase shift of 7r/2. So, you can write

y = 2 sin(x — 2 •

In Example 7, you can find a cosine function with the same graph. The amplitude for this cosine curve is 2. The period is 27r, and there is a right phase shift of 7r. So, you can write

y = 2 cos(x —

Figure 4.46

—1.57

—3

Figure 4.47

3

3

• • • 4

• 3 •

• 2


loathematical Modeling

sone and cosine functions can be used to model many real-life situations, includ-

- I,jIl

tric currents, musical tones, radio waves, tides, and weather patterns. elec

EsmoPLE 8 Finding a Trigonometric Model

Throughout the day, the depth of water at the end of a dock varies with the tides.

The table shows the depths (in meters) at various times during the morning.

Midnight 2 A.M. 4 A.m. 6 A.M. 8 A.M. 10 A.M. Noon

2.55 3.80 4.40 3.80 2.55 1.80 2.27

(time)

1, (depth)

of this example rive CD-ROM an is text.

Changing Tides

tility to here ;. How cl(:.es the graph?

+ 3 cos 2x

a. Use a trigonometric function to model this data.

b.Find the depths at 9 A.M. and 3 P.M.

C. A boat needs at least 3 meters of water to moor at the dock. During what times

in the afternoon can it safely dock?

Solution a. Begin by graphing the data, as shown in Figure 4.48. You can use either a sine

or cosine model. Suppose you use a cosine model of the form

y = a cos(bt - + d.

The difference between the maximum height and minimum height of the graph is twice the amplitude of the function. So, the amplitude is

a = [(maximum depth) - (minimum depth)] = (4.4 - 1.8) = 1.3.

The cosine function completes one half of a cycle between when the maximum and minimum depths occur. So, the period is

p = 2[(time of min. depth) - (time of max. depth)] = 2(10 - 4) = 12

which implies that b = 27r/p 0.524. Because high tide occurs 4 hours after midnight, consider the left endpoint to be c/b = 4, so c 2.094. Moreover, because the average depth is 1(4.4 + 1.8) = 3.1, it follows that d = 3.1. So, you can model the depth with the function

y = 1.3 cos(0.524t - 2.094) + 3.1.

b.The depths at 9 A.M. and 3 P.M. are as follows:

y = 1.3 cos(0.524 • 9 - 2.094) + 3.1 ----- 1.97 meters 9 A.M.

y = 1.3 cos(0.524 • 15 - 2.094) + 3.1 4.23 meters. 3 P.M.

C. Using a graphing utility, graph the model with the line y = 3, as shown in Figure 4.49. From the graph, it follows that the depth is at least 3 meters between 12:54 P.M. (t •""--= 12.9) and 7:06 P.M. (t 19.1).

Note that a sine model for the data in Example 8 is Y = 1.3 sin(0.524t - 0.524) + 3.1. Can you see how to find this?

4.72

5 -

Dep

th (

in m

eter

s)

4 A.M. 8 A.M. Noon Time

Figure 4.48

(12.9, 3) (19.1,3)

o IIIII 24

y= 1.3 cos (0.524t - 2.094) + 3.1

Figure 4.49


al Exercises In Exercises 1-14, find the period and amplitude.

1. y = 3 sin 2x 4

- 6.28

-4

6.28

2. y = 2 cos 3x

3

-3.14 A A V

- 3

3.14

In Exercises 23-26, describe the relationship b the graphs off and g.

23.

- 6.28

-3

3

6.28 -6.28

24. 2

A Vel

-2

).25

5 x 3. y = -

2cos -

2 3

12.56 -18.84

- 4

5. y = -2

sin 7TX 3

2

3.14 -6.28

7. y= -2 sinx

9. y = 3 sin 6x

1 2x 11. y = -

4 cos -

3

13. y = 3 sin 47rx

16. f(x) = cos x

g(x) = cos(x + 7r)

18. f(x) = sin 3x -

g(x) = sin(-

20. f(x) = sin x

g(x) = sin 3x

22. f(x) --- cos 4x

g(x) = -6 + cos 4x

26.

12.56 12.55

III 1

Abdo, Conjecture In Exercises 35-38, use a graphing ity to graph f and g in the same viewing I (Include two full periods.) Make a conjectur the functions.

35. f(x) = sin x

g(x) = cos(x - -7T) 2

37. f(x) = cos x

g(x) = - sincx - 2

4. y = -3 sin :13

4

- 12.56 18.84

-3

3 7 6. y = -- cos -1x

2 2

-3.14 AAA V

6.28

-2

8. y = -cos-2x 5

1 10. y = -

3 sin 10x

12. y = -5

cos -x

2 4

14. y= 2c0s-7rx12

4

15. f(x) = sin x

g(x) = sin(x - 7r)

17. f(x) = cos 2x

g(x) = - cos 2x

19. f(x) = cos x

g(x) = - 5 cos x

21. f(x) = sin x

g(x) = 4 + sin x

25.

-12.56 IA IA VI V

2

-12.56

4

-2

In Exercises 27-34, sketch the graphs of the tions in the same coordinate plane. (Include t periods.)

27. f(x) -= -2 sin x 28. f(x) = sin x

g(x) = 4 sin x g(x) = sin 3

29. f(x) = cos x 30. f(x) = 2 cos 2x

g(x) = 4 + cos x g(x) = - cos 4x

31. f(x) = - -1

sin -x

32. f(x) = 4 sin rrx 2 2

1 x g(x) = 5 - - sin - g(x) = 4 sin 7rx

2 2

33. f(x) = 2 cos x 34. f(x) = -cos x

g(x) = 2 cos(x + 7r) g(x) =

36. f(x) = sin x

g(x) = - coscx

38. f(x) = cos x

g(x) = - cos(x

e about

-12

In Exercises 15-22, describe the relationship between the graphs of f and g.

.III I

IT

E% 0 ha

39. Y

41. )

43. Y

45. Y

47. Y

49. Y

51. y

53. y

In Ex the fu amplit

55. y

57.y

58. y

59.y 60.y 61:y

Graphi d for graph

63.

-1-1 1141

etweet

45.

47. 3" 8 cos(x +

1 - cos 607rx 10 49. Y

53. Y

52. y = 2 cos x - 3

54. y = -3 cos(6x + 7r)

IT cos

(x +

-cos(x - 7r)

-12.56

65.

-6.28

-8

12.56

-2

of the two lune. Enclude two full

sin x

x sin -

3

2 cos 2x

- cos 4x

4 sin 7TX

4 sin 7rx - 1

In Exercises 55-62, use a graphing utility to graph

the function. (Include two full periods.) Identify the amplitude and period of the graph.

-2 sin(4x + 7r) 56. y = -4 sin(-2 x - 3 3

cos(27rx - + 1 2

7TX IT 3 cos(T + -

2) - 3

5 sinew - 2x) + 10

5 cos(rr - 2x) + 6 97

Tyi sin 120771 62. y = - -100 cos 507rt

55. y =

57.y =

58.y =

59.y =

60.y

61.y =

Graphical Reasoning In Exercises 63-66, find a and d for the function f(x) = a cos x + d so that the graph off matches the figure.

63. 9 64. 2

7r\ - COS( X - -

2 /

a graphing util-

iewing window. majecture about

sin x

77. S = 22.3 - 3.4 cos -I" 12.56 6

78. S = 74.50 + 43.75 sin -7rt 6


ionship between

2

ercises 39-54, sketch the graph of the function

, . Use a graphing utility to verify your sketch. han. ttclu-e two full periods.)

(i - 2 sin 4x

39. Y

1. Y

43. Y

cos 27rx

27rx - 2 sin

3

sin(x - -7r )

4

Graphical Reasoning In Exercises 67-70, find a, b, and c for the function f(x) = a sin(bx - c) so that the graph off matches the figure.

40. y= -3 cos 2x 67. 4 68. 3

5 . 7TX 42. y = - sin -

2 4

'7TX

-6.28 1111 6.28 -12.56

84% 12.56

-4 -3 44. y = -10 cos -

6

69. 2 70. 3

1 46. y = - sin(x - 7r)

2 -6.28

6.28 -6.28 A A 6.28

V

( IT

48. y = 6 cos x + -

-2 -3

6 In Exercises 71-74, use a graphing utility to graph y1

50. y = -4 + 5 cos -7rt and y2 for all real numbers x in the interval 12 [- 27r, 27r]. Use the graphs to find the real numbers x

2 7rx 2 - 2 sin

3

2 cos( - - 3 2 4

66. -12.56

- 6

such that y1 = y2.

71. y, = sin x 72. y, = cos x

Y2= 21 Y2 = 1 73. y1 = cos x 74. y1 = sin x

Y2 - Y2 2 2

75. Respiratory Cycle For a person at rest, the velocity v (in liters per second) of air flow during a respira-tory cycle is

v = 0.85 sin -7rt

3

where t is the time (in seconds). (Inhalation occurs when v > 0, and exhalation occurs when v < 0.)

(a) Use a graphing utility to graph v.

(b) Find the time for one full respiratory cycle.

(c) Find the number of cycles per minute.

76. Respiratory Cycle The model in Exercise 75 is for a person at rest. How might the model change for a person who is exercising? Explain.

Sales In Exercises 77 and 78, use a graphing utility to graph the sales function over 1 year, where S is the sales in thousands of units and t is the time in months, with t = 1 corresponding to January. Determine the months of maximum and minimum sales. 12.56

-6.28 6.28

(0.375, 1.65) • •••

0 / ••• •

• • S . • ••• ••

_ (0.125, 2.35)


79. Ferris wheel You are riding a Ferris wheel. Your height h (in feet) above the ground at any time t (in seconds) can be modeled by

h = 25 sin -177.5(t - 75) + 30.

The ferris wheel turns for 135 seconds before it stops to let the first passengers off.

(a) Graph the model.

(b) What are your minimum and maximum heights above the ground?

80. Blood Pressure The pressure P (in millimeters of mercury) against the walls of the blood vessels of a certain person is modeled by

P = 100 - 20 cos -t 3

where t is the time (in seconds). Graph the model. If one cycle is equivalent to one heartbeat, what is the person's pulse rate in heartbeats per minute?

81. Fuel Consumption The daily consumption C (in gallons) of diesel fuel on a farm is modeled by

C = 30.3 + 21.6 sin'

+ 10.9) 2,7rt

where t is the time in days, with t = 1 corresponding to January 1.

(a) What is the period of the model? Is it what you expected? Explain.

(b) What is the average daily fuel consumption? Which term of the model did you use? Explain.

(c) Use a graphing utility to graph the model. Use the graph to approximate the time of the year when consumption exceeds 40 gallons per day.

82. Data Analysis The motion of an oscillating weight suspended by a spring was measured by a motion detector. The data was collected, and the approxi-mate maximum displacements from equilibrium

= 2) are labeled in the figure. The distance y from the motion detector is measured in centimeters and the time t is measured in seconds.

(a) Is y a function of t? Explain.

(b) Approximate the amplitude and period.

(c) Find a model for the data.

(d) Use a graphing utility to graph the model in part (c). Compare the result with the data in the figure.

3

-1

FIGURE FOR 82

83. Data Analysis The normal monthly precipitationni, (in inches) for Seattle, Washington, for month / shown in the table, with t = 1 corresponding

t(,

January. (Source: National Climatic Data Cent4

1 2 3 4 5 6

p 5.4 4.0 3.8 2.5 1.8 1.6

7 8 9 10 11 12

p 0.9 1.2 1.9 3.3 5.7 6.0

(a) Find a trigonometric model for the nom' monthly precipitation in Seattle.

(b) Use a graphing utility to graph the data and the model found in part (a). How well does the model fit the data?

84. Data Analysis The table shows the normal daik high temperatures for Honolulu H and Chicago C(in degrees Fahrenheit) for month t, with t = 1 cone sponding to January. (Source: NOAA)

t 1 2 3 4 5 6

H 80.1 80.5 81.6 82.8 84.7 86.5

C 29.0 33.5 45.8 58.6 70.1 79.6

t 7 8 9 10 11 12

H 87.5 88.7 88.5 86.9 84.1 81.2

C 83.7 81.8 74.8 63.3 48.4 34.0

(a) A model for Honolulu is

H(t) = 84.40 + 4.28 sin(- + 3.86). 6

7rt

Find a trigonometric model for Chicago.

(b) Use a graphing utility to graph the data p

and the model for the temperatures in Honolul

How well does the model fit the data?

0.9

85. The gra

86. The gra of y = s

87. Writing tion y Write a graph fo

88. Writing tion)' = Write a graph fo

89. Writing' tion y = paragrap specifie

90. Writing tion y =-Write a graph fo

91.Explora f (x) = c an odd f and use odd, or n (a) h(x)

(b) h(x) " (c) h(x)

(c) U and Ho

(d) Us tern els

(e) Wh wh

(f) Whi atur

'eh mod

Synthesis

True or Fa whether th answer.

5 6

the normal daily end Chicago C(in with t = 1 cone-)AA)

84.7 86.5

70.1 79.6

6

1.6

12

6.0

2

6

-6


for the normal

the data and the w well does th

48.4 34.0

84.1

11

81.2

12

- 3.86).

Chicago.

h the data points

ures in Honolulu.

Le data?

) Use a graphing utility to graph the data points

and the model for the temperatures in Chicago. How well does the model fit the data?

(co Use the models to estimate the average annual temperature in each city. Which term of the mod-

'. els did you use? Explain.

(e ) What are the periods of the two models? Are they

what you expected? Explain.

to Which city has the greater variability in temper-

ature throughout the year? Which factor of the models determines this variability? Explain.

Synthesis

True or False? In Exercises 85 and 86, determine

whether the statement is true or false. Justify your

wer.

3 . 3x 207r 5. The graph of y = 6- -4 sm To has a period of 3 .

86.The graph of y = - cos x is a reflection of the graph

of y = sin(x + 7r/2) in the x-axis.

87.Writing Use a graphing utility to graph the func-don y = a sin x for a = 1, a =1 and a = -3. Write a paragraph describing the changes in the graph for the specified changes in a.

88.Writing Use a graphing utility to graph the func-tion y = d + sin x for d = 2, d = 3.5, and d = - 2. Write a paragraph describing the changes in the graph for the specified changes in d.

89.Writing Use a graphing utility to graph the func-tion y = sin bx for b = b = 1 and b = 4. Write a paragraph describing the changes in the graph for the specified changes in b.

90.Writing Use a graphing utility to graph the func-tion y = sin(x - c) for c = 1, c = 3, and c = -2. Write a paragraph describing the changes in the graph for the specified changes in c.

91.Exploration In Section 4.2 it was shown that f(x) = cos x is an even function and g(x) = sin x is an odd function. Use a graphing utility to graph h and use the graph to determine whether h is even, odd, or neither.

(a) h(x) = cos2 x (b) h(x) = sin2 x

(c) h(x) = sin x cos x

92. Conjecture Iff is an even function and g is an odd function, use the results of Exercise 91 to make a conjecture about the following.

(a) h(x) = (x)]2

(b) h(x) = [g(x)]2

(c) h(x) = f (x)g(x)

93. Graphical Reasoning The figure shows the graphs of the functions f(x) = 1 - 1.X2 and g(x) -= cos x. Identify the graphs and explain your reasoning.

6

Review

In Exercises 94-97, sketch the graph of the rational function and identify its asymptotes. Use a graphing utility to verify your results.

2 5x - 3 94. f(x) =

11 + x 95. f(x) =

10 x2 - 7x + 12 96. f(x) =

x2 + x 12 97. f(x) =

x + 2

In Exercises 98-101, convert the angle measure from radians to degrees. Round your answer to three deci-mal places.

98. 137r

99. - -7r 2 9

100. 8.57r 101. -0.48

tly precipitation p

n, for month t is

corresponding to tic Data Center)

= t an

3


Graph of the Tangent Function

Recall that the tangent function is odd. That is, tan( = — tan x. Consequently, the graph of y = tan x is symmetric with respect to the origin. You also know from the identity tan x = sin x/cos x that the tangent is undefined when cos x = 0. Two such values are x = ±712 ±1.5708.

X — —IT 2

— 1.57 —1.5 —1 0 1 1.5 1.57 IT —2

tan x Undef. —1255.8 —14.1 —1.56 0 1.56 14.1 1255.8 Undef.

You Should Le n:

• How to sketch the graphs tangent functions

• How to sketch the graphs cotangent functions

• How to sketch the graphs secant and cosecant funet

• How to sketch the graphs damped trigonometric fu tions

(tan x approaches — oo as x approaches — -7112 from the right.

As indicated in the table, tan x increases without bound as x approaches 77-/2 from the left, and it decreases without bound as x approaches — 7r/2 from the right. So, the graph of y = tan x has vertical asymptotes at x = 7/2 and — 7r/2, as shown in Figure 4.50. Moreover, because the period of the tangent function is 7r, verti-cal asymptotes also occur when x = 7r/2 + nv, where n is an integer. The domain of the tangent function is the set of all real numbers other than x = 712 + n7r, and the range is the set of all real numbers.


You can use tangent, cotange secant, and cosecant function model real-life data. For instanc Exercise 70 on page 342 shows you how a tangent function can be used to model and analyze the distance between a television camera and a parade u it.

4

tan x approaches oo as x approaches ir/2 from the left.

Figure 4.50

Period.' 11-

BOMA,' 811 X 0 7 ± /NT

Range: ( — no, 00) Vertical asymptotes:

x =' JT

+ fir

Sketching the graph of a function of the form y = a tan(bx — c) is similar to sketching the graph of y = a sin(bx — c) in that you locate key points that iden-tify the intercepts and asymptotes. Two consecutive asymptotes can be found by solving the equations bx — c = — 712 and bx — c = 7r/2. The midpoint between two consecutive asymptotes is an x-intercept of the graph. The period of the function y = a tan(bx — c) is the distance between two consecutive asymp-totes. The amplitude of a tangent function is not defined. 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.

a = fan

> —3 -Tr 371 Tr

4.6 • Graphs of Other Trigonometric Functions 335

Learn

cotangent, int functions to ta. For instance, ge 342 shows t function can and analze the a television ide unit

sik

d Learri—mm

the graphs of ons the graphs of

,-;tions the graphs of

;ecant functions the graphs of

lometric func_

EgmvipLE 1 Sketching the Graph of a Tangent Function

s ketch the graph of y = tan -2.

Solution B) solving the equations

X = IT — — 2 2

X = 7T

you can see that two consecutive asymptotes occur at x = - IT 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 4.51. Use a

graphing utility to confirm this graph.

— 7 — -i , U -i - 7

Undef. -1 0 1 Undef.

EXAMPLE 2 Sketching the Graph of a Tangent Function

Sketch the graph of y = -3 tan 2x.

Algebraic Solution Graphical Solution

x and 2 2

tan -2

Figure 4.51

A computer animation of this example appears in the Interactive CD-ROM and Internet versions of this text.

Use a graphing utility set in radian mode and dot mode to graph y = -3 tan 2x, as shown in Figure 4.53. Use the zero or root feature or the zoom and trace features of the graphing utility to approximate the key points as shown in the table.

By solving the equations 2x = - ir/2 and 2x = you can see that two consecutive asymptotes occur at x = - IT/4 and x = ir/4. Between these two asymp-totes, plot a few points, including the x-intercept, as shown in the table. Three complete cycles of the graph are shown in Figure 4.52.

X •i _ IT

4 — IT

8 0 TT

8 Ir -4:

-3 tan 2x Undef. 3 0 -3 Undef.

NMI 111111121151111M11111111111=M111111111111 11.1 IIIIMIIIMMEMI111111111111MMEMIN 1111111111111111111•11ffitillii11111111111111111EN 1111111111111111111MMINIMMIIIIIIIIIIIIMMEN MIIIIIIIIIIIIM11111111•11111111111111111111111MIN immuunnumimmumum Illowniummomrs•nummw limiturammutsmarsimirink Illimmaiiiimismaimmumi • moNanomimmunium•••• limmummummasmossom Itimmusammonmasnum

Figure 4.52

x -0.7854 -0.3927 0 0.3927 0.7854

-3 tan 2x Undef. 3 0 -3 Undef.

3.14

y = —3 tan 2x

3.14

-5

Figure 4.53

sin x y = cot x =


By comparing the graphs in Examples 1 and 2, you can see that the graph of y = a tan(bx — c) is increasing between consecutive vertical asymptotes if a > 0 and decreasing between consecutive vertical asymptotes if a < 0. In other words, the graph for a < 0 is a reflection in the x-axis of the graph for a > 0.

Graph of the Cotangent Function

The graph of the cotangent function is similar, to the graph of the tangent func-tion. It also has a period of IT. However, from the identity

you can see that the cotangent function has vertical asymptotes x = nr, where n is an integer, because sin x is zero at these x-values. The graph of the cotangent function is shown in Figure 4.54.

EXAMPLE 3 Sketching the Graph of a Cotangent Function

Sketch the graph of y = 2 cot 3

Solution To locate two consecutive vertical asymptotes of the graph, solve the equations x/3 = 0 and x/3 = IT to see that two consecutive asymptotes occur at x = 0 and

37r.

Then, 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 4.55. Use a graphing utility to confirm this graph. [Enter the function as y = 2/tan(x/3).] Note that the period is 37r, the distance between consecutive asymptotes.

x 0 37r 4

37r 2

97r 4

1

'71- X

2 cot —3

Undef. 2 0 —2 Undef.

= Jtot ,3

> K

I

a

Figure 4.55

Period: 27

.00171aill: all

Range: ally

Vertical asyri

Symmetry'

Figure 4.51

3

a

a

Period: 7

901118ill: all x f ir Range: (— co, co)

Vertical asymptotes: x = 177

Figure 4.54

GraP

The gr the gra

For ins

the Y-c exist. N of the t

have ve is zero

have ye

To sket sketch first ske obtain p graphs

y = sin

Tr a


Graphs of the Reciprocal Functions

graphs of the two remaining trigonometric functions can be obtained from

tile graphs of the sine and cosine functions using the reciprocal identities

1 and sec x =

cos x

For instance, at a given value of 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 of x, the behavior of the secant function is similar to that of the tangent function. In other words, the graphs of

sin x 1 and sec x =

have vertical asymptotes at x = /712 + mr, where n is an integer and the cosine

s zero at these x-values. Similarly,

cos x cot x — and csc x =

sin x 7 77"

have vertical asymptotes where sin x = 0, that is, at x = n7r.

To sketch the graph of a secant or cosecant function, you should 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 4.56.

CSC X Slri X

1

sin x

siri x )1, y= Csc A

tan x — cos x

Period: 2 -

1101118/11: allx

177

Rall98: all y not in (-1, 1)

Vocal asymptotes: x =

Symmetry: origin

Figure 4.56

Period: 27 7r

001113ill: all x * —2

+ fi 7

Range: all y not in (-1, 1) ir

Vertical asymptotes: x = —2

+ n77-

Symmetry: y-axis

set. y cos Y = A 3

X Tr Tr

3

Explorat' -

Use a graphing utility to graph the functions yi = sin x and y2 = csc x = I /sin x in the

4 same viewing window. How are

the graphs related? What hap-

pens to the graph of the cose-cant function as x approaches the zeros of the sine function?

y = 2 sin 5

C> IT 77r 0 < x +

IT < 27r

4

1

sin[x + (7r/4)] y = 2 csc(x + 17.) = 2(

4

0),

(37r o), . .

IT x = --

4' IT 37r

x = x = . . . 4


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 interchanged. For example, a hill (or maximum point) on the sine curve corresponds 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 4.57.

EXAMPLE 4 Comparing Trigonometric Graphs

Use a graphing utility to compare the graphs of

7r y = 2 sin(x + — and y = 2 csc(x + —

4)

• 4

Solution The two graphs are shown in Figure 4.58. Note how the hills and valleys of the graphs are related. For the function y = 2 sin[x + (7r/4)], the amplitude is 2 and the period is 27. By solving the double inequality

Figure 4.57

you can see that one cycle of the sine function corresponds to the interval from x = — 7714 to x = 77r/4. The graph of this sine function is represented by the gray curve in Figure 4.58. Because the sine function is zero at the endpoints of this interval, the corresponding cosecant function

has vertical asymptotes at x = — 7r/4, 37r/4, 77r/4, etc. The graph of the cose-cant function is represented by the black curve.

EXAMPLE 5 Comparing Trigonometric Graphs

Use a graphing utility to compare the graphs of

y = cos 2x and y = sec 2x.

Solution Begin by graphing y1 = cos 2x and y2 = sec 2x = 1/cos 2x in the same viewing window, as shown in Figure 4.59. Note that the x-intercepts of y = cos 2x

correspond to the vertical asymptotes

of the graph of y = sec 2x.

-5

Figure 4.58

—3.14

= cos 2.1

Figure 4.59

—10

y=2c

= sec 2.v

Sc

EXA

An*

Solu Consi

each you k means

Furth

and

the gr

has int


mped Trigonometric Graphs

roduct of two functions can be graphed using properties of the individual

&ions. For instance, consider the function

1(x) x sin x

e product of the functions y = x and y = sin X. Using properties of absolute and the fact that 'sin xl 5 1, you have 0 5 lx1 'sin xl .5 Ixl. Consequently,

- lx1 x5_ sin x lx1

• h means that the graph of f(x) = x sin x lies between the lines y = -x and

x. Furthermore, because

rr

a Tr

A 3 11

11

f(x) x sin x = ±x at x = ± 2

f(x) x sin x = 0 at x =

the graph off touches the line y = -x or the line y = x at x = ir/2 + fir and

has x-intercepts at x = nir. A sketch of f is shown in Figure 4.60. In the function f(x) x sin x, the factor x is called the damping factor.

EXAMPLE 6 Analyzing a Damped Sine Curve

Analyze the graph of

f(x) = e- x sin 3x.

Solution Consider f(x) as the product of the two functions

y = e- x and y = sin 3x

each of which has the set of real numbers as its domain. For any real number x, you know that e- x 0 and 1sin 3x1 5 1. Therefore, le- x1 kin 3x1 5 e- x, which means that

-e - x e- x sin 3x 5 e- x.

Furthermore, because

IT flIT f(x) e- x sin 3x = ±e- x at x = - + —

6 3

3

Figure 4.60

3.14

—nrr

f(x) e- x sin 3x = 0 at x = 3

the graph of f touches the curves y = -e- x and y = nIT

e- 71-

x at x = -6

+ —3

and n71-

has intercepts at x = —3

. The graph is shown in Figure 4.61.

f(x). x s x

air

1T

Figure 4.61


Figure 4.62 summarizes the graphs, domains, ranges, and periods of the six basic trigonometric functions.

Y = tan

in Exercis

State the labeled (a)

(a)

—6.28

3 A

a

- c OS

ii a IT

a

3

(c) Domain: all reals Domain: all x # — + nir .1 2 —1.57 Range: [- 1, 1] Range: (-cc, 00)

Period: 27r Period: 7T

(e) -'-

y = sec x / =

y = cot x -

—9.42

cos x

/

(g)

—9.42 u-

DOM& all x # —2

+ fi 7T

Range: (- 00, -1] and [1, oo) Period: 2,71-

3

3

a

a

A

= S in x

Domain: all mats

Range: [- 1, 1]

Period: 27r

Domain: all x # n7r

Range; (-co, -1] and [1, co) Period: 27 Figure 4.62

Domain: all x 0 nir

Range: (-cc, 00) Period: IT

1. y = sec

3. y = tan

5. y = cot

7. y -

In Exercise (Include tvi verify your

9. 11. y =—

13. y = — I

15. y =—s

17. Y = sec

19. Y csc

21. y c

Writing About Math Combining Trigonometric Functions. Recall from Section 1.4 that functions can be combined arithmetically. This also applies to trigonometric functions. For each of the functions h(x) = x + sin x and h(x) = cos x - sin 3x, (a) identify two simpler func-tions f and g that comprise the combination, (b) use a table to show how to obtain the numerical values of h(x) from the numerical values of f(x)-and g(x), and (c) use graphing utility graphs off and g to show how h may be formed. Can you find functions

f (x) = d + a sin(bx - c) and g(x) = d + a cos(bx - c)

such that f(x) + g(x) = 0 for all x? If so, write a short paragraph explaining how you found the functions.

-5

-5

(f)

-5

5

9.42

-3.93

-5

9.42

(h)

-3.14

5

111111

-5

fl n (e) -9.42

(g)

-9.42

- 6

5

-x

1. y = sec 2

3. y = tan 2x

7TX 5. y = cot -

2 7. y =

2. y = tan

-2

4. y 2 csc x

1 6. y = -

2 sec

8. y = -2 sec 27r)c

(c)

(a) 5 (b)

_6.28 6.28 -3.14

rr - 5

_1.57

6


Exercises

10 Exercises 1-8, match the function with its graph.

Ste the period of the function. [The graphs are

iatbeled (a), (b), (c), (d), (e), (f), (g), and (h).]

5

-5

5

In Exercises 9-30, sketch the graph of the function. (Include two full periods.) Use a graphing utility to verify your result.

9. y = tan x 10. y = tan 2x 11. Y = -2 tan 2x 12. y = -3 tan 7TX

13. Y = sec x 14. y = sec 2x

15. Y = -sec 7TX 16. y = 2 sec 4x

17. Y = sec 7rx - 3 18. y= -2 sec 4x + 2

19. Y csc 20. y = -csc -3

21 -1 x . Y rz-- 2

cot -2

71X 22. y = 3 cot --

i 23. y = sec 2x

7TX 25. y = 2 tan -

4

27. y = csc(- - 4

1 29. y = 2 cot

(x - -2

24. y= - tan 7rx

26. y = sec(x + 7r)

28. y = sec(7r -

1 1 T 30. y = -

4 csc x +

4

31. y = -3

tan -3

1 x 32. y = -2 tan 2.7rx

1 33. y = -2 sec 4x 34. y = -

4 sec ix

36. y = -csc(4x - 7r)

7vc ir) 38. y = 0.1 tan( 7 . + -4

I (irx , a") 40. y = -3 sec - 2

In Exercises 41-44, use a graph to solve the equation in the interval [ - 2 7T, 27r].

41. tan x = 1 42. cot x = - 43. sec x = -2 44. csc x =

In Exercises 45 and 46, use the graph of the function to determine whether the function is even, odd, or neither.

45. f(x) = sec x 46. f(x) = tan x

In Exercises 47-50, use a graphing utility to graph the two equations in the same viewing window. Use the graphs to lend evidence that the expressions are equivalent. Verify the results algebraically.

47. y1 = sin x csc x, Y2 = 1 48. y1 = sin x sec x, Y2 = tan x

cos x

sin x y2 = cot x

50. y1 = sec2 x - 1, y2 = tan2 x

3.14

3.93

In Exercises 31-40, use a graphing utility to graph the function. (Include two Tull periods.) DescHbe- y7IFir

9.42 viewing window.

35. y = tan,/ x - 5-)

4

37. y = I cot(x + 4 2

3.14 1

39. y = - sec(2x - 71-) 2

5 (c)

-6.28

(d)

6.28 -6.28

-5

51. f(x) = x cos x

53. g(x) = lx1 sin x

-5

52. f(x) = Ix sin xl 54. g(x) = Ixl cos x

6.28

',1111.11,1111k1xPrw-

-5

5 I

5m1

36m

71"


In Exercises 51-54, match the function with its graph. Describe the behavior of the function as x approaches zero. [The graphs are labeled (a), (b), (c), and (d).]

(a)

-6.28 8$4,,8 5

6.28

(b)

-6.28 Nnv 5

6.28

69. Distance A plane flying at an altitude of over level ground will pass directly over a radar tenna. Let d be the ground distance from the Elite to the point directly under the plane and let A be angle of elevation to the plane from the ant: Write d as a function of x and graph the functi, Dn the interval 0 < x < IT.

Conjecture In Exercises 55-58, use a graphing util-ity to graph the functions l and g. Use the graphs to make a conjecture about the relationship between the functions.

55. f(x) = sin x + cos(x + -2), g(x) = 0

sin x - cos(x + -2 ), g(x) = 2 sin x

sin2 x, g(x) = 1(1 - cos 2x)

1 COS2 -2 , g(x) = -2(1 + cos irx)

In Exercises 59-62, use a graphing utility to graph the function and the damping factor of the function in the same viewing window. Describe the behavior of the function as x increases without bound.

59. f(x) = e- x cos x 60. g(x) = e 2/2 sin x

61. f(x) = 2-44 cos Trx 62. h(x) = 2.- x2/4 sin x

Exploration In Exercises 63-68, use a graphing util-ity to graph the function. Describe the behavior of the function as x approaches zero.

70. Television Coverage A television camera is reviewing platform 36 meters from the street which a parade will be passing from left to Express the distance d from the camera to a pante. ular unit in the parade as a function of the angie x, and graph the function over the

on

interval

fight on

- 7r12 <x < IT/2. (Consider x as negative when a unit in the parade approaches from the left.)

71. Predatory-Prey Model Suppose the population of a

certain predator at time t (in months) in a region is es-

timated to be P = 10,000 + 3000 sin (211/24) aod

the population of its primary food source (its prey) lc

estimated to be p = 15,000 + 5000 cos 01/24

Use the graph of the models to explain the oscilla-

tions in the size of each population.

56. f(x) =

57. f(x) =

58. f(x) =

63. f(x) = + cos x

65. f(x) = sin x

67. f(x) = tan x

64. f(x) = sin x - -4

- 66. f(x) =

1 cos x x

68. f(x) = coxt x

20,000

•Z 15,000 '

P., 10,000

5,000

10 20 30 40 50 60 70 80 90 Time (in months)

population of a a a region is es-1 (271/24) and rce (its prey) is 0 cos (2704). lin the oscilla-

80 90

Equilibrium +


tude of 5 mile over a radar rom the anrenn ' and let x be th: 'm the antenna he function ever'

camera is on a the street on

m left to right, iera to a partie. in of the angle r the interval egative when a le left.)

Normal Temperatures The normal monthly high 2. temperatures in degrees Fahrenheit for Erie, Penn-

sylvania, are approximated by

y(t) --,-- 54.33 - 20.38 cos —7rt - 15.69 sin —Irt 6 6

and the normal monthly low temperatures are approximated by

L(r) = 39.36 - 15.70 cos —7'1 6

where t is the time (in months), with t = 1 cone-to January. (Source: National Oceanic

and Atmospheric Association)

(a) Use a graphing utility to graph each function. What is the period of each function?

(b) During what part of the year is the difference between the normaT high and low temperatures greatest? When is it smallest?

(c) The sun is the farthest north in the sky around June 21, but the graph shows the warmest tem-peratures at a later date. Approximate the lag time of the temperatures relative to the position of the sun.

73. Harmonic Motion An object weighing W pounds is suspended from a ceiling by a steel spring. 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 = -1 e-'14 cos 4t, t > 0

2

where y is the distance (in feet) and t is the time (in seconds).

(a) Use a graphing utility to graph the function.

(b) Describe the behavior of the displacement func-tion for increasing values of time t.

74. Sales The projected monthly sales S (in thousands of units) of a seasonal product is modeled by

irt S 74 + 3t + 40 sin -6-

where t is the time (in months), with t = 1 cone-sponding to January.

(a) Graph the sales function over 1 year. Use a graphing utility to verify your graph.

(b) Which month should have the greatest number of sales? Which month should have the least number of sales? Justify your answers.

75. Sales The projected monthly sales S (in thousands of units) of a seasonal product is modeled by

S = 52 + St - 28 cos —6

where t is the time (in months), with t = 1 corre-sponding to January.

(a) Graph the sales function over 1 year. Use a graphing utility to verify your graph.

(b) Which month should have the greatest number of sales? Which month should have the least number of sales? Justify your answers.

76. Data Analysis The motion of an oscillating weight suspended by a spring was measured by a motion detector. The data was collected, and the approxi-mate maximum (positive and negative) displace-ments from equilibrium are shown in the graph. The displacement y is measured in centimeters and the time t is measured in seconds.

(a) Is y a function of t? Explain.

(b) Approximate the frequency of the oscillations.

(c) Fit a model of the form y = cibt cos ct to the data. Use the result of part (b) to approximate c. Use the regression capabilities of a graphing util-ity to fit an exponential model to the positive maximum displacements of the weight.

(d) Rewrite the model in the form y = aekt cos ct.

(e) Use a graphing utility to graph the model. Compare the result with the data in the graph.

13 12) (0.7622, 3.76)

• 1(1.5476, 1.16) • \ • ao • 111 2

• / Eboo (1.1549, -2.09)

-8 / (0.3695, -6.78) (1.9403, -0.64)

77. Writing Write a short paragraph describing the specified change in the physical system of Exercise 76.

(a) A spring of less stiffness is used, and so the length of time in each oscillation is greater.

(b) The effect of friction is decreased.

- 14.16 sin.-7rt 6

1111.411111.17


78. Numerical and Graphical Reasoning A crossed belt connects a 10-centimeter pulley on an electric motor with a 20-centimeter pulley on a saw arbor. The electric motor runs at 1700 revolutions per minute.

(a) Determine the number of revolutions per minute of the saw.

(b) How does crossing the belt affect the saw in rela-tion to the motor?

(c) Let L be the total length of the belt. Write L as a function of 0, where 0 is measured in radians. What is the domain of the function? (Hint: Add the lengths of the straight sections of the belt and the length of belt around each pulley.)

(d) Use a graphing utility to complete the table.

4) 0.3 0.6 1.2 1.5

(e) As 0 increases, do the lengths of the straight sec-tions of the belt change faster or slower than the lengths of the belt around each pulley?

(f) Use a graphing utility to graph the function over the appropriate domain.

10 cm 20 cm

Synthesis


79. The graph of y = tan(x

8 2 + 7r) has an asymptote

at x = —37r.

80. The graph of y = —2 csc(x + i.r) has an asymptote

at x = — — 3

81. In the graph of y = 2x sin x, as x approaches — oo, y approaches 0.

82. Writing Describe the behavior of f(x) x approaches 7r/2 from the left and from tin

83. Writing Describe the behavior of f(x) approaches i from the left and from the rig

84. Graphical Reasoning Consider the two f

f (x) = 2 sin x and g(x) = csc x on the (0, IT).

(a) Use a graphing utility to graph f and same viewing window.

(b) Approximate the interval where f> g.

(c) Describe the behavior of each of the fun x approaches IT. How is the behavior of to the behavior of f as x approaches 7r?

85. Pattern Recognition

(a) Use a graphing utility to graph each fun

y1 = sin + —3

sin 37tx 4 (

i 1

4 1 1 y2 =- —

7r sin 7TX + —

3in s 37rx + —

5 sin 57

Identify the pattern in part (a) and find a function y3 that continues the pattern one more t elln. Use a graphing utility to graph y3.

The graphs of parts (a) and (b) approxi periodic function in the figure. Find a fu that is a better approximation.

3

Review

In Exercises 86-89, determine whether the fu one-to-one. If it is, find its inverse.

86. f(x) = —10 87. f(x) = (x — 7)2 +3

88. f(x) = ,/3x — 14 89. f(x) = 5

In Exercises 90 and 91, find the exact values of the Sit

trigonometric functions of the angle O.

2v7 91.

14

Mtery

•, In th,

lions ,6

g related

(b)

(c) mate the nction yi

nction is !It

90. 2

Inver

Recall Hod zo

the test

Figure 4.

Flowev jog to hold.

1.On t

2. On —1

3. On t

So, on called t

The n The ar of a ce means commo sine fu — 7r/2

rh

When [- 1,

When arcs me

You Should Learn:

• How to evaluate inverse sine functions

• How to evaluate other inverse trigonometric functions

• How to evaluate compositions of trigonometric functions


Inverse trigonometric functions can be useful in exploring how two aspects of a real-life problem relate to each other. Exercise 70 on page 353 investigates the rela-tionship between the length of rope from a winch to a boat and the angle of elevation between them.

4.7 • Inverse Trigonometric Functions 345

f (x) tan .t

from the rightli `(x) ese x

the right.

Le two on th ; -e alter\

f and g i the

11111

the functions as Igor of g related lies 7r?

ach function.

sin 57rx)

i find a function more term. Use

ipproximate the nd a function y4

the function is

(x - 7)2 + 3

- 5

inverse Sine Function

Recall from Section 1.5 that for a function to have an inverse, it must pass the

Horizontal Line Test. From Figure 4.63 it is obvious that y = sin x does not pass

the test because different values of x yield the same y-value.

Sin x has an inverse on this interval.

Figure 4.63

However, if you restrict the domain to the interval - 7r/2 x 5. 7r/2 (correspond-

ing to the black portion of the graph in Figure 4.63), the following properties

hold.

1.On the interval [- 7/2, 7/ 2], the function y = sin x is increasing.

2. On the interval [- 7/2, 7/2], y = sin x takes on its full range of values, - 1 < sin x 1.

3. On the interval [- 7/2, 7/2], y = sin x passes the Horizontal Line Test.

So, on the restricted domain - 7/2 5_ x 7/2, y = sin x has a unique inverse called the inverse sine function. It is denoted by

y = arcsin x or y = sin-1 x.

The notation sin-1 x is consistent with the inverse function notation f -1(x). The arcsin x notation (read as "the arcsine of x") comes from the association of a central angle with its intercepted arc length on a unit circle. So, arcsin x means the angle (or arc) whose sine is x. Both notations, arcsin x and sin-1 x, are commonly used in mathematics, so remember that sin-1 x denotes the inverse sine function rather than 1/sin x. The values of arcsin x lie in the interval - 7r/2 arcsin x 7/2. The graph of y = arcsin x is shown in Example 2.

inition of Inverse Sine Function

f> g.

dues of the six

14

The inverse sine function is defined by

y = arcsin x if and only if sin y = x

Where -1 < x < 1 and -7/2 5_ y 7712. The domain of y = arcsin x is H 1, 1] and the range is [- 7/2, 7/ 2].

When evaluating the inverse sine function, it helps to remember the phrase "the aresme of x is the angle (or number) whose sine is x."

y ( IT

5 IT

'rr

4-5

2'

Y=

-7f

rr , a

Figure 4.64 4.64

ar-

-S FR


STUDY Tlp •

As with the trigonometric fune, tions, much of the work with the inverse trigonometric functioRs can be done by exact calcula. tions rather than by calculator approximations. Exact calcula tions help to increase your understanding of the inverse functions by relating them to the triangle definitions of the trigonometric functions.

EXAMPLE 1 Evaluating the Inverse Sine Function

If possible, find the exact value.

a. arcsin(- 2

b. sin -1 2 C. sin-1 2

Solution 1

a. Because sin(- 7r/6) = - -2

for - 7r/2 y 7r/2, it follows that

7F arcsin(- -

1) = - 2

b. Because sin(7r/3) = -"J./2 for - 7r/2 y 7/2, it follows that

sin-1 2

= . Angle whose sine is 3

c. It is not possible to evaluate y = sin-1 x when x = 2 because there is no angle whose sine is 2. Remember that the domain of the inverse sine func-tion is [-1, 1].

EXAMPLE 2 Graphing the Arcsine Function

Sketch a graph of y = arcsin x.

Solution By definition, the equations

y = arcsin x and sin y = x

are equivalent for - 7r/2 y 7r/2. So, their graphs are the same. From the inter-val [- 7r/2, 7r/2], you can assign values to y in the second equation to make a table of values.

Y IT

—2- IT

—4-

IT

- 0 IT — 6

'77- — 4

IT

- z

x = sin y -1 .,/, 1

---i 0

i

1 - 2 2

The resulting graph for

y = arcsin x

is shown in Figure 4.64. Note that it is the reflection (in the line y = x) of the black portion of the graph in Figure 4.63. Use a graphing utility to confirm this graph. Be sure you see that Figure 4.64 shows the entire graph of the inverse sine function. Remember that the range of y = arcsin x is the closed interval [- IT/2, 7712].

Angle whose sine is —1

library of Functions The inverse trigonometric func-tions are obtained from the .4 trigonometric functions in much the same way that the logarithmic function was developed from the exponential function. However, unlike the exponential function, the trigonometric functions are not one-to-one, and so it is neces-sary to restrict their domains to regions that pass the Horizontal Line Test. Consult the Library of Functions Summary inside the front cover for a description of the inverse trigonometric fun_ tions.

Similar , of y =' initions ing thr

The grai 4.66.

A c /Inc

Y

v

Cos x has an inverse

on this interval.

Figure 4.65

Consequently, on this interval the cosine function has an inverse function—the

inverse cosine function—clenoted by

y arccos x or y = c0S-1 X.

Because these equations are equivalent for 0 y 7T, their values are the same,

as shown in the table below.

- I•

Y 0 IT IT IT

-i , 2/r -4 -

57r — 6

IT

x = cos y , i.

-,./- 1 i

, u

1 —2-

,/ , - 1 2 - 2

ir

Domain: [— 1, 1]; R1111118:

Domain: [— 1, 1]; Range: [0, 7r]

Function

Y arcsin x if and only if silly = x

Y arccos x if and only if cos y = x

v arctan x if and only if tan y = x

Range

IT IT -- < V

2 -2

0 y IT

IT IT < y < -

2 2

Jr

Domain: (—no, co); Range: (— 1" — 72) •

7T

Figure 4.66

inition of Inverse Trigonometric Functions In

The graphs of these three inverse trigonometric functions are shown in Figure 4.66.

® \ computer animation of this concept appears in the era dive CD-ROM and Internet versions of this text.

2

ir 2


V

aometric funeI work with the

.ttric functions ract calcula_

calculator 1 -1xact calcula-...ase your the inverse ing them to tions of the ctions.

!ions metric func-from the Lions in much the logarithmic oped from the m. However, tial function, unctions are I so it is neces domains to

e Horizontal the Library of y inside the scription of the ic functions.

Other Inverse Trigonometric Functions

[tic cosine function is decreasing on the interval

ure 4.65.

Similarly, you can define an inverse tangent function by restricting the domain of y = tan x to the interval (- 7112, 7112). The following list summarizes the def-initions of the three most common inverse trigonometric functions. The remain-ing three are discussed in Exercises 80-82 in this section.

0 x 7r, as shown in


EXAMPLE 3 Evaluating Inverse Trigonometric Functions

Find the exact value.

a. arccos —2

b. arccos(— 1) c. arctan 0 d. arctan (— 1)

Solution a. Because cos(7r/4) = ,2, and 7r/4 lies in [0, ir], it follows that

7T arccos —

2 = —4. Angle whose cosine is —

2

b. Because cos ir = — 1, and 77- lies in [0, 7r], it follows that

arccos(— 1) = IT. Angle whose cosine is — 1

C. Because tan 0 = 0, and 0 lies in (-7r/2, 7r/2), it follows that

arctan 0 = 0. Angle whose tangent is 0

d. Because tan (— 7r/4) = — 1 and —7r/4 lies in (-7r/2, 7r/2), it follows that

arctan (— 1) = Angle whose tangent is — 1

STUDY lip

Most graphing utilities do not have function keys for the inverse cosecant function, the inverse secant function, or the inverse cotangent function. To evaluate these functions, you can use the inverse sine, inverse cosine, and inverse tangent functions, respectively. For instance, to evaluate arcsec 3.4, enter the expression as arccos(1/3.4).

You can use the siN-11, (cos-1), and values of trigonometric functions.

keys on your calculator to approximate TAN-1

EXAMPLE 4 Calculators and Inverse Trigonometric Functions

Use a calculator to approximate the value (if possible).

a. arctan(— 8.45) b. arcsin 0.2447 c. arccos 2

Solution

Function Mode Graphing Calculator Keystrokes

a. arctan(— 8.45) Radian 1TAN-11 C) 0 8.45 UI

From the display, it follows that arctan(— 8.45) — 1.453001.

b. arcsin 0.2447 Radian siN-1) 0.2447

From the display, it follows that arcsin 0.2447 0.2472103.

c. arccos 2 Radian cos-1 2

ENTER

ENTER

ENTER

EX

If po

a. ta

Sol a'. B

In real number mode, the calculator should display an error message because the domain of the inverse cosine function is [— 1, 1].

In Example 4, if you had set the calculator to degree mode, the display would have been in degrees rather than in radians. This convention is peculiar to calcu-lators. By definition, the values of inverse trigonometric functions are always in radians.


itT

lities do not s for the unction, the ction, or the function. To ctions, you a sine, inverse a tangent vely. For tte arcsec 3.4,

compositions of Functions

gccal' from Section 1.5 that inverse functions possess the properties

= x

Th inverse trigonometric versions of these properties are given below.

m as

.f -1(f(4) = x.

Properties

if • x land -IT/2 5_ y 7r/2, then

sm(arcsin = x and arcsin(sin = y.

1f _l • x 1 and 0 y 7r, then

cos(arccos = x and arccos(cos y) = y.

If x is a real number and -7r/2 <y < 7r/2, then

tan(arctan = x .and arctan(tan = y.

Keep in mind that these inverse properties do not apply for arbitrary values of x

and y. For instance,

1101.

arcsin(sm = arcsin(- 1) = --7r -3

7r. 2 2 2

. 37r

In other words, the property

arcsin(sin = y

is not valid for values of y outside the interval [-7r/2, 70].

• ‘ Exploration se a graphing utility to graph = arcsin(sin 4. What are the

domain and range of this func-tion? Explain why arcsin(sin 4) ttloes not equal 4.

Now graph y = sin(arcsin and determine the domain and range. Explain why sin(arcsin 4) is not defined.

and

EXAMPLE 5 Using Inverse Properties

If possible, find the exact value.

a. tan[arctan(-5)] .

b. arcsin(sin —3

c. cos(cos-1 7r)

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, 57r/3 does not lie within the range of the arcsine function, - 7r/2 y 7r/2. However, 57r/3 is coterminal with

57r 7F

Which does lie in the range of the arcsine function, and you have

arcsin(sm —3 = arcsin[sin(- -3 )] = . 57r 7r 7r

C. The expression cos(cos-1 7r) is not defined because c05-1 7r is not defined. Remember that the domain of the inverse cosine function is [- 1, 11

y = tan(arccos x

-J

(b)

-2

(a)

Figure 4.68

\ X=.66667 Y=1.1180239

'3


Example 6 shows how to use right triangles to find exact values of compositions of inverse functions. Then, Example 7 shows how to use triangles to convert a trigonometric expression into an algebraic expression. This conversion technique is used frequently in calculus.

EXAMPLE 6 Evaluating Compositions of Functions

Find the exact value of (a) tan(arccos and (b) cos[ arcsin 3 5

Algebraic Solution Graphical Solution 2 2 a. If you let u = arccos 5, then cos u = 5. Because

cos u is positive, u is a first-quadrant angle. You can sketch and label angle u as shown in Figure 4.67(a). Consequently,

2 opp tani arccos —

3 -= tan u =

adj —

2 •

b. If you let u = arcsin (4, then sin u = Be-

cause sin u is negative, u is a fourth-quadrant angle. You can sketch and label angle u as shown in Figure 4.67(b). Consequently,

3 adj 4 cos [ arcsin(

5 — — )1 = cos u =

hyp = —

5

(a (b) Figure 4.67

a. Use a graphing utility set in radian mode to 0, o=aph y = tan(arccos 4, as shown in Figure 4.68(a). Use the value feature or zoom and trace features of the grE utility to find that the value of the composition of nfu c. tions when x = 0.66667 is y = 1.118

b. Use a graphing utility set in radian mode to y = cos(arcsin 4, as shown in Figure 4.68(b). II value feature or zoom and trace features of the gn utility to find that the value of the composition of tions when x = — = — 0.6 is y = 0.8 =

graph se the phing fun.

EXAMPLE 7 Some Problems from Calculus

Write each of the following as an algebraic expression in x.

1 a. sin(arccos 3x), 0 x < b. cot(arccos 3x), 0 x —

1

3 3

Solution If you let u = arccos 3x, then cos u = 3x. Because cos u = 3x/1 = adj/hyp you can sketch a right triangle with acute angle u, as shown in Figure 4.69. From this triangle, you can easily convert each expression to algebraic form.

opp a. sin(arccos 3x) = sin u = — = — 9x2, 0 x < —

1 hyp 3

adj 3x b. cot(arccos 3x) =- cot u =

— 9x2 0 x < —1

opp -,/1. ' 3

A similar argument can be made here for x-values lying in the interval [-I, 0].

3x

Figure 4.69

(3

(b)

(c)

(d)

3. Na fun

6 10 0 2 4 8

2 sin — ▪ = 1 LiI> arcsin 1 =

2

— 0.2 x —1 —0.8 —0.6 —0.4

—0.6 —0.4 — 0.2 —1 —0.8

1 0.4 0.6 0.8 0 0.2

arcsin

arccos

(b) arcsin 0

(b) arccos 0

(b) arctan 1 arctan 3

Numerical and Graphical Analysis Consider the

tunction y= arcsin x.


NONi

(b) Plot the points from the table in part (a) and graph

the function. (Do not use a graphing utility.)

(c) Use a graphing utility to graph the inverse sine function and compare the result with your hand-drawn graph in part (b).

(d) Determine any intercepts and symmetry of the graph.

2. Numerical and Graphical Analysis Consider the function y = arccos x.



(b) Plot the points from the table in part (a) and graph the function. (Do not use a graphing util-ity.)

(c) Use a graphing utility to graph the inverse tan-gent function and compare the result with your hand-drawn graph in part (b).

(d) Determine the horizontal asymptotes of the graph.

In Exercises 4-7, write each trigonometric expression in inverse function form, or vice versa. For example,

mode to graph .68(a). Use the of the graphing °Mimi of fun

-J/2.

mode to graph .68(b). Use the of the graphing osition of func-

4 3.

s(arcsin x 3

4. sin(T6--) = =>

6. arccos = 2 4

7. arcsin(— 1) = — 72 =>

(b) Plot the points from the table in part (a) and graph the function. (Do not use a graphing utility.)

(c) Use a graphing utility to graph the inverse cosine function and compare the result with your hand-drawn graph in part (b).

(d) Determine any intercepts and symmetry of the graph.

3. Numerical and Graphical Analysis Consider the function y = arctan x.

In Exercises 8-15, evaluate the expression using a cal-culator.

2 (b) arcsin — ‘n-.)

2 11. (a) arccos — -

-,h) (

12. (a) arctan( — -0) (b) arctan -0

-li 13. (a) arccos(— 1

) 2 (b) arcsin

2

1 0.8 0.6 0.4 0.2 0

—10 —2 —4 —6 —8

In the

60. Y

62.

1 term

A cos

Use funct

68. f

69.

70.

18. (a)

19. (a)

20. (a)

21. (a)

22. (a)

23. (a)

arccos 0.22

arcsin(- 0.75)

arctan(- 6)

arcsin 0.41

arccos(- 0.51)

arctan 0.98

(b) arcsin 0.45

(b) arccos(- 0.7)

(b) arctan 18

(b) arccos 0.36

(b) arcsin(- 0.125)

(b) arctan 4.7

sin(arctan

cos( arcsin 245-)

sec[arctan(- A

sin[arccos(-

38.

40.

42.

44.

39. sec(arcsin)

41. csc[arctan( -

43. tan[arcsin(- -D]

45. cot(arctan

47. sin(arctan x)

49. sec[arcsin(x -

51. cot( 1

arctan .-; )

( 53. cos arcsin x -

46. cot(arctan x)

48. sin(arccos x)

50. tan( arccos -x)

5

( x )

52. csc arctan -0)

16. 17.

arccos( ), 0


14. (a) arcsin( - 2 (b) arctan - -3

15. (a) arcsin(- 1) (b) arccos 1

In Exercises 16 and 17, determine the missing coordi-nates of the points on the graph of the function.

= arccos x

/6)

1.5

In Exercises 18-23, use a calculator to approximate the value of the expression. (Round your answer to two decimal places.)

In Exercises 24 and 25, use a graphing utility to graph f, g, and y = x in the same viewing window to verify geometrically that g is the inverse of f. (Be sure to properly restrict the domain of f.)

24. f (x) = tan x, g(x) = arctan x

25. f (x) = sin x, g(x) = arcsin x

In Exercises 26-29, use an inverse trigonometric function to write 0 as a function of x.

10

x + 2

27.

29.

x + 1

26.

28.

In Exercises 30-37, use the properties of inv( tions to evaluate the expression.

31. tan(arctan 35)

33. sin[arcsin(- ij.1

( 35. arccos cos 1 -17 2

37. arcsin(sin 7-71 4

e expre, Er result.

In Exercises 46-53, write an algebraic expre is equivalent to the expression. (Hint: Sketc triangle, as demonstrated in Example 7.)

In Exercises 54 and 55, use a graphing utility to graph f and g in the same viewing window to verify that the two are equal. Explain why they are equal. Identif) any asymptotes of the graphs.

54. f (x) = sin(arctan 2x), g(x) = 1 + 4x

.x 2 55. f (x) = tan( arccos -

x) g(x) =

2

In Exercises 56-59, complete the equation.

56. arctan-14

= arcsin( ), x * 0 x

./36 - x 2 57. arcsin =

6

In Exercises 38-45, find the exact value of tt 1.5 sion. Use a graphing utility to verify yot

(Hint: Make a sketch of a right triangle.)

Ese

sion that h a right

1)]

h

2x

30. sin(arcsin 0.7)

32. cos[arccos(- 0.3)]

34. arcsin(sin 37r)

36. arctan(tan 11 6

/r)


of inverse funt‘

n 35)

1(— 0.1)]

DS 72.

1

177r\ 4

te of the expires, fy your result. le.)

1 12

1( —i)] M

expression that Sketch a right

7.)

1.

1(x — 1)]

n-

1

n x-12 )

r

utility to graph verify that the

equal. Identify

2x + 4x2

- X2

don.

0 < x < 6

), Ix — < 2

/0 Exercises 60-67, use a graphing utility to graph

the function.

61. y = arcsin 60.3 2 arccos x 2

62. f arcsin(x — 2) 63. g(t) = arccos(t + 2)

65. f(x) = 7r + arctan x 64. f (x) arctan 2x

66. h(v) tan(arccos v) 67. f(x) = arccos

4

in Exercises 68 and 69, write the given function in

terms of the sine function by using the identity

A = ./A2 + B2 sin wt + arctan — . A cos tot + B sin ag

Use a graphing utility to graph both forms of the

function. What does the graph imply?

68. f -= 3 cos 2t + 3 sin 2t

69. f(t) = 4 cos 7rt + 3 sin m

70. Docking a Boat A boat is pulled in by means of a winch located on a dock 10 feet above the deck of the boat. Let 0 be the angle of elevation from the boat to the winch and let s be the length of the rope from the winch to the boat.

(a) Write 0 as a function of s.

(b) Find 0 when s = 52 feet and when s = 26 feet.

71. Photography A television camera at ground level is filming the lift-off of a space shuttle at a point 750 meters from the launch pad. Let 0 be the angle of ele-vation to the shuttle and let s be the height of the shuttle.

(a) Write 0 as a function of s.

(b) Find 0 when s = 400 meters and when s = 1600 meters.

FIGURE FOR 71

72. Rock Salt Different types of granular substances naturally settle at different angles when stored in cone-shaped piles. This angle 0 is called the angle of repose. When rock salt is stored in a cone-shaped pile 11 feet high, the diameter of the pile's base is about 34 feet. (Source: Bulk-Store Structures, Inc.)

(a) Find the angle of repose for rock salt.

(b) How tall is a pile of rock salt that has a base diameter of 40 feet?

73. Granular Angle of Repose When whole corn is stored in a cone-shaped pile 20 feet high, the diame-ter of the pile's base is about 82 feet.

(a) Find the angle of repose for whole corn.

(b) How tall is a pile of corn that has a base diame-ter of 100 feet?

74. Photography A photographer is taking a picture of a 3-foot painting hung in an art gallery. The camera lens is 1 foot below the lower edge of the painting. The angle f3 subtended by the camera lens x feet from the painting is

3x 13 = arctan X2 + 4, x > 0.

(a) Use a graphing utility to graph as a function of x.

(b) Move the cursor to approximate the distance from the picture when 0 is maximum.

(c) Identify the asymptote of the graph and discuss its meaning in the context of the problem.

59. arccos 2

3 = arcsin( ccos s/ x2 _ 2x + 10

x — 2 = arctan(


0 75. Area In calculus, it is shown that the area of the region bounded by the graphs of y = 0, y= 1/(x2 + 1), x = a, and x = b is

Area = arctan b — arctan a.

Find the areas for the following values of a and b.

(a) a = 0, b = 1 (b) a = — 1, b = 1

(c) a = 0, b = 3 (d) a = — 1, b = 3 •

. 1 57r arcsin — = —

2 6

76. Angle of Elevation An airplane flies at an altitude of 5 miles toward a point directly over an observer. Consider 0 and x as shown in the figure.

(a) Write El as a function of x.

(b) Find 0 when x = 10 miles and x = 3 miles.

5 mi

77. Security Patrol A security car with its spotlight on is parked 20 meters from a long warehouse. Consider 0 and x as shown in the figure.

(a) Write 0 as a function of x.

(b) Find 0 when x = 5 meters and when x = 12 meters.

Synthesis

. 57r 1 78' sin 6 — 2

=>

79. arctan x — arcsin x arccos x

80. Define the inverse cotangent function by restri the domain of the cotangent function to the inte (0, 7r), and sketch the inverse function's graph,

81. Defme the inverse secant function by restricting tk domain of the secant function to the interval [0, 7r/2) and (7r/2, id, and sketch the inverse Nile tion's graph.

82. Define the inverse cosecant function by restrietitt the domain of the cosecant function to the interval' [— 7r/2, 0) and (0, '7121 and sketch the inverse fun( tion's graph.

83. Use the results of Exercises 80-82 to evaluate thc following without using a calculator.

(a) arcsec (b) arcsec 1

(c) arccot( ) (d) arccsc 2

In Exercises 84-89, prove the identity.

84. arcsin(—x) = — arcsin x

85. arctan( = — arctan x

86. arccos(—x) = 7r — arccos x

1 7r 87. arctan x + arctan —

x = x> 0

IT 88. arcsin x + arccos x =

2

Review

In Exercises 90-93, evaluate the sine, cosine, and tan.

gent of the angle without using a calculator. m

90. 840° 91. —585°

92. 177r

93. 197r

3 —

4

In Exercises 94-97, sketch the graph of the function.

(Include two full periods.) Use a graphing utility to

verify your result.

89. arcsin x = arctan — x2

So, c

tan A

So, a =

cos A

4

Figure 4.70

Solution Because C solve for e

APPIIC

In this se

and C angles b

EXAM

Solve the

Recall fro the horizon the angle f an angle of

Observer

True or False? In Exercises 78 and 79, determine whether the statement is true or false. Justify your answer.

1 94. y = —

2 sin(x + 7r) 95. y = 2 — cos(x

96. y = tan(x + 11=2 1

97. y = 4 cot —2x

Figure 4.71

Observer Horizontal I Angle of •

depression

4.8 • Applications and Models 355

n by restrictinu to the intetv:,

n s graph. '

>r restricting the

) the intervals Le inverse fune_

by restricting to .the intervals ne inverse func.

to evaluate the

1

2

mine, aid tan-ator.

i ti°ns Involving Right Triangles

•APFlca

this section, the three angles of a right triangle are denoted by the letters A, B, III

C (where C is the right angle), and the lengths of the sides opposite these

:ILL:iles by the letters a, b, and c (where c is the hypotenuse).


Solve the right triangle shown in Figure 4.70.

A b = 19.4

Figure 4.70

Solution Because c = 90°, it follows that A + B = 90° and B = 90° — 34.2° = 55.8°. To

solve for a, use the fact that

opp a tan A = = —

adj b

So, a = 19.4 tan 34.2 13.18. Similarly, to solve for c, use the fact that

adj b c=

Recall from Section 4.3 that the term angle of elevation denotes the angle from the horizontal upward to an object and that the term angle of depression denotes the angle from the horizontal downward to an object. An angle of elevation and an angle of depression are shown in Figure 4.71.

You Should Learn:

How to solve real-life prob-

lems involving right triangles • How to solve real-life prob-

lems involving directional bearings

• How to solve real-life prob-lems involving harmonic motion


You can use trigonometric func-tions to model and solve real-life problems. For instance, Exercise 60 on page 365 shows you how a trigonometric function can be used to model the harmonic motion of a buoy.

^->, a = b tan A.

cos A = = hyp c

So, c =19.4—

23.46. cos 34.2°

cos A*

Observer

Figure 4.71

f the futxtio11.

hing utility to

— cos(x + Tr)

rot 2

Object

Angle of elevation

• Horizontal Object

c= 110 f


EXAMPLE 2 Finding a Side of a Right Triangle

A safety regulation states that the maximum angle of elevation for a rescue lad-der is 72°. If a fire department's longest ladder is 110 feet, what is the maximum safe rescue height?

Solution A sketch is shown in Figure 4.72. From the equation sin A = a/c, it follows that

a = c sin A = 110 sin 72° 104.6.

So, the maximum safe rescue height is about 104.6 feet above the height of the fire truck.

Trig Oil

surve‘;

A bearin north-so

Figure 4'

EXAMPLE 3 Finding a Side of a Right Triangle Aik Figure 4.72

At a point 200 feet from the base of a building, the angle of elevation to the bot-tom of a smokestack is 35°, arid the angle of elevation to the top is 53°, as shown in Figure 4.73. Find the height s of the smokestack alone.

Solution Note from Figure 4.73 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

a = 200 tan 35°.

Now, in the larger right triangle, use the equation

tan 53° = a + s

11101111

to conclude that s = 200 tan 53° — a. So, the height of the smokestack is

s = 200 tan 53° — a = 200 tan 53° — 200 tan 35° 125.4 feet. Figure 4.73

EXAMPLE 4 Finding an Angle of Depression

A swimming pool is 20 meters long and 12 meters wide. The bottom of the pool has a constant slant so that the water depth is 1.3 meters at the shallow end and 4 meters at the deep end, as shown in Figure 4.74. Find the angle of depression of the bottom of the pool.

Solution Using the tangent function, you see that

tan A = —opp

= —2.7

= 0.135. adj 20

1.3 m

A

So, the angle of depression is Figure 4.74

A = arctan 0.135 0.13419 radian 7.69°.

200

20 m

Angle of depression

53°

'II

5,

A

The angle bearing of t

N78.1

Pinally froll

c sin

20 n

_

Figure 4.76

Solution In triangle can be det

b = 2

In triangle

tan A

(a)

Figure 4.75

EXAMP

A ship le hour). At the ship's

35'

Distance from port

N 78.18° W. Bearing

Finally, from triangle ACD, you have sin A = b/c, which yields

b 20 sin 36° C — 57.4 nautical miles. sin A sin 11.82°

(b) (c)


TrigonoMetry and Bearings

s,oeying and navigation, directions are generally given in terms of bearings.

I; bearing measures the acute angle a path or line of sight makes with a fixed

ortn_south line, as shown in Figure 4.75. For instance, the bearing of S 350 E in

1; iatire 4.75(a) means 35 degrees east of south.

350

(a) Figure 4.75

EXAMPLE 5 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. the ship changes course to N 54° W, as shown in Figure 4.76. Find

the ship's bearing and distance from the port of departure at 3 P.M.

54°

40 nm = 2(20 nm)

Figure 4.76

Solution In triangle BCD, you have B = 90° — 54° = 36°. The two sides of this triangle can be determined to be

b -= 20 sin 36° and d = 20 cos 36°.

In triangle ACD, you can find angle A as follows.

20 sin 36° tan A — 0.2092494

d + 40 20 cos 36° + 40

A •-=-•-• arctan 0.2092494 0.2062732 radian 11.82°

The angle with the north—south line is 90° — 11.82° = 78.18°. Therefore, the bearing of the ship is

3

tee of pression

1.3m

2.7 In

45° 80.

— 200 ft —01

S 35° E N 80°W N 45° E

—10 cm

- 0 cm -0 cm

10 cm 10 cm


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 bobbing up and down on the end of a spring, as shown in Figure 4.77. Suppose that 10 centimeters is the maximum distance the ball moves vertically upward or downward from its equilibrium (at-rest) posi-tion. 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 fric-tion or air resistance, the ball would continue to move up and down in a uniform and regular manner.

Equilibrium Maximum negative Maximum positive displacement displacement

Figure 4.77

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.

EXAM

Write t

4.77 vvi

Becaus

Moreov 4, you h

Am

Peni

Conseq

d=

Note th tially m

Freq

One illus the wave The way Shown in bob is att from the monic m

Figure 4.78

x

Figure 4.79


Olp nt that moves on a coordinate line is said to be in simple harmonic

l'uotion if its distance d from the origin at time t is given by either

------ a sin wt or d = a cos wt

where a and ware real numbers such that &i > 0. The motion has amplitude I al,

period 2ir I a), and frequency co/27r.

EXAMPLE 6 Simple Harmonic Motion

Write the equation for the simple harmonic motion of the ball illustrated in Figure

4.77, where the period is four seconds. What is the frequency of this motion?

Solution Because the spring is at equjlibrium = 0) when t = 0, you use the equation

d = a sin wt.

Moreover, because the maximum displacement from zero is 10 and the period is

4, you have

Amplitude = I al = 10

Period = 4 7r 2

Consequently, the equation of motion is

d= 10 sin —7r t. 2

Note that the choice of a = 10 or a = — 10 depends on whether the ball ini-tially moves up or down. The frequency is

co Frequency = —

27r

7r/2 27r

1 = —

4 cycle per second.

One illustration of the relationship between sine waves and harmonic motion is the wave motion that results when a stone is dropped into a calm pool of water. The waves move outward in roughly the shape of sine (or cosine) waves, as shown in Figure 4.78. 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 har-monic motion, as shown in Figure 4.79.


EXAMPLE 7 Simple Harmonic Motion

Given the equation for simple harmonic motion

d = 6 cos 7rt 4

a. Use the maximum feature of the graphing utility to estimat, maximum displacement is 6.

that the maximum displacement from the point of equiiib to 37r/4 3 rium y = 0 is 6, as shown in Figure 4.80(a).

b. Frequency = cycle per unit of 27r 27r 8 b. The period is the time for the graph to complete one Cycle

time which is x 2.667, as shown in Figure 4.80(a). you can

37r estimate the frequency as follows. c. d = 6 cos[74 (4)] = 6 cos 37r = 6(— 1) = —6

d. To fmd the least positive value of t for which d = 0, solve the equation

d = 6 cos —37rt = 0 4

to obtain

3'n- 7r 37r 57r 2 10 2

— t = —, —2

, —2

, • • • WEIP t — , 2, , • 4 3 3

So, the least positive value of t is t = i.

find (a) the maximum displacement, (b) the frequency, (c) the value of d when t = 4, and (d) the least positive value of t for whieh d = 0.

Algebraic Solution Graphical Solution

The given equation has the form d = a cos cot, with Use a graphing utility set in radian mode to graph a = 6 and co = 37r/4.

37r y = 6 cos-4 x.

a. The maximum displacement (from the point of equilibrium) is given by the amplitude. So, the

Writing About Math Radio Waves

-a (a)

Figure 4.80

y= 6 cos 4

—4.7 \

Maximum =6 X=2.6667

Exercis

the figure I

places.)

A = 2-

3. g 71

5. 0 == 6.

7. b = 18

9. A == 1

10.B = 6 5

a

In Exercis triangle sh answers to

11. 0 = 52

12. 0 = 18

13. 0= 41

14. 0 = 72

1 Frequency —

2.667 0.375 cycles per unit of time

c. Use the value or trace feature to estimate that the value ol y when x = 4 is y = —6.

d. Use the zero or root feature to estimate that the least pdA tive value of x for which y = 0 is x 0.6667, as shown Figure 4.80(b).

15. Length ated by horizon.

(a) Wril

(b) Use

y = 6 cos —3rx 4

4.7 4.7

Zero X=.6667 =0

-a (b)

Many different 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. The letters AM stand for amplitude modulation. An FM radio signal has its frequency modified in order to carry sound, hence the term frequency modulation. Of the two graphs at the left, one shows an AM wave and the other shows an FM wave. Which is which? Write a short paragraph explaining your reasoning.

(b)

(a)

II

10° 20° 30° 40° 50°

0 80° 60° 65° 70° 75.

(c) The angle measure increases in equal increments in the table. Does the length of the shadow change in equal increments? Explain.

17. Height A ladder of length 20 feet leans against the side of a house. The angle of elevation of the ladder is 0.

(a) Write the height h of the ladder as a function of 0.

(b) Use a graphing utility to complete the table.

10° 20° 30° 40° 50°

4.8 *Applications and Models 361

graph

utility to estima !, point of equili (a).

mplete one cycle, 4.80(a). You can

mit of time un

that the value of 1111

hat the least posi-5667, as shown in

-x

Exercises 1-10, solve the right triangle shown in In the

ognre below. (Round your answers to two decimal

places.)

A = 25°, b = 10 2. B = 56°, c = 15

3. B = 710, b = 24 4. A = 7.4°, a = 40.5

5. a 6, b = 16 6. a = 25, c = 35

7. b 18, c = 42 8. b = 1.72, c = 8.35

9. A 12° 15 c = 430.5

10.B = 65° 12', a -= 14.2

In Exercises 11-14, find the altitude of the isosceles triangle shown in the figure below. (Round your answers to two decimal places.)

11.0 = 52°, b = 4 inches

12.0 = 18°, b = 10 meters

13.0 = 41.6°, b = 14.2 feet

14.0 = 72.94°, b = 3.36 centimeters

(c) The angle measure increases in equal incre-ments in the table. Does the length of the shadow change in equal increments? Explain.

16. Length of a Shadow A shadow of length L is cre-ated by an 850-foot building when the sun is 0° above the horizon.

(a) Write L as a function of 0.


-8

15. Length of a Shadow A shadow of length L is cre-ated by a 60-foot silo when the sun is 0° above the horizon.

(a) Write L as a function of 0. (b) Use a graphing utility to complete the table.

18. Height The length of a shadow of a tree is 130 feet when the angle of elevation of the sun is 0°.

(a) Write the height h of the tree as a function of 0.


0 10° 15° 20° 25° 30°

0

II

•

4000 mi 'Angle of depression

Earth

4150 mi

door b4,

ackboai.„

"°bien., es.

an ericii.


19. Height From a point 50 feet in front of a church, the angles of elevation to the base of the steeple and the top of the steeple are 350 and 47° 40 respec-tively.

(a) Draw right triangles that give a visual represen-tation of the problem. Label the known quantities and the unknown height of the steeple.

(b) Use a trigonometric function to write an equa-tion involving the unknown quantity.

(c) Find the height of the steeple.

20. Height From a point 100 feet in front of a public library, the angles of elevation to the base of the flag-pole and the top of the flagpole are 28° and 39° 45', respectively. The flagpole is mounted on the front of the library's roof. Find the height of the pole.

21. Depth of a Submarine The sonar of a navy cruiser detects a nuclear submarine that is 4000 feet from the cruiser. The angle between the water level and the submarine is 31.5°. How deep is the submarine?

24. Angle of Elevation The height of an out ketball backboard is 121 feet, and the I casts a shadow 17i- feet long.

(a) Draw a right triangle that represents the Label the known and unknown quantiti

(b) Use a trigonometric function to write tion involving the unknown quantity.

(c) Find the angle of elevation of the sun.

25. Angle of Depression A television satellite is tid‘ eling in a circular orbit 150 miles above the

surface of the earth. Find the angle of depression from the television satellite to the horizon. Assume that the radius of the earth is 4000 miles.

Orbit

26. 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 21- miles offshore.

27. Airplane Ascent When an airplane leaves the way, its angle of climb is 18° and its speed is 275 per second. Find the plane's altitude after 1 minute

28. Airplane Ascent How long will it take th Exercise 27 to climb to an altitude of 10 16,000 feet?

e plane in 000 feet?

31.

32.

33.

Nay how hov fron

Nav bear man mile

Sun acro The poin bear

a th

22. Height of a Kite A 100-foot line is attached to a kite. When the kite has pulled the line taut, the angle of elevation to the kite is approximately 50°. Approximate the height of the kite.

23. Angle of Elevation An engineer erects a 75-foot vertical cellular-phone tower. Find the angle of ele-vation to the top of the tower from a point on level ground 95 feet from its base.

29. Mountain Descent • A sign on the roadway at the

top of a mountain indicates that for the next 4 mile,

the grade is 9.5°. Find the change in elevation f car descending the mountain.

30. Mountain Descent A road sign at the top of s

mountain indicates that for the next 4 miles the glad'

is 12%. Find the angle of the grade and the ehallgel' elevation for a car descending the mountain.

35. Navi souti-

port,


[an outdoor b

d the backb Oard

ents the probie quantities. 111' o write an

Nua-

ie sun. 4 1 satellite is tray„ tbove the surface iression front the Assume that the

;le of ression

;le of depression feet above water rifles offshore.

.e leaves the run-speed is 275 feet after 1 minute.

take the plane in ; of 10,000 feet?

.t roadway at the the next 4 miles n elevation for a

at the top of a

4 miles the grade md the changeln lountain.

Navigation An airplane flying at 550 miles per

hour has a bearing of N 58° E. After flying 1.5 hours, how far north and how far east has the plane traveled

from its point of departure?

Navigation A ship leaves port at noon and has a bearing of S 290 W. If the ship sails at 20 knots, how

many nautical miles south and how many nautical

miles west will the ship have traveled by 6:00 P.m.?

surveying A surveyor wishes to find the distance 33.

across a swamp. The bearing from A to B is N 32° W.

The surveyor walks 50 meters from A, and at the

point C the bearing to B is N 68° W. Find (a) the

bearing from A to C and (b) the distance from A to B.

34.Location of a Fire Two fire towers are 30 kilome-ters apart, tower A being due west of tower B. A fire is spotted from the towers, and the bearings from A and B are E 14° N and W 34° N, respectively. Find the distance d of the fire from the line segment AB.

W<IkE

4111111

- - ' • d , 34°( •

30 km

35.Navigation A ship is 50 miles east and 35 miles south of port. If the captain wants to sail directly to port, what bearing should be taken?

36. Navigation A plane is 160 miles north and 85 miles east of an airport. If the pilot wants to fly directly to the airport, what bearing should be taken?

37. Distance Between Ships An observer in a light-house 350 feet above sea level observes two ships directly offshore. The angles of depression to the ships are 4° and 6.5°. How far apart are the ships?

(not to scale)

38. Distance Between Towns A passenger in an air-plane flying at an altitude of 10 kilometers sees two towns directly to the left of the plane. The angles of depression to the towns are 28° and 55°. How far apart are the towns?

•5 '128°

•

10 km

39. Altitude of a Plane A plane is observed approach-ing your home and you assume its speed is 550 miles per hour. If the angle of elevation of the plane is 16° at one time and 57° 1 minute later, approximate the altitude of the plane.

40. Height of a Mountain While traveling across flat land, you notice a mountain directly in front of you. The angle of elevation to the peak is 2.5°. After you drive 18 miles closer to the mountain, the angle of elevation is 10°. Approximate the height of the mountain.

2.


mz mt tan a = 1 + m2mi

where m1 and m2 are the slopes of L1 and L2 , respec-tively. (Assume me% * —1.)

47. Geometry A regular pentagon (a Pentz gon congruent sides and angles) is inscribed in n

— Circle radius 24 inches. Find the length of the sides pentagon.

48. Geometry A regular hexagon (a hexagon With eoli gruent sides and angles) is inscribed in s circle radius 24 inches. Find the length of the Sid

es of hexagon.

41. 3x — 2y = 5

x+ y = 1

42. 2x + y = 8

x — 5y = —4 Trusses In Exercises 49 and 50, find the 1 all the unknown members of the truss.

engths of

43. Geometry Determine the angle between the diago-nal of a cube and the diagonal of its base, as shown in the figure.

49.

35° 10 10

50.

6 ft

6 ft

FIGURE FOR 43 FIGURE FOR 44 ft-0-1 F 36 ft

Harmonic Motion In Exercises 51-54, for the simple harmonic motion described by the trig( nometric function, find (a) the maximum displacement, (b) tlu frequency, and (c) the least positive value of t for which d = 0. Use a graphing utility to verify your results.

52. d = cos 20711

54. d = 4 sin 7927 T.t

Harmonic Motion In Exercises 55-58, find a model

for simple harmonic motion satisfying the specified

conditions.

Period

sec

(

i

sec

.8 sec

10 sec

59. Tuning Fork A point on the end of a tuning 0

moves in the simple harmonic motion de ciibed

d = a sin wt. Find (o given that the tuning tb

for middle C has a frequency of 264 vibi ations pe

second.

44. Geometry Determine the angle between the diago-nal of a cube and its edge, as shown in the figure.

45. Wrench Size Express the distance y across the flat sides of a hexagonal nut as a function of r, as shown in the figure.

X-0-1

46. Bolt Circle The figure shows a circular sheet of diameter 40 centimeters. The sheet contains 12 equally spaced bolt holes. Determine the straight-line distance between the centers of the bolt holes.

40 cm

51. d = 4 cos 87rt

53. d = sin 1407t

55.

56.

57.

58.

Displacement (t = 0)

0

0

3 in.

2 ft

Amplitude

8 cm

3m

3 in.

2 ft

(a) U

(b)

(c) D

Nunz high age b level five i coast

Geometry In Exercises 41 and 42, find the angle a between two nonvertical lines L1 and L2. The angle a satisfies the equation


pentagon [bed in a ciro f the sides of

exagon With 0 ed in a circle f the sides of

Wave Motion A buoy oscillates in simple har-monic motion 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 it is at its high

point at t = 0.

d the lengths 3.

Springs A weight stretches a spring 1.5 inches. The weight is pushed 3 inches above the point of equilibrium and released. Its motion is modeled by

y -4 cos 16t, t > 0 4, for the simple

trigonometri acement, (b) the

value of t fo to verify your

is 207rt

58, find a mo ng the spec

7927rt

where y is in feet and t is in seconds.

(a) Use a graphing utility to graph the function.

(b) What is the period of the oscillations?

(c) Determine the first time the weight passes the point of equilibrium (y = 0).

Numerical and Graphical Analysis A 2-meter-high fence is 3 meters from the side of a grain stor-age bin. A grain elevator must reach from ground level outside the fence to the storage bin. The objec-five is to determine the shortest elevator meeting the constraints.

le Period

2 sec

6 sec

1.8 sec

10 sec

of a tuning fork

ion described by

the tuning fork

64 vibrations per

(a) Complete four rows of the table.

0 Li L2 Li + L2

0.1 2

323.0

sin 0.1 cos 0.1

0.2 2 3

13.1 sin 0.2 cos 0.2

(b) Use a graphing utility to generate additional rows of the table. Use the table to estimate the minimum length.

(c) Write the length L as a function of 0.

(d) Use a graphing utility to graph the function. Use the graph to estimate the minimum length. How does your estimate compare with that of part (b)?

63. Data Analysis The times S of sunset (Greenwich Mean Time) at 40° north latitude on the 15th of each month are: 1(16:59), 2(17:35), 3(18:06), 4(18:38), 5(19:08), 6(19:30), 7(19:28), 8(18:57), 9(18:09), 10(17:21), 11(16:44), 12(16:36). The month is repre-sented by t, with t = 1 corresponding to January. A model (where minutes have been converted to the decimal part of an hour) for this data is

S(t) = 18.09 + 1.41 sin(716 + 4.60).

(a) Use a graphing utility to graph the data points and the model in the same viewing window.

(b) What is the period of the model? Is it what you expected? Explain.

(c) What is the amplitude of the function? What does it represent in the model? Explain.

64. Numerical and Graphical Analysis The cross sec-tions of an irrigation canal are isosceles trapezoids where the length of three of the sides is 8 feet. The objective is to find the angle 0 that maximizes the area of the cross sections.

8 ft

Equilibrium High point

3.5 ft

Low point

On over

roach


Base I Base 2 Altitude Area

8 8 + 16 cos 100 8 sin 10° 22.1

8 8 + 16 cos 20° 8 sin 20° 42.5

(a) Complete six rows of the table.

(b) Use a graphing utility to generate additional rows of the table. Use the table to estimate the maxi-mum cross-sectional area.

(c) Write the area A as a function of O.

(d) Use a graphing utility to graph the function. Use the graph to estimate the maximum cross-sectional area. How does your estimate compare with that of part (b)?

65. Data Analysis The tabl9 shows the average sales S (in millions of dollars) of an outerwear manufac-turer for each month t, where t = 1 represents Jan-uary.

t 1 2 3 4 5 6

S 13.46 11.15 8.00 4.85 2.54 1.70

t 7 8 9 10 11 12

S 2.54 4.85 8.00 11.15 13.46 14.30

(a) Create a scatter plot of the data.

(b) Fmd a trigonometric model that fits the data. Graph the model on your scatter plot. How well does the model fit?

(c) What is the period of the model? Do you think it is reasonable given the context? Explain your reasoning.

(d) Interpret the meaning of the model's amplitude in the context of the problem.

66. Numerical and Graphical Analysis Consider the shaded region outside the sector of the circle of radius 10 meters and inside the right triangle.

10

(a) Write the area A of the region as a function of O. Determine the domain of the function.

(b) Use a graphing utility to complete the ta

61 0 0.3 0.6 0.9 1.2 1.5

A

(c) Use a graphing utility to graph the functi the appropriate domain.

(d) What does the area approach as 7r/2?

Synthesis

True or False? In Exercises 67 and 68, determine whether the statement is true or false. Justify yo answer.

67. In the right triangle shown below, a = tan 41

A 22.56 C

68. N 240 E means 24 degrees north of east.

Review

In Exercises 69-72, solve the equation. (Round your answer to three decimal places.)

300 69. e2x = 54 70. -= 100

1 + e- x

71. ln(x2 + 1) = 3.2 72. 1og8x + log& -

In Exercises 73-76, use a calculator to approximate the value of the expression. Round your answer to two

decimal places.

73. arccos 0.13 74. arcsin 0.96

75. arcsin(- 0.11) 76. arctan(- 12)

app

22.56

- l)=

what

section

0 How

O How'

o How t

section

fl How t(

o How t

O How a

O How t

Section

O How t

O How t

O How t

O How t

Section

O How t

O How t

O How t

Section

O How ti cosine

O How to O How to

Section

O How to O How to O How to

Section 4

O How to O How to

Section z' F-1 How to I O How to

C3 How to

23. 171° 21. 8° 22. 94°

17. 40°

19. -110°

18. 190°

20. -405°

2.

30. 25. .11 32. - 327.93°

al In Exercises 1-4, estimate the angle to the near-est one-half radian.

In Exercises 5-8, (a) sketch the angle in standard posi-tion, (b) determine the quadrant in which the angle lies, and (c) list one positive and one negative cotermi-nal angle.

In Exercises 9-12, find the complement (if possible) and supplement of each angle.

7F 9. -8

3 7F 11. To

In Exercises 13-16, convert the measure from radians to degrees. Round your answer to two decimal places.

5 7r 37r 13. ,T 14. -

15. -3.5 16. 1.55

In Exercises 17-20, (a) sketch the angle in standard position, (b) determine the quadrant in which the angle lies, and (c) list one positive and one negative coterminal angle.

137r 10.

18

27r 12.

21

In Exercises 25-28, convert the measure to dec. dec. degree form. Round your answer to

places.

25. 135° 16'65"

27. 5° 22' 53"

26. - 2340 40" 28. 280

two

08 '50"

In Exercises 29-32, convert the measure to Dom s form.

29. 135.290

31. -85.36°

In Exercises 33-36, convert the measure from degree to radians. Round your answer to four decimal plae

34. - 16.5°

36. 84° 15'

33. 480°

35. -330 45 '

37. Find the radian measure of the central ang cle with a radius of 12 feet that intercepts length 25 feet.

38. Find the radian measure of the central ang cle with a radius of 60 inches that intercept length 245 inches.

39. Find the length of the arc on a circle with a radius 20 meters intercepted by a central angle of 138°.

40. Find the radius of a circle with a central angle of IT

that intercepts an arc of length 8.5 kilometers.

41. Difference in Latitudes Assuming that sphere of radius 6378 kilometers, what is ence in latitude of two cities, one of wh kilometers due north of the other?

42. Angular Speed A car is moving at a miles per hour, and the diameter of its about 2i feet.

(a) Find the rotational speed of the wheels tions per minute.

(b) Find the angular speed of the wheels per minute.

earth is the cliff ch is 6

in radians


In Exercises 21-24, find the complement Of and supplement of each angle.

1.

3.

P Ossibi

24, 49„

e of a an arc

e of a s an arc

ate of 2 wheels

in revol

S

‘s

p7-4-125 ft-

• Review Exercises 369

ent (if possible)

24, 49e

sure to decimal Lo two decimal

6. - 234° 40" 8. 280° 8' ,,

sure to Wm 's,'

0. 25.8'

2. -327.93°

re from degrees decimal places.

4. -16.5°

6. 84° 15'

-al angle of a cir-ercepts an arc of

-al angle of a cir-tercepts an arc of

with a radius of .ngle of 138°.

Ltral angle of 7r/3 dlometers.

g that earth is a

/hat is the differ-of which is 600

at a rate of 28

of its wheels is

wheels in revolu-

vheels in radians

61.

"pit

43. t 3

circle that corresponds to the real number t. in Exercises 43-46, find the point (x, y) on the

44. t = -37r 4

46. t = - -47 57r 3 45, t 6

in Exercises 47-50, evaluate (if possible) the six igonometric functions of the real number.

tr 77"

48. t = -4

50. t = 27r

In Exercises 51-54, evaluate the trigonometric func-

tion using its period as an aid.

11 ir 51, sin 52. cos 47r

4

( 177r 53. sin - - 6

) 137 54. cos(

7) 3 )

In Exercises 55-58, use a calculator to evaluate the trigonometric function. Round to two decimal places.

55. cot 2.3 56. sec 4

57. cos -57r 3

58. tan( 1177") 6 )

In In Exercises 59-62, find the exact values of the six trigonometric functions of the angle 0 given in the figure.

59. 2

0 10

12

62.

In Exercises 63 and 64, use trigonometric identities to transform one side of the equation into the other.

63. csc 0 tan 0 = sec 0

cot 0 + tan 0 64. = sec2

cot 0 0

In Exercises 65-68, use a calculator to evaluate each function. Round your answer to four decimal places.

65. (a) cos 84° (b) sin 6°

66. (a) csc 52° 12' (b) sec 54° 7 '

67. (a) cos -4 7F

(b) sec 71.

68. (a) tan() (b) cot-

20

69. Altitude of Elevation Find Find the altitude of the triangle shown.

70. Width of a River An engineer is trying to deter-mine the width of a river. From point P, the engineer walks downstream 125 feet and sights to point Q. From this sighting, it is determined that 0 = 62°. How wide is the river?

131 In Exercises 71-76, find the six trigonometric functions of the angle 0 (in standard position) whose terminal side passes through the given point.

72. (- 4, - 6)

74. (4, -8)

75. i) 76. (x, 44, x > 0

77r - 6

27r - 3

60. 10

12

15 71. (12, 16)

73. (- 7, 2)

S x

108. f(x) = 2 cos x 110. f(x) = Z co

SaleS ity to

IS the mOnt Deter sales.

131.

132.

f (x) graph

133.

-9.42

In Ex tion.

135.

137. f

139. f

141. f

143.

144. f

145. f

147. f(

149. f

107. f(x) = 3 sin x

109. f(x) = sin x

111. 112.

25.12

113. 114.

y = -3.4 sin 2x

-6.28 628

6.28

the film. In Exercises 115-126, sketch the graph of tion.

-12.56

-4 -3

(Lr -2) 4'


In Exercises 77-80, find the remaining five trigono-metric functions of 0 satisfying the given conditions.

77. sec 0 = tan 0 < 0

78. tan 0 = - 1, sin 0 > 0

79. sin 0 = cos 0 < 0

80. cos 0 = sin 0 > 0

In In Exercises 107-110, sketch the graph function.

of tilt

In Exercises 111-114, find the period and ar tiPlituk

In Exercises 81-86, evaluate the trigonometric func-tion without using a calculator.

7T 81. tan -

77- 82. sec -

4 3

83. sin (- -537r) 84. cot(_r56

)

85. cos 4950 86. csc 2700

In Exercises 87-90, use a calculator to evaluate the trigonometric function. Round to two decimal places.

87. tan 33° 88. csc 105°

89. sec 121T

90. sin(- 7r) 5 9

In Exercises 91-94, find the reference angle for the given angle.

91. 264° 92. 635°

93. -67r - 94. 177T

3

5

In Exercises 95-102, evaluate the sine, cosine, and tangent of the angle without using a calculator.

95. 2400 96. 315°

97. -210° 98. - 3150

11 7T 99. - 100 .

6

4

101. - 102. - -7r

2

3

In Exercises 103-106, find the point (x, y) on the unit circle that corresponds to the given real number t. Use the result to evaluate sin t, cos t, and tan t.

103. -27r 104. -77T

3

4

105. -77r 106. -37T

6

4

4

6.28

115. f(x) = 3 cos 2= 116. f(x) = -2 sin Trx v

117. f(x) = 5 sin -2; 118. f(x) = 8 cos1\ 4

119. f (x) = - -5

cos-x

120. f(x) = - Sin - 2 4 2 4

121. f(x) = sin(x - 7r) 122. f(x) = 3 cos(x +

123. f(x) = 4 sin(x - 712 ) 124. f(x) = sin 77-x - 3

x 7r)125. f(x) = -3 cos (-2 - -4

126. f(x) = 4 - 2 cos(4x + 7r)

In Exercises 127-130, find a, b, and c for f(x) = a cos(bx - c) so that the graph of f matches

the graph shown.

127. 128. 3

12.56 -6.28

156. f(x) = 2x cos x 155. f(x) = ex sin 2x

157. f(x) = x sin 7TX

158. f(x) = 2.5e- x/4 sin 27x

arcsin 0.63

arccos (-0.12)

arctan 21

arctan 6.4

(b)

(b)

(b)

(b)

arccos 0.42

arcsin (-0.94)

arctan (-12)

arctan 0.81

9.42 -9.42

163.

164.

165.

166.

(a)

(a)

(a)

(a)

x + 3 x + 1

• Review Exercises 371

he graph

= 2 cos x _ - 4 cos x

and amplitude,

n 2 2

Jtx 2

5

6.28

-5

ph of the fun

= -2 sin 7rx

= 8 cos(--4

1 . 717C =

= 3 cos(x + 7r)

= sin 7TX - 3

5, and c for )11 of f matches

4

130.

6.28 -3

in Exercises 131 and 132, use a graphing util-

sa/es graph the sales function over one year, where S

it y to he sales (in thousands of units) and t is the time (in

i

ths) ;t with t = 1 corresponding to January. mon, Determine the months of maximum and minimum

sales.

131. s = 48.4 - 6.1 cos-731 6

132. S = 56.25 + 9.50 sin 771 6

In Exercises 133 and 134, find a and b for

f(x) = a tan bx so that the graph of f matches the

graph shown.

133. 3

YV

2.35

In Exercises 135-154, sketch the graph of the func-tion.

135. f(x) = 4

irx 137. f(x) = - 14 tan

139. f(x) = tan(x - 12-r)

141. f(x) =- 3 cot

143. f(x) = cot(x -

144. f(x) = 4 cot(x + IT4

150. f(x) = csc(2x + 7)

151. f(x) = 2 sec(x - 77-)

152. f(x) = -2csc(x -

153. f(x) = csc(3x -

154. f(x) -= 3 csc(2x +

In Exercises 155-158, sketch the graph of the function and the damping factor of the function by hand. Use a graphing utility to verify your graph.

1311 In Exercises 159-162, evaluate the expression without using a calculator.

159. (a) arcsin 1 (b) arcsin 4

160. (a) arcsin( - /1 2

(b) arcsin 2

161. (a) arccos 2 (b) arccos( - -2

-,/j)

162. (a) arctan( - 13-) (b) arctan(- 1)

In Exercises 163-166, use a calculator to approximate the value of the expression. (Round your answer to three decimal places.)

In Exercises 167 and 168, use an inverse trigonomet-ric function to write 0 as a function of x.

168.

6.28 145. f(x) = 41 sec x 147 f(x) = zti csc 2x

149. f(x) = sec(x -

= csc x

= sec 2=

20

136. f(x)

138. f(x)

140. f(x)

142. f(x)

146. f(x)

148. f(x)

A A A V 3

-3

134.

-2.35

= 2 + 2 tan

1 irx = - cot -

2 2

= 4 tan 71" X

= tan(x + 4

167.

172. csc(arcsin 10x) 2

171. sin(arccos 4 — x2

) In t his

of the

since 1 data on averag

readin sous:

D1))),In°

1974

329.7

1987

349.0

a. Ent Dra a lit

b. Fin data

Que 1. TN

le Ye

.

pa , de


In Exercises 169-172, write an algebraic expression for the given expression.

170. tan(arccos —x) 2

MI 173. Railroad Grade A train travels 3.5 kilometers on a straight track with a grade of 10 10 What is the vertical rise of the train in that distance?

3.5 km

177. Wave Motion A buoy oscillates in snip pie 1,4,

monic motion as waves go past. At a giver " is noted that the buoy moves a total of 6 fc its low point to its high point, returning to point every 15 seconds (see figure). Write E

.1

ltS high

et (raiii

tion that describes the motion of the buoy if at t it is at its high point.

n High point .:, i

ii

i

169. sec[arcsin(x — 1)]

6 ft 1°10'

Low point

174. Distance Between Towns A passenger in an airplane flying at an altitude of 37,000 feet sees two towns directly to the left of the airplane. The angles of depression to the towns are 32° and 76°. How far apart are the towns?

Town 1 Town 2

37,000 ft

175. Navigation An airplane flying at 490 miles per hour has a bearing of N 46° W. After flying 1.8 hours, how far north and how far west has the plane traveled from its point of departure?

176. Distance 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.

Synthesis

True or False? In Exercises 178-181, del ermine whether the statement is true or false. Justify your answer.

178. sin 60°

= sin 2° 179. tan[(0.8)2] tan2(08) sin 30°

180. y = sin 0 is not a function because

sin 30° = sin 150°.

181. tan--=37T —1 4

182. In your own words, explain the meaning c angle in standard position, (b) a negative a coterminal angles, and (d) an obtuse angle.

183. A fan motor turns at a given angular speed. Hott

does the speed of the tips of the blades change di fan of greater diameter is installed on the motor'

Explain.

f (a) an ngle, (r)

317 arctan(— 1) —

- 4

Equilibrium

The Interactive CD-ROM and

°

In

o

T. They also offer Chapter Pre-1 sts (

previous chapters) and Chapter post Tests, both of which have randomly

versions of this text provide ai the Chapter Tests and Cumulative

test key skills and concepts co

generated exercises with

diagrvered ost

swers

capabilities.

FIGURE FOR 3

—18.84

FIGURE FOR 18


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.

1. Consider the angle of magnitude 57r/4 radians.

(a) Sketch the angle in standard position.

(b) Determine two coterminal angles (one positive and one negative).

(c) Convert the angle to degree measure.

2. A truck is moving at a rate of 90 kilometers per hour, and the diameter of its wheels is 1 meter. Find the angular speed of the wheels in radians per minute.

3. Find the exact values of the six trigonometric functions of the angle 0 shown in the figure.

4. Given that tan 6+ = find the other five trigonometric functions of 0.

5. Determine the reference angle 0' of the angle 0 = 290° and sketch 0 and 0' in standard position.

6. Determine the quadrant in which 0 lies if sec 0 < 0 and tan 0> 0.

7. Find two values of 0 in degrees (0 < 360°) if cos 0 = (Do not use a calculator.)

8. Use a calculator to approximate two values of 0 in radians (0 < 27r) if csc 0 = 1.030. Round the result to two decimal places.

9. Find the five remaining trigonometric functions of 0, given that sec 0 = and tan 0 < 0.

In Exercises 10-15, graph the function through two full periods without using a graphing utility.

10. g(x) = -2 sin(x - 11. f (a) = tan 2a

12. f(x) = sec(x - 7r) 13. f(x) = 2 cos(ir - 2x) + 3

14. f(x) = 2c5+ + 15. f(x) = cot(x -

In Exercises 16 and 17, use a graphing utility to graph the function. If the function is periodic, find its period.

16. y = sin 27rx + 2 cos irx 17. y = 6e- °.12t cos(0.25t), 0 t 32

18. Find a, b, and c for the function f(x) = a sin(bx + c) so that the graph of f matches the graph at the right.

19. Find the exact value of tan(arccos i) without using a calculator.

In Exercises 20-22, sketch the graph of the function

20. f(x) = 2 arcsin(x) 21. f(x) = 2 arccos x 22. f(x) = arctan 2

23. A plane is 160 miles north and 110 miles east of an airport. If the pilot wants to fly directly to the airport, what bearing should he take?

24. From a 100-foot roof of a condominium on the coast, a tourist sights a cruise ship. The angle of depression is 2.5°. How far is the ship from the shore line?

x 0

9

6

3 :0, -5)

I I Irx

3 6 9

6

2 4 6 8 10 3

I 1 I -9 -6 -3

34. 35.

Answers to Odd-Numbered Exercises and Tests A95

17. 4 - 5i 18. + 6-1f0i 19. -1± 12i

5

20. 29 + 26i

21. f(x) = x5 - 5x4 + 19x3 - 71x2 + 48x + 144

22. 1, -3,±2i; f(x) = (2x - 1)(x + 3)(x + 2i)(x - 2i)

23.

Domain: x -5,2

x-intercept: 0)

Vertical asymptotes: x = -5, x = 2

Horizontal asymptote: y = 0

24. 6 25. -4 26. 10 27. -3 28. 6.733

29. 8772.934 30. 0.162 31. 51.743

32. 33.

-40

-50

- 60

- 70

- 80

- 90

x because ues of y.

;raph of y.

; the graph of

r, the graph of

10

-9

-12

Domain: 0 < x < 273

(c) 76 ft by 197 ft

49. 5.549(1.412)x

60

10

Chapter 4

Section 4.1 (page 291)

1.2 3.-3

5. (a) Quadrant I

7. (a) Quadrant IV

9. (a) Quadrant HI

11. (a)

(b) Quadrant TH

(b) Quadrant II

(b) Quadrant H

(b)

13. (a) (b) -3 - 3

- 6

-9

36. 1.892 37. 0.872 38. -7.484 39. -3.036

40. In x2

1 41. 3 In x + -2

ln(x - 5) 42. 1.242 + 5

43. 3.811 44. 12.8 45. 5.021 46. 401.429

47. 16.558 257r 237r

15. (a) 12 12

actor.

87r 47r (b)

57r 37r is 287r 327r

17. (a) 4 4 (") 15 15

48. (a) A(x) = 273x - x2

(b) 20,000

300

•

ectio

1. sin 0

-II -5 -4 -3

12

628

109.

--- .••••11.191

A96 Answers to Odd-Numbered Exercises and Tests

71- 27r 19. (a) Complement: Supplement: -

3 6

(b) Complement: none; Supplement: -7r 4

21. (a) Complement: none; Supplement: none

(b) Complement: none; Supplement: lir

23. 2100 25. -45°

27. (a) Quadrant II (b) Quadrant IV

29. (a) Quadrant III (b) Quadrant I

31. (a) (b)

85. 47r meters = 12.57 meters 87. 1141.02 miles 89. 0.094 rad 5.39° 91. T52- rad 23.87°

93. (a) 540° = 9.42 rad (b) 9000 = 15.71 rad

(c) 1260° 21.99 rad

95. (a) 448.2 rev/min

(b) 2816 rad/min

97. 20.167r in./sec

99. False. A radian is larger: 1 rad = 57.3°.

101. True. The sum of the angles is IT, or 180°.

103. Two angles in standard position are coterminal ane they have the same initial and terminal sides.

507r 105. -

3 square meters

107. (a) A = 0.4r2, r> 0; s = 0.8r, r> 0

8

cos

tan 0

csc

sec 0

cot 0

3.

33. (a) (b) 2

11. (0, -

The area function changes more rapidly for /->l because it is quadratic and the arc length function is linear.

13. sin

cos (b) A = 500, 0 < < 27r; s = 100, 0 < 0 < 27r

3 tan

35. (a) 412°, -308° (b) 324°, -396°

37. (a) 660°, -60° (b) 590°, - 130°

39. (a) Complement: 66° (b) Complement: none

Supplement: 156° Supplement: 54°

41. (a) Complement: 11° (b) Complement: none

Supplement: 101° Supplement: 30°

57T 7T 471-

43. (a) -71. (b) -6

45. (a) - -9

(b) - -3 6

47. 2.007 49. - 3.776 51. 11.205 53. - 0.014

55. (a) 270° (b) -210° 57. (a) 420° (b) -39°

59. 25.714° 61. 562.5° 63. -756°

65. - 114.592° 67. (a) 64.75° (b) - 124.5°

69. (a) 85.308° (b) -408.274°

71. (a) 280°36' (b) - 115° 48'

73. (a) 257° 49' 52" (b) - 205° 7' 8"

75. -56 rad 77. 41 rad 79. T83 rad 81. 2g rad

83. 147r inches 43.98 inches

17. sin(

cos(

tan(

21. sin

23.

a (

st

c

i

o

nn

s

. 3

cos

3 tan-

67. Answers will vary. 69. It is an even function.

71. f-1(x) = -V-4(x - 1) 73. f-1(x) = 2 - x, x < 2

75.

-4

77.

79. $108.63 81. (a) 300 (b) 570

tan(-1 = cot(-11 3 3 - 3

87r 27r 1 29. sin 571- = sin ir = 0 31. cos = cos =

33. cos(-37r) = cos(- 7r) = -1

35. sin(--4-97T) = sin T.77 -- 37. 'a` - -1 (b) -3 3

39. (a) -1 (b) -5 41. (a) t (b) -t

43. 0.7071 45. 1.0378 47. -0.1288

49. 1.3940 51. -1.4486 53. (a) -1 (b) -0.4

55. (a) 0.25, 2.89 (b) 1.82, 4.46 57. 0.7943

59. (a) 0.2500 foot (b) 0.0177 foot (c) -0.2475 foot

61. 0.0707 = cos 1.5 2 cos 0.75 = 1.4634

63. False. sin(- t) = -sin t means that the function is odd, not that the sine of a negative angle is a negative number.

65. (a) y-axis (b) sin t1 = sin(ir - t1)

(c) cos(7r - t1) = -cos ti


1. sin 6 = csc 0 =

cos 0 = sec 0 =

tan 0 = cot 0 =

••11.

115.


IT 7F 25. sin -2

= 1 csc -2 = 1

cos -2

= 0 sec -2 is undefined.

IT 7r -2 tan -

2 is undefined. cot = 0

-4 3

27. sin --11- = --2

( csc( 3 --7r) = - 213-- 3

-3 -4 -5

1 cos_-) = - sec(--1 = 2

3 2 3

(page 300) 5 3. sin 0 = --13

Section 4.2

al angles if 1. sin 0 =

COS 0 = - 177 COS 0 = ii 8

tan 0 = - -8 tan 0 = --I2 5

17 13

15

CSC 0 = d CSC 0 = -

8 sec 0 = -g sec 0 = --17

COt 9 = --I5 COt 0 = -2 8

110

7(-. 4 1) 9 (1

4)

- 1 - 2 ' 2 2' 2

11. (0, -1)

77- -4 r for r> 1 13. sink = -

2 15. sinE:)

1 = - 2

function is

4 2 cosql = L23 IT -fi

tan -4

= 1 tan(--:) = /3-3 7T

17. sin(_ -7411 = I-22 19. sin 6 = --2

117r 1

117r 4 cos(--7141 = -j- -22 cos =

6 2

117r 4 tan(-L:) = 1 tan =

6 3

i I I I , A 21. 2 3 4 5 sin (-- -'27r) = 1

(-3;) cos- =0

tan(- -2

371 is undefined.

311- -4, 23. sin -4- = csc -311. = 12

4

37r 12 37r cosT = -, . sec 7 -1. = --J

37T tan -3.7 = -1

4 cot 71: = -1

< 27r


8 3-1-ff 3. sin 0 = fi 5. sin 0 -

13

15 2-,/1. - cos 0 = cos 0 =

17 13

8 3 15

tan 0 = tan 0 = i

17 ./IJ csc 0 = csc 0 =

8 3

17 sec 0 = 15 sec 0 =

2

15 2 cot 0 = cot 0 =

8 3

1 3 7. sin 0 = 9. sin 0 = 3 5

2-12-, , 4 cos 0 = cos a = 3

3

./, tan 0 = 3 tan 0 =

4 4

5 csc 0 = 3 csc 0 =

3

3-12- • 5 sec 0 - sec 0 =

4 4

cot 6 =2..J cot 0 = 3

The triangles are similar The triangles are similar and corresponding sides and corresponding sides are proportional. are proportional.

-11T. 11. cos 0 =

6

tan 0 = 1 1

6 csc 0 = 3

6-I1J sec 0 -

cot e = 5

4

1 cos 0

tan 0 =

4

csc 0 - 15

-1F5 cot 0 -

15

4

13. sin 6 =

1. V13

15. sin 0 = 10

./170 cos 6 = lo

csc 6 = 3

sec 6

1 cot 6 = -3

4-1T7 17. sin 6 = 97

9-10 cos 0 = 97

tan 0 = -4 9

csc 0 = 4

-1§77 sec 0 =

9

1 19. (a)../ (b) -

2 (c)-- (d) -

2 3

1 212 21. (a) 3 (b) (c) (d) 3

23. (a) 4 (b) (c) ±-15 (d) -1

4 4

25.-31. Answers will vary.

1 33. (a)

2 ( ) -13- 3

35. (a) 1 (b) 2

37. (a) (b) 2 39. (a) 0.4226 (b)

41. (a) 1.3499 (b) 1.3432

43. (a) 5.0273 (b) 0.1989

45. (a) 1.1884 (b) 0.5463

ir 47. (a) 300 = -6 (b) 300 = -6

7r 71-

49. (a) 600 = -3 (b) 450 = -4

ir ir 51. (a) 60° = -3 (b) 450 = -4

53. (a) 55° 0.960 (b) 89° 1.553

55. (a) 50° = 0.873 (b) 25° = 0.436

57. 35-,./ 59. 61. 11.5 3

.4226

63. (a)

65. 160.0

67. (a)

(b) si

(c) 9.

69. 425.3

71. 6.57 c

sin 75

cos 7:

tan 7

csc 7

cot

sec ,0

75. True,

79.

cos

cos 0

0 and

81.

x-int

y-intt

• Quad


63. (a) (b) = 3 135

(c) 270 feet

t-K- 132 3

Not drawn to scale

65. 160.03 feet

67. (a) (9 20

Section 4.4 (page 320) 1. (a) sin 0 = 15 (b) sin 0 = -T7

4 COS 0 = COS 0 = 187 3

tan 0 =

tcansc ==1--1157 csc 0 =

sec 0 = sec 0 = - 1-87

cot= cot 0 = A 1

- 3. (a) sin 0 = - (b) sin 0 = -2

31 3 cos 0 = - - 2 cos 0 = -- 2-

3- 1 (b) sin 0 = = -

20 6

(c) 9.590 (d) 9.590

tan 0 = 3

csc 0 = -2

sec 0 = 3

tan 0 = -1

csc 0 =

sec 0 = -

1 4

77. True for all 0

S

69. 425.3 meters; 431.9 meters

71. 6.57 centimeters

73. sin 750 0.97

cos 750 0.26

tan 75° 3.73

csc 750 1.04

sec 75° •-=--• 3.86

cot 750 0.27

75. True, sec x = csc(90° - x)

79.

2

cot 0 = cot 0 = -1

5. sin 0 = T5- 7. sin 0 = 12 24

COS 0 = COS 0 = 24 tan 0 = tan 0 = - y

csc 8= csc e = -fi

sec 0 = -T25 sec 0 =

cot 0 = cot =

i 9. sm = 11. sn - • 29 85

b) 0.4226

0 0° 20° 40° 60° 80°

cos 0 1 0.9397 0.7660 0.5000 0.1736

sin(90° - 0) 1 0.9397 0.7660 0.5000 0.1736 2-,/2§. 2.-,

cos 0 = cos 0 = 29 85

9 tan 0 =_ 5

2 tan 0 = --

2 cos 0 = sin(90° -

0 and 90° - 0 are complementary angles.

81. 83. csc 0 = csc 0 = 5 9

sec 0 = sec 0 = 2 2

2 2 cot 0 _ = cot 0 _ =

5 9

x-intercept: -9

y-intercept: -9

x-intercept: -5i

y-intercept: 2

85. Quadrant II 87. Quadrant I

13. Quadrant III

19. sin 0 =

cos 0 =

tan 0 =°

csc 0 =

sec 0 =

cot 0 =

15. Quadrant II 17. Quadrant I

21. sin 0 = 17 cos 0 = 17

tan 0 = 8 CSC 0 = -7157

sec 0 = 8 COY 0 = -15


23. sin 9 = 25. sin 9 = 0

1 cos 9 =

47. 0' = 3.5 - IT 49. 0' = 3.68 - 71- 13.5384

tan 0 = -

2-13. csc 0 = 3 sec 0 = -2

101.

103. 0

105. 0 107. F

109. 0

cot 0 =

cos 0 = -I

tan 0 = 0

csc 0 is undefined.

sec 9 = -1

cot 0 is undefined. sin 0

sin(1 2/5-

27. sin 0 = - 2 29. sin 0 = - 5

15-

cos 0 =

s/2 51. sin(225°) = -

cos(225°) = --

2

53. sin( - 750°) =- 2

cos(-7501 = 2

tan 0 = -1

csc 0 =

sec 0 = -

cot 0 = -1

tan(225°) = 1 _ tan 0 = 2

csc 9 = - -2

sec 0 = -

1 cot 0 = -

2

tan( - 750°) =

31. -1 33. 0 35. 1 37. Undefined

39. 0' = 28° 41. 0' = 65°

,T3 55. sin(- 240°) = 57. sin T5Ir = ---y-

cos( - 240) = -.1 57T 1 cos T = -2-

tan( - 240°) = - -,/ 5 7T tan -

3 = -

-ii 1 59. sin( IT sin---) = - 61. si

117T n ,i. = - 27

117r cos-)( :f = cos -,T. = - - 2-

7T ./N 117r tan_)( = --T- tan 7, F. = -1

63. sin( - ) = 1 65. 0.1736 67. 1.7 21

cos( -Z) = C

43. 0' = 68° 7T 45. 0' = -3

( -L67) tan = -

69. -0.3420 71. 5.7588 73. 0.6052

75. 0.3640 77. -2.9238

79. (a) 30° = 150° = %-rr (b) 210° = 330°

81. (a) 60° = frr, 120° = (b) 135° = , 315°

SE

123. si

- 3 57r 77r 27r

83. (a) 150° = - 210° = - 6 6

(b) 120° = 240

85. 54.99°, 125.01° 87. 33.12°, 213.12°

89. 144.78°, 215.22° 91. 208.64°, 331.36°

93. 10°, 350° 95. 50°, 230° 97. -4 99. --

5 2

1.

3.

5.

7.

31. 33.

35. 2

6.28 6.28

-2

2 37.

6.28 6.28

-2

f

-3

Asymptote at x = 1

x-intercept at x = 2

119. x = 0.002

7

Answers to Odd-Numbered Exercises and Tests Al 01

O.5384

2

sn.

2

3

2

1.7321

117r 30°

7/r 15° =

47r , 240°

2

101. l•

103. (a) 25.2°F (b) 65.1°F (c) 50.8°F

105. (a) 12 miles (b) 6 miles (c) 6.9 miles

107. False. 0' 180° -

109. (a)

0 _

0° 20° 40° 60° 80°

sin 0 _

0 0.3420 0.6428 0.8660 0.9848

sin(180° - 0) 0 0.3420 0.6428 0.8660 0.9848

(b) sin 0 = sin(180° -

111. Answers will vary.

113.

4

3 tan 0 =

csc 0 = 3

sec 0 = 8

123. sin 0 =

cos ,-, = 24

CSC 0 = 24

sec 0=

cot 0 =

Section 4.,5 (page 330)

1. Period: 7r; Amplitude: 3

3. Period: 47r; Amplitude: I

5. Period: 2; Amplitude:


9. Period: ' Amplitude: 3 3


13. Period: Amplitude: 3

15. g is a shift of f sr units to the right.

17. g is a reflection of f about the x-axis.

19. g is a reflection in the x-axis and five times the amplitude of f.

21. g is the graph of f shifted 4 units up.

23. g has twice the amplitude of f

25. g is a horizontal shift off 7T units to the right.

27. Y 29.

Asymptote at y = 0

y-intercept at y =

117. x = -1.752

121. sin 6 = 73

8/7 cos 6 =

73

115.

2

(2,0)

2 3 4 5 6

-2

2 71.

2 73.

0.017

Al 02 Answers to Odd-Numbered Exercises and Tests

20

-4

Amplitude: 5

Period: IT

61.

- 0.017

-0.02

Amplitude: pi o-

Period: -26

63. f(x) = 4— 4 cos x 65. f(x) = 1 — 6 cos x

67. f(x) = —3 sin(2x) 69. f(x) = sin(x —

-2

IT 57r 77r 117r x =

6

-2

IT 77r x = +— ±—

_ 4' _ 4

75. (a) 2

H.0%4 -2

3.14

6.28

6.28

55. 57.

47. y 49.

51. 53.

45.

39. 41.

43.

AAAAA

3.14 3.14

-4

Amplitude: 2

. IT Period: —

2

4

Amplitude: 1

Period: 1

59.

- 3.14

0.02

6.28

6.28

(b) 6 seconds

(c) 10 cycles minute

12.57

77. 28

12

6.28 6.28

89.

91. 2

3.14

135

Period changes

(a) Even

(b) Even

(c) Odd

Maximum sales: June

Minimum sales: December

79. (a) so

1111 lllllll

(b) 7

•

• •

•

1111 12


-2

(b) Minimum height: 5 get

Maximum height: 55 feet

81. (a) 365 days. The cycle is one year.

(b) 30.35 gallons per day; Both terms of the model; Average of the minimum and maximum of the model

(c) 60

35

93. g is the periodic function and f is the parabola.

95. 97.

-40-30-20-10 I /10 20 30 40

/

Asymptotes: x = 0, y = 5 Asymptotes: x = -2, y = x - 9

99. -20° 101. -27.502°

Consumption exceeds 40 gallons per day from the beginning of May through part of September.

83. (a) p = 2.55 cos( 71 -12 + 3.45 6

The model fits the data fairly well.

85. True

87. 4

a=-3 -: I a= _ a= 4

-4

Amplitude changes


1. (g), 43r 3. (f), - 5. (b), 2 7. (e), 277-

2

-3 2 -2

9. y 11.

seconds

D cycles per dilute

6.28 6.28

OS X

14.1 HFH

57.

-6.28

Equal

f-30

3.

-25.13

Apprc

i5.

-18.85

Appro

7.

-3.14 L.,

Appro

9. d = 5 36

-36

L. As the decrew of pred

31. 4

14.1

33.

6.28 6.28

1.57 1.57

-4 Equal


-4

-2

3

4.71 4.71

35.

-4.71 4.71

3.14 3.14

17. 19.

23.

J 21.

1

- x

25. 27.

37.

- 3

- 8

39.

47. 2

3 3

49. 4

77r 37r 57r 41. - - 7T - 43. - - - - -

4 4 4 4 3 3 3'3 47r 277 27r 47r

Equivalent

51. (d); as x approaches 0, f (x) approaches 0.

53. (b); as x approaches 0, g (x) approaches 0.

55. 2

45. Even

-2

Not equivalent; y, is undefined at x = 0.

6.28 6.28

-4

-6

29.

M

2 2

2 57.

6.28 6.28

61. 6

9 6 -9

-6

6

f—> 0 as co

-3

0 as co

63.

25.13 - 25.13

2 65.

18.85 - 18.85

6 67.

-3.14 314

1 (b) y3 = —

4 kin(7rx) + —

3 sin(37rx)

+ —1

sin(5x) + —1

sin(77rx)] 5 7

2

JL (c) y4 =

1 —4

kin(7rx) + — sin(37rx) + —1

sin(57rx) 7r 3 5

87. Not one-to-one

1 1 + —

7 sin (7 7rx) + — sin (97rx)]

9

89. Yes. f -1(x) = x3 + 5


73. (a) 06

vAm, -06

12.57

(b) Not periodic and damped; goes to 0 as t increases.

75, (a)

120-

IlIIIIIIlII I.,t r 2 4 6 8 10 12

Month (1 44 January)

(b) According to the graph, the greatest number of sales occurs when t = 6, or in June. The least number of sales is in January.

77. (a) 0 < c < 8 (b) 0.22 < b < 1

79. True 81. True

83. As x approaches 7F from the left, f approaches oo. As x approaches 77- from the right, f approaches — oo.

85. (a)

3 -3

-2

90 -

•B 4; 60-

10

4

-3 3

- 6

Approaches 1

69. d = 5 cot x 36

3.14

- 36

71. As the predator population increases, the number of prey decreases. When the number of prey is small, the number of predators decreases.

3

-2

3

_dm]

- 2

Equal

59. 3

-6

Approaches ±oo

•

Approaches 1

7r

AAP


x - 1 -0.8 -0.6 -0.4 -0.2

y -1.5708 -0.9273 -0.6435 -0.4115 -0.2014

x 0 0.2 0.4 0.6 0.8 1

y 0 0.2014 0.4115 0.6435 0.9273 1.5708

1. (a)

3 3


9,77 ,/2777 91. sin 0 = csc 0 =

277 9

14-IZ7 cos 0 = sec 0 =

277 14

(c) (d) y +a. 2

" 10

9 tan 0 = A

14 cot 0 =

7. sin(-i) = -1

9. (a) 1.047 (b) 1.571 11. (a) 2.356 (b) -0.785

13. (a) 2.094 (b) 0.785 15. (a) -1.571 (b) 0

17. 7r -27r3 -- 24 ' 19. (a) -0.85 (b) 2.35

21. (a) 0.42 (b) 1.20 23. (a) 0.78 (b) 1.36

7r 5. arctan(-1) = -71

27. 0 arccos -4 (b)

25.

1.5

29.

x+ 1 31. 35 0 = arctan 10

7T 5 13 37. 39.

4 74. 41. -

12

47. x

149.

.1x2 + 1 ,/2x - x2

- (x - h)2

53.

r

55.

2

-2

x = 0

57. 59. x - 2

61.

3. (a)

(c)

(d) Intercept: (0, 0); Symmetric about the origin

x - 10 -8 -6 -4 -2

y -1.4711 -1.4464 -1.4056 -1.3258 -1.1071

x 0 2 4 6 8 10

y 0 1.1071 1.3258 1.4056 1.4464 1.4711

2

(b)

-2

2

1.57

3

1.5

33. -0.1 35. I I 2

45. - 11 6

-1.57 -2

Answers to Odd-Numbered Exercises and Tests Al 07

rx 1 2 -2 -I

-4

5

67.

-5 0

69. 5 s1n17rt + arctan

Pf\i The two forms are equivalent.

71. (a) U = arctan (b) 0.49 rad, 1.13 rad

73. (a) 0.45 rad (b) 24.39 feet

75. (a) -7T (b) -IT (c) 1.25 (d) 2.03 4 2

77. (a) 0 = arctan -x

(b) 0.24 rad, 0.54 rad 20

79. False. Inverse trigonometric functions do not have the same relationships that trigonometric functions have.

81. 83. (a) 4

(b) 0

I ) x 2ir

19. (a)

35° Soft

21. 2090 feet

27. 5099 feet

(b) h = 50(tan 47°40' - tan 35°) (c) 19.9 feet

23. 38.3° 25. •---••15.5°

29. 0.66 miles


91. sin(-585°) =

cos( - 585°) = - - 2-

tan(-585°) = -1

95.

93. ( 19/r

sin------) = _ 2

cos(

=

tan( _19/r) =

97.

Section 4.8

1. a 4.66

C 11.03

B = 65°

7. a 37.95

A = 64.62°

B = 25.38°

11. 2.56 inches

(page 361)

3. a 8.26

c = 25.38

A = 19°

9. a 91.34

b 420.70

B = 77°45'

13. 6.30 feet

. c 17.09

A = 20.56°

B 69.44°

15. (a) L = 60 cot 0

0 10° 20° 30° 40° 50°

L 340 165 104 72 _

50

• (c) No. Cotangent is not a linear function.

17. (a) h = 20 sin 0

(b)

(b)

0 60° 65° 70° 75° 80°

h 17.3 18.1 18.8 19.3 19.7

_ 2

1

36

4 :cos -x

IT

35' 2

- 0.785

b) 0

3.14

628

3.14


31. 437 miles north; 700 miles east

33. (a) N 580 E (b) •--=-• 68.82 meters 35. N 55°W

37. 1933.3 feet 39. 17,054 feet 3.23 miles

41. 78.69° 43. 35.3° 45. y =

47. =,28.2 inches 49. a 12.2, b 7

67. False. a = 22.56

69. x = 1.994 tan(48.1°)

71. x = ±4.851 73. 1.44 radians or 82.53°

75. -0.11 radian or -6.32°

51. (a) 4

53. (a)

(b) 4 (c)

(b) 70 (c) T1470 55. y= 8 sin irt

Review Exercises (page 368)

1. 0.5 rad 3. 4.5 rad

5. (a)

57. y = 3 cos(1°711 9

59. co = 5287r X

(b) Quadrant I

337r (C) - - 317r

16 16

61. (a) 0.5 (b) -ir seconds 8

(c) -7r

seconds 3.14 32

--0 5 7.(a)

63. (a) 22

(b) Quadrant HI

217r 397r (c) s'

- 15

7:9 15

I I I I I I I 1 1 L 12 14

(b) 12 months. Yes, 1 period is 1 year.

(c) 1.41 hours. 1.41 represents the maximum change in 9. Complement: -311-• Supplement: -77T 8 ' 8

time from the average time (d = 18.09) of sunset. 7 ir

11. Complement: -7r

•' Supplement: -10 5 13. 128.57°

15. -200.54°

17. (a)

65. (a) 6 .i.6-, 41 15

• • . i : a : • • g.'D 2 9 g .2 • • ... F: 6 - fl • •

= 3 .- •••

1 1 1 1 1 1 1 1 1 1 1 1) 1 2 4 6 8 10 12

Month (1 <4 January)

(b) Quadrant I

(c) 400°, -320°

40°

(b)

19. (a)

1 1 1 1 1 1 1 1 1 1 / 2 4 6 8 10 12

Month (1 <4 January)

63. Ans

67. (a) 0

71. sin

cos 0

tan 0

(b) Quadrant III

(c) 250°, -470°

S = 8 + 6.3 cos(-6 ); The model is a good fit.

(c) 12 months. Sales of outerwear is seasonal.

(d) Maximum displacement of 6.3 from the average sales

73. sin 0

of 8

cos 0

tan 0

-320°


TT

16

it ifi

397r 15

8.57°

15,4l. 75. sin 0 = 77. sin 0 =

241 6

4-./14. 5 cos 0 - 241

cos 0 =

15 tan 0 = 7-t tan 0 =

5

ifeT1 6-11.1. csc 0 = csc 0 =

15 11

, 5-VT.1. sec 0 = cot 0 = 4 11

4 cot 0 =

79. cos 0 = 8

3./5-3 tan 0 -

55 8

csc 0 = -3

sec 0 = 55

cot 0 = 3

s/2 81. I§ 83. 2- 85. - 87. 0.65

ir 89. 3.24 91. 84° 93.

-,/ 1 95. sin 240° = - 97. sin(-210°) = i

1 -./ cos 240° = - i cos( - 210°) = -.T-

-,/ tan 240° = ,/5 tan( - 210°) = - -T

99. sin(-7) = _4 101. sin(-11) = - 1

cos(-9," IT) = .1-/2-1 cosq = 0

(7) tan- = -1 tan(-1 is undefined. 2

103. ( - 2' 23 )

105. (- 23' - 2)

. 2.17- _ -./-- 7/r 1 sin T s in = - 2- -

27r 1 77T -1 cos T = - cos iT = --

77r -"J. tan 2-7r = - -,/3 tan -w- =

3

21. Complement: 82°; Supplement: 172°

23. Complement: none; Supplement: 9° 25. 135.28°

27. 5.38° 29. 135°17'24" 31. -85°21'36"

33. 8.3776 35. -0.5890 37. 2.083 rad

39. 48.171 meters 41. 0.094 rad 5.39°

43. (_1 sij.) 45(- 1 2' 2

. 7) 2 '2

47. sin = 49. = 77r 1

77r cos = - cos(-) = -

7/r -,/§ tan = tan(-217) =

csc -6 = -2 CSCg IT) 2

= - ; /. -3 77r

77r 2-./ sec = 3 sec( - = -2

cot -6

= cot(-2i T) = 29 . 7,77.

117 3/r 51. sin = sin =

53. sinET) = sin(-7) = -

5. -./ 59. sin 0 - 61. sin 0 =

61 9

6T 4 cos 0 =

61 cos 0 =

5 ./63 fan 0 = tan 0 =

4

uit 1 ./6T 9-V,33 csc 0 = csc 0 =

5 65

ja. 9 sec 0 = sec 0 = ,7t

6

6 4. 1/ cot 0 = 3 cot e =

65

63. Answers will vary. 65. (a) 0.1045 (b) 0.1045

67. (a) 0.7071 (b) 1.4142 69. 9.2 meters

rttIII 71. sin 0 = csc 0 =

470° cos 0 = 5 sec 0 = 3

tan 0 = cot 0 =

73. sin 0 - csc 0 = 53 2

cos 0 = sec 0 = 53 7

2 7 tan 0 = --

7 cot 0 = --2

55. -0.89 57. 1

155.

-4 -3

159.

163. (a

165. (a

169.

173. A

175.

177. y

179. F

181. Far

183. In

Chapt

1. (a)

•

2. 3000

3. sin 0

cos 0

tan 0

csc

sec 0

cot 0

147. 149. 123. 125.

153. 151. 129. f (x) = -4 cos(2x - 127. f (x) = - 2 cos(x - -1-r) 4


4 137. 3

2 A 27t

x

141.

133. f(x) = tan(x)

135.

139.

107. 109.

111. Period: 2; Amplitude: 5

113. Period: 'n-; Amplitude: 3.4

115. 117.

4,1 -4,1 27r

119. 121.

3 2

-27c 2rz

3

2

-

( Th -271

143. 145.

4

3

131. 56

12 40

Maximum sales: June; Minimum sales: December

Jr

1

3

4

2

-4 -3 -2 -I -5

-10

5. 0' = 70°

cos0 = -

.117

6

tan 0 -

6.111

csc 0 = 11

5.111 cot 0 =

11

6. Quadrant III 7. 150°, 210° 8. 1.33, 1.81

11J. 9. sin 0 =

6

137r 37r (b) ,

4 4 1. (a)

(c) 225°

15.

2n 2


155. 157.

15

- 15

'n• ir 57r 159. (a) -

2 (b) Does not exist 161. (a) -

4 (b) -

6

163. (a) 1.137 (b) 0.682

165. (a) -1.488 (b) 1.523 167. 0 = arcsin(x + 3) 16

169 1-x2 + 2x

171 214 - 2x2

. . -x2 + 2x 4 - x2

173. About 0.071 kilometer or 71 meters

175. 612 miles north and 634 miles west

177. y = 3

Cos( 27t)

15 )

179. False. The expressions are not equal.

181. False. 37r/4 is not in the range of the arctangent function.

183. Increases. The linear velocity is proportional to the radius.

Chapter Test (page 374) 27

12. 3' 13.

4 -

22, 2. 3000 radians per minute

41f7 3. sin 0 -

17

-7C

111157 4. sin 0 - ±

157

1T.7 61157 cos 0 = cos 0 - ±

17 157

1157 tan 0 = -4 csc 0 = ±

11

1f7 1157 csc = sec 0 = ±

4 6

6 sec 0 = -1f7 cot 0 = -

11

1 cot 0 = - -

4

85.

87. es

95. 5

99. 0

103. In

113. Fal

115. sin

tan

CSC

sec

cot

The

1.5

sec x = 2

cot x = 3

csc x = 3

tan 0 = - 1

csc 0 = -

cot 0 = - 1

24 7

7 25 24 25 25 - 7

Y I - Y2 5. sin x =

cos x =

csc x =

cot x =

7. cos 4) = - 1

tan 4) = 0

csc is undefined.

cot is undefined.


18. y = - 2 sin(-2 - -4)

19. - 2 x

20.

2

22.

2

23. S 34.5° W 24. 2290 feet 0.434 mile

1 cot x - - 2 cot 0 = -4

13. cos 0 = 0

tan 0 is undefined.

csc 0 = - 1

sec 0 is undefined.

15. 1, 1 17. no, 0 19. (d) 21. (a)

23. (e) 25. (b) 27. (f) 29. (e)

31. cos x 33. cos2 cf) 35. cos x

37. 1 39. -tan x 41. cot x 43. 1 + sin y


63. cos2 x 65. sin2 x tan2 x 67. sec4 x

69. sin2 x - cos2 x 71. 1 + 2 sin x cos x

73. tan2 x 75. 2 csc2 x 77. 2 sec x

1.0

4

17. 6

16.

6

32

-4 -2

2 9. sin x = -

3

cos x = - - 3

3 csc x =

sec x = - 3

5

11. sin 0 =47

17

cos 0 = -1117 17

csc 0 = 4

sec 0 = - -i17 Period: 2

6

21.

81. 3(sec x + tan x)

x 0.2 0.4 0.6 0.8

y1 0.1987 0.3894 0.5646 0.7174

Y2 0.1987 0.3894 0.5646 0.7174

79. 1 + cos y

83.

1. tan x =

x 1.0 1.2 1.4

y1 0.8415 0.9320 0.9854

y2 0.8415 0.9320 0.9854

Chapter 5 Section 5.1 (page 381)

3. cos 0 = -2

Trigonometric Functions - Birmingham Public Schools

Documents

Transcript of Trigonometric Functions - Birmingham Public Schools