Their Measure - University of Utahaksamit/Chap2Zill.pdfTheir Measure E Introduction We begin our...

Their Measure

E Introduction We begin our study of trigonometry by discussing angles and twomethods of measuring them: degrees and radians. As we will see in Section 3.1, it is theradian measure of an angle that enables us to define trigonometric functions on sets ofreal numbers.

D AngLes An angle is formed by two half-rays, or half-lines, which have a commonendpoint, called the vertex. We designate one ray the initial side of the angle and theother the terminal side. It is useful to consider the angle as having been formed by a rotation from the initial side to the terminal side as shown in FIGURE 2.1.1(a). An angle is said to

Terminal be in standard position if its vertex is placed at the origin of a rectangular coordinateside system with its initial side coinciding with the positive x-axis, as shown in Figure 2.1.1(b).

D Degree Measure The degree measure of an angle is based on the assignment of360 degrees (written 3600) to the angle formed by one complete counterclockwise rotation, as shown in FIGURE 2.1.2. Other angles are then measured in terms of a 360° angle,with a 1° angle being formed by of a complete rotation. If the rotation is counterclockwise, the measure will be positive; if clockwise, the measure is negative. For example, the angle in FIGURE 2.1.3(a) obtained by one-fourth of a complete counterclockwiserotation will be

(360°) = 90°.

Shown in Figure 2.1.3(b) is the angle formed by three-fourths of a complete clockwiserotation. This angle has measure

(—360°) = —270°.

(a) 900 angle (b) —270° angle

FIGURE 2.1.3 Positive measure in (a);negative measure in (b)

O—360,

+ 3600

FIGURE 2.1.4 Threecoterminal anges

O Coterminat AngLes Comparison of Figure 2,1.3(a) with Figure 2.1.3(b) showsthat the terminal side of a 90° angle coincides with the terminal side of a —270° angle.When two angles in standard position have the same terminal sides we say they arecoterminal. For example, the angles 0, 0 + 360°, and 0 — 360° shown in FIGURE 2.1.4

are coterminal. In fact, the addition of any integer multiple of 360° to a given angleresults in a coterminal angle. Conversely, any two coterminal angles have degree measures that differ by an integer multiple of 360°.

Ancjtes and Coterminat Anqtes

For a 960° angle:(a) Locate the terminal side and sketch the angle.(b) Find a coterminal angle between 0° and 3 60°.(c) Find a coterminal angle between —360° and 0°.

Terminalside

Vertex

ialside

(a) Two half-rays

V

Initialside

(b) Standard position

FIGURE 2.1.1 Initia andtermina sides of an ange

V

4

360°

FIGURE 2.1.2 Angle of360 degrees

90 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY

olution (a) We first determine how many full rotations are made in forming thisgle. Dividing 960 by 360 we obtain a quotient of 2 and a remainder of 240. Equivalentlye can write

960 = 2(360) + 240.

hus, this angle is formed by making two counterclockwise rotations before completig = of another rotation. As illustrated in FIGURE 2.1.5(a), the terminal side of 960°es in the third quadrant.) Figure 2.1.5(b) shows that the angle 240° is coterminal with a 960° angle.) Figure 2.1.5(c) shows that the angle — 120° is coterminal with a 960° angle.

y y

9600

240°

/ 220:

(a) (b) (c)

FIGURE 2.1.5 Ang[es in (b) and (c) are coterminal with the angLe in (a)

I Minutes and Seconds With calculators it is convenient to represent fractions ofgrees by decimals, such as 42.23°. Traditionally, however, fractions of degrees werepressed in minutes and seconds, where

1° = 60 minutes (written 60’)* (1)

1’ = 60 seconds (written 60”). (2)

rexample, an angle of 7 degrees, 30 minutes, and 5 seconds is expressed as 7°30’5”.me calculators have a special DMS key for converting an angle given in decimalgrees to Degrees, Minutes, and Seconds (DMS notation), and vice versa. The folwing examples show how to perform these conversions by hand.

EXAMPLE 2 Usinq (1) and (2

86.23° = 86° + 0.23°86° + (0.23)(60’)

= 86° + 13.8

onvert:

i) 86.23° to degrees, minutes, and seconds,

) 17°47’13” to decimal notation.

iilution In each case we will use (1) and (2).

i) Since 0.23° represents j of 1° and 1° = 60 we have

ow 13.8’ = 13’ + 0.8’, so we must convert 0.8’ to seconds. Since 0.8’ represents ofand l’= 60”, we have

ence, 86.23° 86°13’48”.

lie use of the number 60 as a base dates back to the Babylonians. Another example of thee of this base in our culture in the measurement of time (1 hour = 60 minutes andminute = 60 seconds).

86° + l3’+ 0.8’= 86° + 13’+ (0.8)(60”)= 86° + 13’+ 48”.

2.1 Angles and Their Measure 91

(b) Since 1° = 60’,itfollowsthat 1’= ()°.Simi1arly, I” = ()‘= Thuswehave

17°47’13” = 17° + 47’ + 13”= 17° + 47()° + l3()°

17° + 07833° + 00036°= 17.7869°.

D Radian Measure Another measure for angles is radian measure, which is generally used in almost all applications of trigonometry that involve calculus. The radianmeasure of an angle 0 is based on the length of an arc on a circle. If we place the vertexof the angle U at the center of a circle of radius r, then 0 is called a central angle. As weknow, an angle 0 in standard position can be viewed as having been formed by the initial side rotating from the positive x-axis to the terminal side. The region formed by theinitial and terminal sides with a central angle 0 is called a sector of a circle. As shownifl FIGURE 2.1.6(a), if the initial side of 0 traverses a distances along the circumference ofthe circle, then the radian measure of 0 is defined by

5

In the case when the terminal side of 0 traverses an arc of length s along the circumference of the circle equal to the radius r of the circle, then we see from (3) that themeasure of the angle 0 is 1 radian. See Figure 2.1.6(b).

Terminal

Terminal side

side r/ \ rS /o

r / Initial r / Initial_____// side __,,i/ side

(a) (b)

FIGURE 2.1.6 CentraL angLe in (a); angLe of 1 radian in (b)

The definition given in (3) does not depend on the size of the circle. To see this, allwe need do is to draw another circle centered at the vertex of 0 of radius r’and subtended arc length s See FIGURE 2.1.7. Because the two circular sectors are similar theratios s/r and s7r’ are equal. Therefore, regardless of which circle we use, we obtain thesame radian measure for 0.

In equation (3) any convenient unit of length maybe used for s and r, but the sameunit must be used for both s and r. Thus,

FIGURE 2.1.7 Concentrics(units of length)circLes 0(in radians) =r(units of length)

appears to be a “dimensionless” quantity. For example, ifs = 6 in. and r = 2 in., thenthe radian measure of the angle is

4 in.0 = —----

2,2 In.

where 2 is simply a real number. This is the reason why sometimes the word radians is omitted when an angle is measured in radians. We will come back to this idea in Section 3.1.


One complete rotation of the initial side of 0 will traverse an arc equal in length tothe circumference of the circle 2irr. It follows from (3) that

s 27rrone rotation = — = — 2 radians.

r r

We have the same convention as before: An angle formed by a counterclockwise rotation is considered positive, whereas an angle formed by a clockwise rotation is negative. In FIGURE 2.1.8 we illustrate angles in standard position ofr/2, —r/2, , and 3radians, respectively. Note that the angle of ir/2 radians shown in (a) is obtained byone-fourth of a complete counterclockwise rotation; that is

radians) = — radians.

The angle shown in Figure 2.1.8(b), obtained by one-fourth of a complete clockwise rotation, is —ir/2 radians. The angle shown in Figure 2.1.8(c) is coterminal with the angleshown in Figure 2.1.8(d). In general, the addition of any integer multiple of 2ir radiansto an angle measured in radians results in a coterminal angle. Conversely, any two coterminal angles measured in radians will differ by an integer multiple of 2r.

y+ y yty

(a) (b) (c) (d)

FIGURE 2.1.8 AngLes measured in radians

______________

A Coterminat AngLe

Find an angle between 0 and 2- radians that is coterminal with 0 = 1 17r14 radians.Sketch the angle.

Solution Since 2r < 11 ‘w/4 < 3ir, we subtract the equivalent of one rotation, or 2radians, to obtain

Alternatively, we can proceed as in part (a) of Example 1 and divide: llir!4 =

2r + 3ir/4. Thus, an angle of 31T/4 radians is coterminal with 0, as illustrated inFIGURE 2.1.9.

0 Conversion FormuLas Although many scientific calculators have keys that convert between degree and radian measure, there is an easy way to remember the relationship between the two measures. Since the circumference of a unit circle is 27r, onecomplete rotation has measure 2 radians as well as 360°. It follows that 360° = 2rradians or

If we interpret (4) as 180(10) = ir (1 radian), then we obtain the following two formulasfor converting between degree and radian measure.

1lr llr 8w 3r

180° rradians. (4)

32r

FIGURE 2.1.9 CoterminaLang’es in ExampLe 3

EXAMPLE 3

2.1 AngLes and Their Measure 93

CONVERSION BETWEEN DEGREES AND RADIANS

10 —fl—radian (5)180

1 radian= (\0

(6)\ITJ

Using a calculator to carry out the divisions in (5) and (6), we find that

10 0.0174533 radian and 1 radian 57.29578°.

F’

TABLE 2.1.1

Degrees 0 30 45 60 90 180Radians 0

:EXAMPLE4 Conversion Between Degree and Radians

Convert:

(a) 20° to radians, (b) 7-/6 radians to degrees, (c) 2 radians to degrees.

Solution (a) To convert from degrees to radians we use (5):

20° = 20(1°) = 20( —radian 1=—radian.\\180 J 9

(b) To convert from radians to degrees we use (6):

71T 71T . 7irul8ON°—radians = — (1 radian) = —I — j 210°.

6 6 6\rJ

(c) We again use (6): approximate answerrounded to twodecimai places

ul8ON° 1360N°2radians = 2(lradian) 2( =1—) 114.59°.

1 1

Table 2.1.1 provides the radian and degree measure of the most commonly usedangles.

D TerminoLogy You may recall from geometry that a 90° angle is called a rightangle and a 180° angle is called a straight angle. In radian measure, r/2 is a rightangle and IT is a straight angle. An acute angle has measure between 0° and 90° (orbetween 0 and ir/2 radians); and an obtuse angle has measure between 90° and 180°(or between ir/2 and IT radians). Two acute angles are said to be complementary iftheir sum is 90° (or ii-/2 radians). Two positive angles are supplementary if their sumis 180° (or IT radians). The angle 180° (or radians) is a straight angle. An anglewhose terminal side coincides with a coordinate axis is called a quadrantal angle. Forexample, 90° (or IT/2 radians) is a quadrantal angle. A triangle that contains a rightangle is called a right triangle. The lengths a, b, and c of the sides of a right trianglesatisfy the Pythagorean relationship a2 + b2 = c2, where c is the length of the sideopposite the right angle (the hypotenuse).


EXAMPLE 5 Complementary and Supplementary Angles

i) Find the angle that is complementary to 0 74.23°.

) Find the angle that is supplementary to q = rr/3 radians.

olution (a) Since two angles are complementary if their sum is 90°, we find the angletat is complementary to 0 74.23° is

90° — 0 = 90° — 74.23° = 15.77°.

) Since two angles are supplementary if their sum is 7T radians, we find the angletat is supplementary to = /3 radians is

i. 3 ir 2rradians.

I Arc Length In many applications it is necessary to find the lengths of the arc subnded by a central angle 0 in a circle of radius r See FIGURE 2.1.10. From the definitionf radian measure given in (3),

50 (in radians) = —.

r

y multiplying both sides of the last equation by r we obtain the arc length formularO. We summarize the result.

HEOREM 2.1.1 Arc Length Formula

or a circle of radius r, a central angle of 0 radians subtends an arc of length

srO. (7)

FIGURE 2.1.10 Lengthof arcs determinedby a centra ange 0

Finding Arc Length

md the arc length subtended by a central angle of(a) 2 radians in a circle of radius 6 inches,

) 30° in a circle of radius 12 feet.

olution (a) From the arc length formula (7) with 0 = 2 radians and r 6 inches, weave s = rO 2 6 = 12. So the arc length is 12 inches.

) We must first express 30° in radians. Recall that 30° rI6 radians. Then from therc length formula (7) we have s = r0 = (12)(/6) = 2r. So the arc length is

6.28 feet.

EXAMPLE 6

Answers to selected odd-numbered problemsExercisesbegin on page ANS-7.

i Problems 1—16, draw the given angle in standard position. Bear in mind that theick of a degree symbol (°) in an angular measurement indicates that the angle isteasured in radians.

1 60° 2 —120° 3 135° 4 150°5. 1140° 6. —315° 7. —240° 8. —210°

I Studenis ofien apply ihe arc length formulaincorrectly by using degree measure.Remember .c = rO is valid only if 0 ismeasured in radians.

2.1 Angtes and Their Measure 95

17r48.

51. —4

9. 10. 11. 12. —

13. — 14. —3- 15. 3 16. 4

In Problems 17—20, express the given angle in decimal notation.

17. 10°39’l7” 18. l43°7’2” 19. 5°10’ 20. 10°25’

In Problems 2 1—24, express the given angle in terms of degrees, minutes, and seconds.

21. 210.78° 22. 15.45° 23. 30.810 24. 110.5°

In Problems 25—32, convert from degrees to radians.

25 10° 26 15° 27 45° 28 215°

29. 270° 30. —120° 31. —230° 32. 540°

In Problems 33-40, convert from radians to degrees.

2r 117r 27r 5-33. — 34. 35. — 36.

9 6 3 12

37. 38. 7r 39. 3.1 40. 12

In Problems 41—44, find the angle between 0° and 360° that is coterminal with the given

angle.

41. 875° 42. 400° 43. —610° 44. —150°

45. Find the angle between —360° and 0° that is coterminal with the angle in

Problem 41.46. Find the angle between —360° and 0° that is coterminal with the angle in

Problem 43.

In Problems 47—52, find the angle between 0 and 2ir that is coterminal with the given

angle.

_2 49. 5.3’w

50. —52. 7.5

53. Find the angle between —2ir and 0 radians that is coterminal with the angle in

Problem 47.54. Find the angle between — 2r and 0 radians that is coterrninal with the angle in

Problem 49.

In Problems 55—62, find an angle that is (a) complementary and (b) supplementary to

the given angle, or state why no such angle can be found.

55. 48.25° 56. 93° 57. 98.4° 58. 63.08°

59. 60. 61. 62.


63. Find both the degree and the radian measures of the angle formed by (a) three-fifths of a counterclockwise rotation and (b) five and one-eighth clockwiserotations.

64. Find both the degree and the radian measures of the obtuse angle formed by thehands of a clock (a) at 8:00, (b) at 1:00, and (c) at 7:30.

65. Find both the degree and the radian measures of the angle through which thehour hand on a clock rotates in 2 hours.

66. Answer the question in Problem 65 for the minute hand.67. The Earth rotates on its axis once every 24 hours. How long does it take the

Earth to rotate through an angle of (a) 240°and (b) 7r16 radians?68. The planet Mercury completes one rotation on its axis every 59 days. Through

what angle (measured in degrees) does it rotate in (a) 1 day, (b) 1 hour, and(c) 1 minute?

69. Find the arc length subtended by a central angle of 3 radians in a circle of(a) radius 3 and (b) radius 5.

70. Find the arc length subtended by a central angle of 30° in a circle of (a) radius 2and (b) radius 4.

71. Find the measure of a central angle 0 in a circle of radius 5 if 0 subtends an arc oflength 7.5. Give 0 in (a) radians and (b) degrees.

72. Find the measure of a central angle 0 in a circle of radius 1 if 0 subtends an arc oflength ir/3. Give 0 in (a) radians and (b) degrees.

73. Show that the area A of a sector formed by a central angle of 0 radians in a circleof radius ris given byA = r20. See Figure 2.1.10. [Hint: Use the proportionality property from geometry that the ratio of the area A of a circular sector to thetotal area rr2 of the circle equals the ratio of the central angle 0 to one completerevolution 2’ir.]

74. What is the area of the red shaded circular band shown in FIGURE 2.1.11 if 0 is measured(a) in radians and (b) in degrees? [Hint: Use the result of Problem 73.]

MisceLLaneous AppLications

75. Angular and Linear Speed If we divide (7) by time t we get the relationshipv = rw, where v = sit is called the linear speed of a point on the circumferenceof a circle and w = 0,/t is called the angular speed of the point. A comrnunications satellite is placed in a circular geosynchronous orbit 35,786 km above thesurface of the Earth. The time it takes the satellite to make one full revolutionaround the Earth is 23 hours, 56 minutes, 4 seconds and the radius of the Earth is6378 km. See FIGURE 2.1.12.

(a) What is the angular speed of the satellite in rad/s?(b) What is the linear speed of the satellite in kmis?

76. Pendulum Clock A clock pendulum is 1.3 m long and swings back and forthalong a 15-cm arc. Find (a) the central angle and (b) the area of the sectorthrough which the pendulum sweeps in one swing. [Hint: To answer part (b), usethe result of Problem 73.)

77. Sailing at Sea A nautical mile is defined as the arc length subtended on the surface of the Earth by an angle of measure 1 minute. If the diameter of the Earth is7927 miles, find how many statute (land) miles there are in a nautical mile.

78. Circumference of the Earth Around 230 BCE Eratosthenes calculated the circumference of the Earth from the following observations. At noon on the longestday of the year, the Sun was directly overhead in Syene, while it was inclined7.2° from the vertical in Alexandria. He believed the two cities to be on the samelongitudinal line and assumed that the rays of the Sun are parallel. Thus he con-

FIGURE 2.1.12 Satellite inProblem 75

Planet Mercury inProbLem 68

FIGURE 2.1.11 Circular band inProblem 74

Satellite

2.1 AngLes and Their Measure 97

‘exandria — -

.- -

stades from2 00Rays

- -(Sun

Syene

FIGURE 2.1.13 Earth inProblem 78

cluded that the arc from Syene to Alexandria was subtended by a central angle of

7.2° at the center of the Earth. See FIGURE 2.1.13. At that time the distance fromSyene to Alexandria was measured as 5000 stades. If one stade 559 feet, find

the circumference of the Earth in (a) stades and (b) miles. Show thatEratosthenes’ data gives a result that is within 7% of the correct value if the polar

diameter of the Earth is 7900 miles (to the nearest mile).79. Circular Motion of a Yo-Yo A yo-yo is whirled around in a circle at the end of

its 100-cm string.

(a) If it makes six revolutions in 4 seconds, find its rate of turning, or angular

speed, in radians per second.(b) Find the speed at which the yo-yo travels in centimeters per second; that is

its linear speed.80. More Yo-Yos If there is a knot in the yo-yo string described in Problem 79 at a

point 40 cm from the yo-yo, find (a) the angular speed of the knot and (b) thelinear speed.

81. Circular Motion of a Tire An automobile with 26-in, diameter tires is traveling

atarateof55 mi/h.(a) Find the number of revolutions per minute that its tires are making.

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

82. Diameter of the Moon The average distance from the Earth to the Moon asgiven by NASA is 238,855 miles. If the angle subtended by the Moon at the eye

of an observer on Earth is 0.52°, then what is the approximate diameter of the

Moon? FIGURE 2.1.14 is not to scale.

Right TriangLe Trigonometry

Sideb opposite

aSide adjacent

FIGURE 2.2.1 Defining the trigonometric function of the angLe U

Introduction As we said in the introduction to this chapter, the word trigonotn

etry refers to the measurement of triangles. In this section we define the six trigonometricfunctions—sine, cosine, tangent, cotangent, secant, and cosecant—as ratios of the

lengths of the sides of a right triangle. The names of these functions are abbreviated assin, cos, tan, cot. sec, and csc, respectively.

We begin with some terminology and a definition.

O TerminoLogy In FIGURE 2.2.1 we have drawn a right triangle with sides labeled

a, b, and c (indicating their respective lengths) and one of the two acute angles denoted by0. From the Pythagorean theorem we know that a2 + b2 = c2. The side opposite the rightangle is called the hypotenuse; the remaining sides are referred to as the legs of the triangle. The legs labeled a and b are, in turn, said to be the side adjacent to the angle 0


Yo-yo inProblems 79 and 80

FIGURE 2.1.14 The curved red arc represents theapproximate diameter of the Moon

Hypotenuse

C

I-

id the side opposite the angle 0. We will also use the abbreviations hyp, adj, and oppdenote the lengths of these sides.

We next define the six trigonometric functions of an acute angle.

)EFINITION 2.2.1 Trigonometric Functions

‘he trigonometric functions of an acute angle 0 in a right triangle are

. opp b adj asinO=—=— cosO=—=—

hyp C hyp C

opp b adj atan0=—-=— cot0=—=— (1)

adj a opp bhyp c hyp c

sec0=—— csc0=—=—adj a opp b

Domains The domain of each of these trigonometric functions is the set of all•ute angles. In Section 2.4 we will extend these domains to include angles other thanute angles. Then in Chapter 3 we will see how the trigonometric functions can befined with domains consisting of real numbers rather than angles.

The values of the six trigonometric functions depend only on the size of the angleand not on the size of the right triangle. To see this, consider the two right trianglesown in FIGURE 2.2.2. Because the right triangles have the same acute angle 0 they arenilar and thus the ratios of the corresponding sides are equal. For example, from thed triangle in Figure 2.2.2(a) we have

• opp bsin 0 = = —,

hyp C

[iereas in the smaller blue triangle in Figure 2.2.2(b) we find that

• opp b’sin 0 —,.

hyp c

it since the red triangle is similar to the blue triangle we must have

other words, we get the same value for sin 0 regardless of which right triangle we usecompute it. A similar argument can be made for the remaining five trigonometricnctions.

nd the exact values of the six trigonometric functions of the acute angle 0 in the rightangle shown in FIGURE 2.2.3.

ilution Matching Figure 2.2.3 with Figure 2.2.1 we see that the side opposite thegleO has length 8 and the side adjacent toO has length 15; that is, b = 8 and a = 15.om the Pythagorean theorem the hypotenuse c hyp is

32 + 152 = 289 and so c = = 17.

C

b

a

(a)

lb

(b)

FIGURE 2.2.2 Similar triangles

b’ b

C’ C•

EXAMPLE I Values of the Trigonometric Functions

hyp8

15

FIGURE 2.2.3 Right triangle inExample 1

2.2 Right Triangle Trigonometry 99

Thus from (1) the values of the six trigonometric functions are

opp b 8 adj a 15sinO=—=—=—, cosO=—=—=—,

hyp C 17 hyp C 17opp a 8 adj b 15

tanO__=—=—-— cotO=—=—=—,adj b 15 opp a 8hyp c 17 hyp c 17

secO=—=—=— cscQ=—=—=—.adj a 15 opp b 8

D Quotient and ReciprocaL Identities There are many important relationshipsamong the trigonometric functions. The basic ones listed next are referred to as thefundamental identities and should be memorized.

Quotient Identities:

sinQ cosQtanO = , cotO =

cosO sinO

Reciprocal Identities:

1 1 1secO = , cscO

—--—,cotO

cosO sinO tanO

Identities (2) and (3) can be obtained from Definition 2.2.1. For example, the first of thequotient identities is verified as follows:

sinQ = opp/hyp = opp= tan ü

cosO adj/hyp adj

The others can be verified in a similar manner. Using these identities, we can find thevalues of all six trigonometric functions once we know the values of sinO and cos 0.

EXAMPLE2

Using (2) and (3)

Given sin 0 and cos 0 , find the values of the remaining four trigonometric functions.

Solution From the fundamental identities, we have4sinO 4

tan0 = = = — from (2)cosO 3

1 15secO = — = — —

cos0 31 15

cscO = = = — 4—from(3)sinO 4

1 13cot0 = - = — = —.

tan0 4

Although we computed cot 0 using the reciprocal identity in (3), we could have also computed cot 0 using the quotient identity in (2).

EXAMPLE 3 Using a Right TriangLe

Given cosO = andtan0 = 2V’. Find sin 0.

Solution We can obtain sin 0 by multiplying the first identity in (2) by cos 0:


1 2Vsin0 = cos0tan0 = —. 2V’ =

The next example illustrates that ifjust one trigonometric function value of an acuteangle is known, it is possible to find the other five function values by drawing anappropriate triangle.

EXAMPLE 4 Using a Right Triangle

[f 0 is an acute angle and sin 0 = , find the values of the other trigonometric functions

Solution We sketch a right triangle with an acute angle 0 satisfying sin 0 = , by makingpp = 2 and hyp = 7 as shown in FIGURE 2.2.4. From the Pythagorean theorem we have

22 + (adj)2 = 72 so that (adj)2 = 72— 22 = 45.

fhus, adj = = 3\/5The values of the remaining five trigonometric functions are obtained from the

iefinitions in (1):

hyp 7csc0 = = —.

opp 2

] Cofunctions The use of the terminology sine and cosine, tangent and cotangent,;ecant and cosecant is a result of the following observation. As shown in FIGURE 2.2.5, ifhe two acute angles of a right triangle ABC are labeled a and /3 and a is the length ofhe side opposite a, b is the length of the side opposite ,13, and c is the length of the side)pposite the right angle, then by Definition 2.2.1,

a bsin a — cos [3, cos a = — = sin [3,

a I,tana = = cot/3, cota = — = tan/’3

3ecause the sum of the angles in any triangle is 1800 (or ir radians), the acute angles amd /3 in a right triangle are complementary. Thus the cosine of an acute angle equalshe sine of the complementary angle, the cotangent of an acute angle equals the tangent)f the complementary angle, the cosecant of an acute angle equals the secant of the:omplementary angle, and conversely. For this reason we say that sine and cosine,angent and cotangent, and secant and cosecant are cofunctions of one another. We can;ummarize the discussion in one simple sentence:

• Cofunctions ofco,nplementaiy angles are equal. (4)

adjFIGURE 2.2.4 Right triangle inExample 3

A b CFIGURE 2.2.5 Acute anglesci’and/3in a righttriangle

] Cofunction Identities If a and /3 are the acute angles in the triangle in Figure 2.2.5,hen

or

adj 3\/cosO =

hyp 7

opp 2 2V’tan0=—= =

adj 3\/ 15

hyp 7 7vsecO=—= =

adj 3\/ 15

adj 3\/cotO = — =

opp 2

B

a

seca = - = csc/3, csca = = secl3.

2.2 Right Triangte Trigonometry 101

Because cos f3 = sin c we obtain

sin(— 13).

The last expression is one of six cofunction identities.

Cofunction Identities:

cosO sin( — o) cotO tan( — cscO = sec( — e)

sinO cosQ- — tanO = cot( — secO = csc( — o)

or equivalently

cosO = sin(90° — 0) cotO tan(90° — 0) cscO sec(90° — 0) (6)sinO cos(90° — 0) tanO = cot (90° — 0) secO = csc(90° — 0).

In (5) and (6) it is understood that 0 is measured in radians and degrees, respectively.

EXAMPLE 5 Using (5) and (6)

From (5):

,, complementary angles ,

(a) cot = tan(— ) = tan

(b) cos sinQ— ) = sin.

From (6):

(c) csc27° sec(90° — 27°) = sec63°

(d) cot 15° = tan(90° — 15°) = tan75°.

D Pythagorean Identities If just one trigonometric function value of an acute

angle is known, it is possible to find the values of the other five functions without using

the relationships in (1). Because the triangle in Figure 2.2.1 is a right triangle, the

Pythagorean theorem relates the lengths of the sides of the triangle by

If we divide this last result by c2, we obtaina2/c2 + b2/c2 = I or

Similarly, if we divide a2 + b2 = c2 by a2 and b2 we obtain, in turn,

and


a2 + b2 = C2.

(aN2 (bN2I—I +1—I =1\\cJ \\cJ

(bV1+1— =

/ \7(C -

(aN2 (CN+1=1—

\\b) \b

(7)

(8)

(9)

ig the appropriate definition in (1) in the results (7), (8), and (9) yield another setnportant identities.

aagorean Identities:

sin2O + cos2O = 1 (10)1 + tan2O = sec2O (11)cot2O + 1 = csc2O. (12)

), (10), and (11), the square of the trigonometric functions are written (sin 0)2 = sin2O,7 2

= cos0, (tanO) = tan 0, and so on.

12FIGURE 2.2.7 TriangLe forProbLem 2

IXAMPLE 6 Using (11)

is an acute angle and tan 0 = \/3 find the value cos 0.

[tion There are several ways of solving this problem. One way is to use theagorean identity (11):

sec20=tan20+l=(V)2+1=5+1=6

so secO = \/. Because secO 1/cosO we have cosO = 1/sec 0. Therefore= 1/V = V/6.

NOTES FROM THE CLASSROOM

we will see in Section 3.4, all the identities given in this section hold for any anglenot just acute angles).

Exercises Answers to selected odd-numbered problemsbegin on page ANS-7.

oblems 1—10, find the values of the six trigonometric functions of the angle Giniven triangle.

FIGURE 2.2.6 TriangLe forProbLem 1

_

FIGURE 2.2.8 TnangLe forProbLem 3 FIGURE 2.2.9 TnangLe for

ProbLem 4

5

3

2.2 Right Triangle Trigonometry 103

>12

FIGURE 2.2.10 Triangle forProblem 5

7.

9.

1.2

FIGURE 2.2.12 TriangLe forProbLem 7

y

FIGURE 2.2.14 TriangLe forProblem 9

2 311. sinO = , cos0 =

2 3\/13. sin 0 = —, cos 0 =

1 115. sinO = , tanO = —

85 5

17. csc0 = —, sec0 = —

1 319. cos0 = —, cscO

3 2V


10.

FIGURE 2.2.13 Triangle forProbLem 8

S


12. sinO = cos 0 =

14. sin0 = cos 0 =

16. cos0 = cotO =

2

18. sin 0 = , cot 0 = 7

20. sin 0 = , cot 0 = 7v

In Problems 2 1—28, find the value of the remaining trigonometric functions by drawing

an appropriate triangle.

1221. sin0 =

13

23. sec 0 = 24. csc 0 = ViO

25. tan 0 = —26. cot 0 = —

5 7

27. sec 0 = —28. tan 0 = 3

29. Ifcos75° — v’), find the exact value of sinl5°.

30. Ifcos75° = — V’),findtheexactvalueofsec75°.31. If tan(ir/8) = — 1, find the exactvalue of cot(3ir/8).32. If tan(ir/8) = — 1,findtheexactvalueoftan(3’ff/8).

x

In Problems 11—20, use the identities given in this section to find the values of the four

remaining trigonometric functions at the acute angle 0.

22. cos0 =

7


[n Problems 47—54, given that cos 300 = V/2. Use the identities of this section tolind the exact value of the given trigonometric function. Do not use a calculator.

47. sin 30°49. tan 60°51. sec 30°53. cos 30°tan 30°

____

Trigonometric Functions ofSpeciaL AngLes

E Introduction The angles 30°(r/6 radian),45°(/4 radian), and 60°(/3 radians)ire considered to be special angles because they occur so often in the study of trigonomtry and its use in calculus. Thus, it is to your advantage to know the exact values of the;ine and cosine of each of these angles. In the discussion that follows, we derive theseialues using some results from plane geometry.

I.

J VaLues of sin 45° and cos 45° To find the values of the sine and cosine functionsa 45° angle, we consider the isosceles right triangle with two equal sides of length 1

;hown in FIGURE 2.3.1. From plane geometry we know that the acute angles in this triangleire equal; therefore, each acute angle measures 45°. We can find the length of thelypotenuse by using the Pythagorean theorem:

(hyp)2 = (1)2 + (1)2 = 2 gives hyp

Ehus from(1) of Section 2.2 we obtain

adj 1cos45

hyp \/ 2

FIGURE 2.3.1 Isoscelesright triangle

In Problems 33-46, use the identities of this section to find the exact value of the giventrigonometric expression. Do not use a calculator.

33. 3sin- + 3cos2—

35. 1 + cos2l8° + sin2 18°

37. tan — sec2

sin 100 sin 100

sin 80° — cos 100

41. 5cot4l°cot49°

43. sin 28° cot 28° csc 62°

45. sin 10°cos 80° + cos 10°sin80°

34. sin35° + sin255°

36. 1 + tan233° — sec233°

38. —4csc213° + 4cot213°

40. sec20° — csc70°

42. cosl1°sec11°.T IT IT

44. lOsin—cot—sec—3 3 3

46. tan 30°cot60° — sec30°csc60°

48. cos 60°50. cot 30°52. csc 30°54. tan 30° + cot 60°

hyp=

opp 1 Vsin45

hyp \/ 2(1)

(2)

2.3 Trigonometric Functions of Special Angles 105

D Values of sin 300 and cos 30° To find the values of the trigonometric functions

of 300 and 60° angles, we consider the equilateral triangle AOB with sides of length 2

shown in FIGURE 2.3.2(a). From plane geometry we know that the three angles of an

equilateral triangle each measure 60°. As shown in Figure 2.3.2(b), if we bisect the

angle at 0, then CO is the perpendicular bisector of AB. It follows that

LAOC = LAOB = i(60°) = 30°,

AC AB = (2) = 1 and LACO = 90c.

A

2

B

(a)

FIGURE 2.3.2 Equiaterat triange in (a); two congruent right trianges in (b)

If we apply the Pythagorean theorem to the red right triangle ACO in Figure 2.3.2(b),

we get ()2 + 12 = 22. Solving for CO we get CO = \/. Therefore, from the right

triangle ACO and(l) of Section 2.2, we obtain the following values:

0PP 1

sin30hyp 2

adjcos30° = =

hyp 2

D VaLues of sin 60° and cos 60° Now using the 60° angle in the red right triangle

ACO in Figure 2.3.2(b) we identify opp = V’, adj = 1, and hyp 2.

Therefore

sin 60° = = -hyp 2

0adj 1

cos60 =—=—.

hyp 2

U Cofunctions We did not have to use a right triangle to derive the values in (5) and

(6). Recall, from Section 2.2 we showed that cofunctions of complementary angles are

equal. Thus, (5) and (6) follow immediately from the results in (3) and (4):

sin 60° = cos 30° ——

cos60° sin 30° —.

2

A

2

2

B

(b)


EXAMPLE 1 Values of the Other Trigonometric Functions

‘ind the values of tan(IT/6), cot(ir/6), sec(ir/6), and csc(ir/6).

;olution The angle 30° is equivalent to r/6 radian. Using the quotient and reciprocallentities in Section 2.2 along with the results in (3) and (4) we get

We leave finding the values tan 0, cot 0, sec 0, and csc 0 for 0 = 7T/4 and 0 = ir/3s exercises. See Problems 1 and 2 in Exercises 2.3.

Table 2.3.1 summarizes the values of the sine, cosine, and tangent functions that weavejust determined for the special angles 30°, 45°, and 60°. As mentioned in the introuction to this section, these function values are used so frequently that we feel thatey should be committed to memory. Knowing these values and the fundamentalIentities we discussed earlier will enable you to determine any of the trigonometricinctions for these special angles.

= sin(r/6) — 1/2 — 1 —

an6 cos(/6)

— /2 —— 3

iT 1cot—= = =-‘;

6 tan(ir/6) 1/v’iT 1 1 2 2V”i

sec--= cos(ir/6) V/2 3

iTcsc—=

6 sin(ir/6) 1/2

TABLE2.3.1

0(degrees)

0(radians)

EXAMPLE 2 Find the Exact Values

md the exact value of the given trigonometric expression.

i) sin2— cos

lution In each case we will use the information in Table 2.3.1.

-rsin— — cos— = —

) cos30° tan60° = =

(b) cos30°tan60°iT iT

(c) 2 + 4sin— — 6cos—3 6

sin 0 cosU tan0

iT 1 \/ \/30° — — —

6 2 2 3

iT v/45° — 1

4 2 2

60°IT

31

— ±_LJ_ I2422 2

)

2.3 Trigonometric Functions of Specia Anges 107

0 Use of a CalcuLator Approximations to the values of the trigonometric functions

can be obtained using a scientific calculator. But before using a calculator to find trigono

metric function values of angles measured in radians, you must set the calculator in

the radian mode. If the angles are measured in degrees, then you must select the degree

mode before making your calculations. Also, if the angles are given in degrees, minutes,

and seconds, they must be converted to decimal form first. Scientific calculators have

keys labeled, and 5j for computing the values of these functions. To obtain

the values of csc, sec, or cot, we can use the.

or keys with the reciprocal

key The following example illustrates the process.

EXAMPLE 3 Using a CaLcuLator

Use a calculator to approximate each of the following.

(a) sin45° (b) cos8°l5’

Solution (a) First we make sure that the calculator is set in degree mode. Then we

enter 45 and use the key to obtain

which is a seven-decimal-place approximation to the exact value \//2 given in (1).

(b) Since the angle is given in degrees and minutes, we must first convert it to decimal

form: 8°15’= 80 + () = 8.25°. Now with the calculator set in degree mode, we

enter 8.25 and use the j key to obtain

(c) Since degrees are not indicated, we recognize that this angle is measured in radians.

To evaluate sec 0.23, we will use the fundamental identity sec 0 = 1/cos 0. With the

calculator set in radian mode, we enter 0.23, use the key, and then take the reciprocal

of the result by pressing the key. Thus we have

(d) We observe that this angle is measured in radians and set the calculator accordingly.

We first enter r, divide by 7, use the key, and then the key to obtain

Exercises Answers to seected odd-numbered probLems

I begin on page ANS-7.

In Problems I and 2, use the results of this section to find the values of tan 0, cot0,

sec 0, and csc 0 for the given angle.

1. 45°


(c) sec 0.23

sin 45° 0.707 1068,

(d) cot

cos 8°15’= cos 8.25° 0.98965 14.

1sec0.23 1.0270458.

cos 0.23

cot— =

____

— 2.0765214.7

tan —

7

2. ir/3

In Problems 3—22, find the exact value of the given trigonometric expression. Do notuse a calculator.

IT3. cos2— 4. tan2—

3 65. sec45°csc45° 6. sin60°cos30°

•IT IT IT IT7. sin — cot 8. 6 sec — csc —

9. 9 sec 450 csc 450 10. tan 600 cot 30°

11. sincos + cosjsin 12. coscos — sinsin

13. 6 tan 30° + 7 tan 60° 14. 3 sin — 5 cos

15. tan45° — cot45° 16. sec2 + 4csc24 3

178sin(Ir/4)

182 — \/sin(ir/4)

sec(ir/3) cos(r/4)

19. sin23O° + cos245° 20. 2 + cot23O° — lOcsc23O°

21tan(ir/4) — tan(Ir/6)

22tan(7T/3) + tan(ir/4)

1 + tan(w/4)tan(n-/6) 1 — tan(ir/3)tan(w/4)

In Problems 23—32, use a calculator to find the approximate values of the six trigonometricfunctions of the given angle. Round your answer to four decimal places.

23 17° 24 82°25 143° 26 3475°27. 71°30’15” 28. 46°15’8”

29. 30. -

31. 0.6725 32. 1.24

For Discussion

33. Without a calculator, find the exact value of the product

2w 3ir 89wtan— tan— tan— tan—.

180 180 180 180

Trigonometric Functions ofGeneraL Angles

Introduction Until now we have defined the trigonometric functions only foracute angles. However, many applications of trigonometry involve angles that are notacute. Consequently, it is necessary to extend the definition of the six trigonometricfunctions in (1) of Section 2.2 to general angles. Naturally, we want the extended definition to agree with the earlier definition whenever the angle is acute. To accomplishthis we proceed in the following manner.

2.4 Trigonometric Functions of Generat Anges 109

y Let 0 be an acute angle in standard position_and choose a point P(x, y) on the ter

minalsideofO.Ifweletr = d(O,P) Vx2 +y2,weseeinFIGIJPE2.4.1 thatx,y,andP(x, Y) r are the lengths of the sides of a right triangle. With y = opp, x = adj, and r hyp,

r andwehavefromDefinition2.2.1,

sinO = cosO and tan0 . (I)

The expressions in (1) provide us with a model on which to base our extended definiFIGURE 2.4.1 An acute ange tion for any angle 0 in standard position, such as those illustrated in FIGURE 2.4.2.

y y

P(x, y)

r x x

x

y P(x, y) P(x, y) ‘*.

(a) (b) (c)

FIGURE 2.4.2 Ang’es that are not acute

We now have the following definition of the trigonometric functions of a gen

eral angle.

DEFINITION 2.4.1 Trigonometric Functions

Let 0 be any angle in standard position,_and let P(x, y) be any point other than (0, 0)

on the terminal side of 0. If r = V’x2 + y2 is the distance between (0, 0) and P(x, y),

then the trigonometric functions are defined to be

y xsinO=— cosO=—

tan0= cot0= (2)

sec0=— csc0=—x y

provided no denominator is 0.

It can be shown by using similar triangles that the values of the six trigonometric

functions depend only on the angle 0 and not on which point P(x, y) is chosen on the

terminal side of 0. The justification of this statement is like the one made for acute

angles on page 99.

D Domains A trigonometric function defined in (2) will be undefined if its denom

inator is zero. Since P(x, y) (0, 0), then r = Vx2 + y2 is never zero. Thus the

domains of the sine and the cosine functions consist of all angles 0. However, the tan

gent and the secant functions are undefined if the terminal side of 0 lies on the y-axis

because thenx = 0. Therefore, the domains of tan 0 and sec 0 consist of all angles 0 except

The angles are odd multiples of ir/2. those having radian measure ±r/2, ±3ir/2, ±5’rr/2, and soon. Using set notation,


ndthefactthatanoddintegercanbewrittenas2n+1,naninteger,thedomainsofietangentandthesecantfunctionsare

{ojo(2n+l)r/2,n=0,±1,±2,...}{o(o(2n+1)900,110,±1,±2,..

‘hecotangentandthecosecantfunctionsarenotdefinedforangleswiththeirterminalidesonthex-axisbecausetheny=0.Thusthedomainsofcot0andcsc0consistof11angles0exceptthosehavingradianmeasure0,±r,±2,±3,andsoon;thatis,Theanglesareintegermultiplesofif.

00n7T,n=0,±1,±2,...}or{00180°n,n=0,±1,±2,...}.

SinceritfollowsthatIxrandIIr,orequivalently,x/rI1andi/ri1.Therefore,asbefore

lsinOI1andIcosOl1.(3)

imilarly,becauseIr/xI1andr/y1wehave

cscoi1andIsecOl1.(4)

‘heinequalitiesin(3)and(4)holdforevery0inthedomainofeachofthesefunctions.

______________

ValuesoftheTrigonometricFunctions

mdtheexactvaluesofthesixtrigonometricfunctionsoftheangle0if0isinstandardositionandtheterminalsideof0containsthepointP(—3,1).

olutionTheterminalsideoftheobtuseangle0issketchedinFIGURE2.4.3.Withthelentificationsx=—3,y=1,and

‘ehavefrom(2),

Vi

_

sin0=

=

tanO=—-—

=

—33v1id sec0==—,

—33

-33Vicos0=-=—_____

10—3

cot0=—j-—

=—3,

csc0=

yi

P(-3,I)

FIGURE2.4.3Angle0inExample1

EXAMPLE2

00

2)1T\

P(O,—1)

FIGURE2.4.4Angle0in

Example2

EXAMPLE1

x

ValuesoftheTrigonometricFunctions

—1 =—l1

==0

—1

mdthevaluesofthesixtrigonometricfunctionsof0if0=—7r/2.

lutionFirstweplace0instandardpositionasshowninFIGURE2.4.4.Accordingtoefinition2.4.1,wecanchooseanypointP(x,y)ontheterminalsideof0.Forconmience,weselectP(0,—1)sothatx=O.y=—1,andr

=\4v2+y2=l.Thus,

.Irsini—

coti—

owever,theexpressionstan0v/xandsec0=rtvareundefinedforD—/2ncex=0.

AlgebraicSignsDependingonthequadrantinwhichtheterminal_sideof0lies,eorbothcoordinatesofP(x,v)maybenegative.Sincer=Vx2+y2isalways

cos( csc(

x

2.4TrigonometricFunctionsofGeneraLAngles111

positive, each of the six trigonometric functions of 0 has negative as well as positive val

ues. For example, sinO — y/r is positive if the terminal side of 0 lies in quadrants I or

II (where y is positive), and sin 0 = v/r is negative if the terminal side of 0 lies in quad

rants III or IV (where)’ is negative). FIGURE 2.4.5 summarizes the algebraic signs of the

six trigonometric functions defined in (2). For convenience, if the terminal side of 0

lies in quadrant II, we will refer to 0 as a quadrant II angle or say that 0 is in quadrant II.

We will use similar terminology when we refer to angles with terminal sides in quad

rants I, Ill, or IV.

II Icos8<O sin9>O cos6>O sinO>OtanO<O cot8<O tan6>O cot8>OsecO<O csc8>O secO>O csc>O

cos6<O sin9<O cosO>O sin8<OtanO>O cot8>O tanO<O cotO<OsecB<O cscO<O sec8>O cscO<O

III IV

FIGURE 2.4.5 Agebraic signs of the six

trigonometric functions

EXAMPLE 3 Using Figure 2.4.5

—+‘--= 1r r

/ \ / \/X\ fy\

or I—I +1—I =1.\rJ J

sin20 + cosO2 = 1.

In which quadrant does the terminal side of 0 lie if sin 0 > 0 and tan 0 < 0?

Solution From Figure 2.4.5 we see that the sine function is positive for angles in quad

rants I and II and the tangent function is negative in quadrants II and IV the terminal

side of 0 must lie in quadrant II.

S U Pythagorean Identities—Revisited The reciprocal, quotient, and Pythagorean

P(, ) identities for acute angles given in Section 2.2 also hold for general angles. For exam

ple, to derive the Pythagorean identities let 0 be any angle in standard position. As showr

r in FIGURE 2.4.6, we let P(x, y) be any point other than the origin on the terminal side ol

0.Ifweagainletr = d(O, P) = Vy2,thenwehavex2 + y2 =r2.Dividingbotl

sides of the last equation by r2 givesFIGURE 2.4.6 An arbitraryange 0

Recognizing that x/r = cos 0 and y/r = sin 0, we obtain the basic Pythagorean identit3

In (5) we have followed the convention that sin2 0 be written first. If both sides o

(5) are divided, in turn, by cos2 0 and sin2 0 we obtain

and

Formulas (5), (6), and (7) are identical to (10), (11), and (12) in Section 2.2. But unlik

the latter formulas, the trigonometric functions in (5), (6), and (7) are

• valid for all angles for which the functions are defined, and

• the values of the functions can have negative values.

We will encounter the Pythagorean identities one more time (in Chapter 3) whei

we show that each of the trigonometric functions can be defined for real numbers instea

of angles.


1 + tan20 = sec2O

cot20 + 1 = csc26.

(5

(6

(7

+(1)2

=

sinO = 1 — — = —.

99

[ 2\/sinO = —

= —1•

Using (5)

liven that cos 0 and that 0 is a quadrant IV angle, find the exact values of themaining five trigonometric functions of 0.

olution Substituting cos 0 = into (5) gives

ince the terminal side of 0 is in quadrant IV sin 0 is negative. Therefore, we must (See Figure 2.4.5.]ect the negative square root of:

low using

e find that the values of the remaining four functions are

tanO=

= —2,

secO = 3,

sin 0tan0

cos 01

secO =

cos 0

1cotO =

tanO

csc0 =sin0

1cot0 = = —,

-2V 4

1 3Vcsc0

—2v’/3 4

sec2O = 1 + (_2)2 =

sec0 = —\/..

Using (6)

iven tanG = —2 and sinO > 0, find the exact values of the remaining five trigonometricinctions of 0.

lution Letting tanG —2 in the identity I + tan2 0 = sec2 0, we find

rice tanG is negative in quadrants II and IV and sinG is positive in quadrants I and II,e terminal side of 0 must lie in quadrant II. Thus we must take

:om sec 0 = 1/cos 0, it follows that

sing tanG = sin 0/cos 0, we obtain

ien

1 1cos0=—=

secO —\/ 5

2\/sinG = cosG tanG —-----)(_2) =

1 1csc0

= = 2V3)5 =

I 1 1cotO = — = = —

tanG —2

EXAMPLE 4

EXAMPLE 5

2.4 Trigonometric Functions of GeneraL AngLes 113

In Section 2.3 we found exact values for the six trigonometric functions of the spe

cial angles of 30°, 45°, and 60° (or 71/6, 71/4, and 71/3, respectively, in radian measure).

These values can be used to determine the exact trigonometric function values of cer

tain nonacute angles by means of a reference angle.

FIGURE 2.4.7 illustrates this definition for angles with the terminal sides in each of the

four quadrants.

FIGURE 2.4.7 An angle B (red) and its reference angle 0’ (bLue)

EXAMPLE 6 Reference AngLes

Find the reference angle for each angle 0.

Solution (a) From FIGURE 2.4.8(a) we see that 0’ 40°.

(b) From Figure 2.4.8(b), 0’ — 0 = — 271/3 = 71/3.

(c) FromFigure2.4.8(c),0’= 0 — 180° = 210° — 180° 30°.

(d) Since 0 —971/4 is coterminal with

971 71

—-a- + 2ir =

we find that 0’ = 71/4. See Figure 2.4.8(d).

9,

DEFINITION 2.4.2 Reference Angle

Let 0 be an angle in standard position such that its terminal side does not lie on a

coordinate axis. The reference angle 0’ for 0 is defined to be the acute angle formed

by the terminal side of 0 and the x-axis.

-V

8=0’

(a)

8

V

e5

(c)(b)

x>Jç

(d)

(a) 0 = 40°271

(b) 0 = (c) 0 210°971

(d) 0=

V y

1’

(a)

FIGURE 2.4.8 Reference angLes in ExampLe 6

(b) (c) (d)


I1 Property of Reference AngLes The usefulness of reference angles in evaluat

ing trigonometric functions is a result of the following property: 4The absolute value ofany trigonometricfunction ofan angle 0 equals the valueofthatfimnction for the reference angle 0’.

Forinstance, sin0 = sinO’, cosol = cosO’, and so on.We will verify the foregoing property for the sine function. If the terminal side of

9 lies in quadrant I, then 0 0’ and sinO is positive, so

sin0’= sin0 = sin0.

rom FIGURE 2.4.9, we see that if 0 is a quadrant II, III, or IV angle, then we have

sin0’= -- = = sin0,

Nhere P(x, v) is any point on the terminal side of 0 and r

y y y

P(x, v)

y=IyIIr

.. x

(a) (b) (c)

FIGURE 2.4.9 Reference angles

We can now describe a step-by-step procedure for determining a trigonometricunction value of any angle 0.

FINDING THE VALUE OF A TRIGONOMETRIC FUNCTIONSuppose 0 represents any angle.

(i) Find the reference angle 0(ii) Determine the value of the trigonometric function for 0

(iii) Select the correct algebraic sign of the value in (ii) by consideringin which quadrant the terminal side of the angle 0 lies.

EXAMPLE 7 Finding VaLues Using Reference AngLes

(b) 0 = 210°97T

(c) 0 =4

md the exact values of sin 0, cos 0. and tan 0 for each of the following angles.27r

) 0 =

olution We use the procedure just discussed along with Table 2.3.1 of Section 2.3.

i) In part (b) of Example 6 we found the reference angle for0 = 2r/3 to be 0’ = ir/3.low we know that sin(7r/3) = v/2, cos(/3) = 1/2, and tan(r/3) =

ecause 0 = 2r/3 is a quadrant II angle, where the sine is positive but the cosine andie tangent are negative, we conclude

2r \/s1n—— =

2irand tan— = —\/.

3

2 1cos—— = —i-,

2.4 Trigonometric Functions of General Angles 115

.f19irN 1Tsini —-——— I = —sin— = —

\ 4/ 4 2

‘rcos_T) = cos

iT= —tan-a- = —1.

y

FIGURE 2.4.10 Solutions inExample 8

Finding AngLes

Find all angles 0 satisfying 0° 0 < 3600 such that sin 0 =

Solution From what we know about the special angles 30°, 60°, and 90°, we know that

o = 30° is one solution. Using 30° as a reference angle in the second quadrant, as shown

in FIGURE 2.4.10, we find 0 = 150° as a second solution. Since the sine function is neg

ative for angles in quadrants III and IV, there are no additional solutions satisfying

0°0<360°.

.iEXAMPLE 9 Finding AngLes

Find all angles 0 satisfying 0 0 < 2ir such that cos 0 = —

Solution Since the given value of the cosine function is negative, we first determine the

reference angle 0’ such that cos 0’ = \//2. From Section 2.3 we know that 0’ = ir/4.

Since the cosine function is negative for angles in quadrants II and III, we position the

reference angle 0= ir/4 as shown in FIGURE 2.4.11. We then obtain 0 = 3iT/4 and

0 = 5ir/4 as solutions.

FiGURE 2.4.11 Solutions in Example 9

(b) Referring to part (c) of Example 6, we see that the reference angle is 0’ = 30°.

Using the property of reference angles and the fact that the terminal side of 0’ 210°

lies in quadrant III, we get

sin2lO° = —sin30° = ——,

cos 210° —cos 30° —

see Figure 2.4.5 for the

2 correct algebraic signs

tan 210° tan 30° =

(c) From part (d) of Example 6 we know that the reference angle 0’ = iT/4. Since

0 = —9ir/4 is a quadrant IV angle, it follows that

EXAMPLE 8

3,r4

(a) (b)


NOTES FROM THE CLASSROOM

In this section we have purposely avoided using calculators. For acomplete understanding of trigonometry, it is essential that youmaster the concepts and be able to perform without the aid of acalculator the types of calculations and simplifications we have discussed. The exercise set that follows should be worked without theuse of a calculator.

P1

Exercises Answers to se’ected odd-numbered probemsbegin on page ANS-7.

47e recommend that you do not use a calculator in solving any ofthe problems thatfohlow.

n Problems 1—10, evaluate the six trigonometric functions of the angle 0 if 0 is in stan-lard position and the terminal side of 0 contains the given point.

1. (6, 8)3. (5, —12)5. (0,2)7. (—2, 3)9. (—v’,—l)

Li. sin0 <Oandtan0 >0.3. tanO <0 and sec0 <0.5. cotO > 0 and sinO > 0.7. sinO > Oandcos0 <0

.9. sinO = , 0 is in quadrant II1. tan 0 = 3, 0 is in quadrant III3. csc 0 = —10, 0 is in quadrant IV5. sinO = —, cosO > 07. tan0 = 8, sec0 > 0

cos9 > Oandsin0 <0secO <Oandcsc0 <0csc0 > Oandcot0 <0tan0 <Oandcsc0 >0

20. cos 0 = —, 0 is in quadrant II22. cot 0 = 2, 0 is in quadrant III24. sec 0 = 3, 0 is in quadrant IV26. cos0 —, sin0 <028. sec0 = —4, csc0 <0

9. If cos 0 = , find all possible values of sin 9.0. If sin 0 = —, find all possible values of cos 0.1. If 2 sinO — cos 0 = 0, find all possible values of sin 0 and cos 0.2. Ifcot0 = , find all possible values ofcsc0.3. If sec 0 = —5, find all possible values of sinO and cos 0.4. If 3 cos 0 = sin 0, find all possible values of tan 0, cot 0, sec 0, and csc 0.

2. (—1,2)4. (—8, —15)6. (—3,0)8. (5,—I)

10. (V/)

n Problems 11—18, find the quadrant in which the terminal side of an angle 0 lies if 0atisfies the given conditions.

12.14.16.18.

I

n Problems 19—28, the value of one of the trigonometric functions of an angle 0 isiven. From the given value and the additional information, determine the values of theive remaining trigonometric functions of 0.

2.4 Trigonometric Functions of Genera Anges 117

35. Complete the following table.

0 (degrees) 0 (radians) sin 0 cos 6 tan 0

0° 0 0 1

300 IT/6 1/2450 ir/4 \//2 \//2

60° /3 ‘.//2 1/2

90° ir/2 1 0 —

120° 2/3 \//2 —1/2 —v’s135° 3ir/4

150° 5ir/6

180° 7T

210° 7i-/6 —1/2 —V’/2

225° 5/4

240° 4r/3

270° 37T/2

300° 5!3

315° 7ir/4

330° 11IT/6

360° 2

36. Complete the following table.

0 (degrees) 0 (radians) csc 6 sec 0 cot 0

0° 0 — 1 —

30° ir/6 2 2\//3

45° T/4

60° Tr/3 2V’/3 2 v/390° /2 I — 0

120° 2ir/3

135° 3r/4

150° 5ir/6

180°

210° 7/6

225° Sir/4

240° 4ir/3

270° 3rr/2

300° 5ir/3

315° 7-/4330° 1l7r/6

360° 27r


In Problems 37—52, find the exact value of the given expression.

37. cos5r

13r39. cot—

41. sin(_)

csc (—i)sec(—120°)sin 1500

tan 405°cot (—720°)

9r40. tan—

23ir42. cos—-—

23ir44. tan—T

46. csc495°48. cos(—45°)50. sin3l5°52. sec(—300°)

In Problems 53—58, find all angles 0, where 0 0 < 360°, satisfying the givencondition.

53. tanO =

55. coso =

57. csc0 = —1

2

54. sin0 = ——

2\/56. sec6 =

31

58. cotO=

In Problems 59—64, find all angles 0, where 0 0 < 2r, satisfying the given condition.

59. sinOO61. sec0 = —\/63. cot0 = —V

MisceLLaneous AppLications

60.cos0 = —162. csc0 = 264. tan0 = 1

65. Free Throw Under certain conditions the maximum height y attained by abasketball released from a height h at an angle a measured from the horizontalwith an initial velocity v0 is given by

y = h + (v sin2a)/2g,

where g is the acceleration due to gravity. Compute the maximum height reachedby a free throw if h = 2.15 m, v0 = 8 m/s, a = 64.47°, and g = 9.81 rn/s2.

66. Putting the Shot The range of a shot put released from a height h abovethe ground with an initial velocity v0 at an angle 0 to the horizontal can beapproximated by

v0cosSR

= g(v0 sin + \/vö sin2 + 2gh),

where g is the acceleration due to gravity.(a) If v0 = 13.7 m/s, 4 = 40°, and g = 9.81 m/s2,compare the ranges

achieved for the release heights h = 2.0 m and h = 2.4 m.(b) Explain why an increase in h yields an increase in R if the other parameters

are held fixed.(c) What does this imply about the advantage that height gives a shot-putter?

38.

43.

45.47.49.51.

&t

h

Free throw

2.4 Trigonometric Functions of General Angles 119

67. Acceleration Due to Gravity Because of its rotation, the Earth bulges at the

equator and is flattened at the poles. As a result, the acceleration due to gravity

actually varies with latitude 0. Satellite studies have shown that the acceleration

due to gravity sat is approximated by the function

g1 978.0309 + 5.18552 sin20 — 0.00570 sin220.

FIGURE 2.4.12 Line throughorigin in Probem 71

CONCEPTS REVIEW

Initial side of an angleTenninal side of an angleStandard position of an angleCoterminal anglesMinutesSecondsDegree measure of an angleCentral angleRadian measure of an angleAcute angleComplementary anglesObtuse angle

For Discussion

Straight angleQuadrantal angleSupplementary anglesRight angleArc lengthConversion:

degrees to radiansradians to degrees

Reference angleRight triangles:

side adjacentside opposite

hypotenuseTrigonometric functions:

of acute anglesof general angles

Quotient identitiesReciprocal identitiesPythagorean identitiesCofunctionsCofunction identities

CHAPTER 2 Review Exercises Answers to selected odd-numberedproblems begin on page ANS-8.

A. True/False

In Problems 1—10, answer true or false.

1. sin(-rr/6) = cos(/3)2. sin(r/2) = sin(5ir/2)

sin = 30°__tan7r = 0cscOl 1sin2O + sin2(90° —0) = 1The angles 120° and —240° are coterminal.If tan 0 = , then sin 0 2 and cos 0 = 5.If sec (3 = \/7, then cos 0 =

30’ is equivalent to 0.5°.


y

(a) Find g1 at the equator (0 = 0°), (b) at the north pole, and (c) at 45° north

latitude.

68. Is there an angle 0 such that cos 0 = ? Explain.

69. Is there an angle 0 such that 2 csc 0 1? Explain.

70. Discuss how it is possible to determine without the aid of a calculator that both

sin 4 and cos 4 are negative.71. Let L be a nonvertical line that passes through the origin and makes an angle 0

measured counterclockwise from the positive x-axis. Prove that the slope in of

the line L is tan 0. See FIGURE 2.4.12.

You should be able to give the meaning ofeach ofthefollowing concepts.

3.4.5.6.7.8.9.

10.

27. sinO cos6 = 4, tano = coto = cscO

29. 3(\/—

31. v’ — 1

33.3 35.2

37. —1 39. 0

41. 5 43.

45.1 47.

49. \/

53. 3

Exercises 2.3 Page 108

15.

rathans

17. 10.6547° 19. 5.17°

21. 210°46’48” 23. 30048360

25. ir/18 27. r/429. 3ir/2 31. —23/18

33. 400 35. 1200

37. 225° 39. 177.62°

41. 155° 43. 110°

45. —205° 47. 7ir/4

49. 1.3r 51. 2 — 4 2.28

53. —-/455. (a) 41.75° (b) 131.75°

57. (a) The given angle is greater than 900.

(b) 81.60

59. (a) /4 (b) 3r/461. (a) The given angle is greater than Tr/2.

(b) ir/3

63. (a) 216°; 1.2ir (b) —1845°; —10.25r

65. because the hour hand moves counterclockwise: —60°, —7r/3

67. (a) 16h (b) 2h

69. (a) 9 (b) 15

71. (a) 1.5 (b) 85.94°

75. (a) 0.000072921 rad/s (b) 3.074641 km/s

77. 1.15 statute miles

79. (a) 3irrad/s

(b) 300IT cm/s81. (a) 711.1 rev/mm

(b) 4468 rad/min

1. tan45° = 1, cot45° = 1, sec45° V’, csc45° =

3.3 5.2 7.

9. 18 ii. 3(’v’ +

13. 9V’ 15. 0 17. 2\/

19.3 21. 2—V’

23. sinl7° = 0.2924,cosl7° = 0.9563,tanl7° = 0.3057,

cotl7° = 3.2709,secl7° = l.0457,cscl7° 3.4203

25. sin 14.3° = 0.2470, cos 14.3° = 0.9690, tan 14.30 = 0.2549,

cot 14.30 3.9232. sec 14.3° = 1.0320, csc 14.3° 4.0486

27. sin7l.504l7° = 0.9483,cos7l.50417° = 0.3172,

tan7l.50417° = 2.9894,cot7l.50417° = 0.3345,

sec7l.504l7° = 3.1522,csc7l.50417° = 1.0545

29. sin() = 0.5878, cos3) = 0.8090, tan() = 0.7265,

cot() = 1.3764, sec() = 1.2361, csc3) = 1.7013

31. sinO,6725 0.6229, cos0.6725 = 0.7823,

tan0.6725 0.7963.cot0.6725 1.2558,

secO.6725 = l.2783,csc0.6725 1.6053

Exercises 2.2 *696 103

Exercises 2.4 rPage 117

1. sinO =3,coso 3,tanO ‘3,cot9 = 3,seco =4,csco =33. sinO = cosO = , tanO = 3, cotO 3.

secO = V’Tö, cscO =

5. sinO =4,coso =‘,tanO =cot0 °,

secO = 7] cscO = 47. sinO = 3, cosO = , tanO = , cotO = 2V,

secO = cscO 3

9. sinO = _,cosQ ,tanO =,cotO oo,

secO=—- ,cscO

11. tan6 = 4, cotO = 4, secO = ,cscO =

13. tanO=5,cot& = ,secB =-,csc =315. cosO = cotO = 8, secO , csc8 = V’

17. sinO = 3, cosO = 3, tanO = 3, cotO 319. sinO = tanO = 2V, cotO = -. secO = 3

21. cos6 = , tanO = , cotO = fs, secO = , cscb = 3423. sinO =4,coso =o2,tanO =,cotO = \/,cscO = 2

25. sinO cosO = cotO = 4. secO , cscO=

1. sinG = 3, cosO = 4, tanG = 3, cscO = 3, secO 4,coto = 37 5 7 3

3. sinG = —5,cosO = y,tanG = —°,cscO = —js,

secO = ,cotO = —

5. sinG = 1, cosG = 0, tanG is undefined, cscG = 1, secO is

undefined, cotG = 0

7. sinG = cosO = —‘,tanG = —4, cscO =

secO = —-f, cotG = —49. sinG = —*,cosG = tanG = csc6 =

secG = _E6,cotO =

11. III 13. II

15. I 17. II

19. cosG = tanG = —,cscG = 4, secO=

cotG = —vT

21. sinG = —°,cosG —-f3,cscG = ,secG= —‘s/iô

cotO = 323. sinG = —, cosG = 344j-1, tanG —, secG =

cotG = —3vTi

25. cosO = tanG = —-j, cscO —5, secO =

cotG =

27. sinG = cos6=5,cscO =,secG = V,cotG = 429. ±

31. sinG = ±, cosG = ±

33. cosO = —3, sinG ±6

ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS ANS-7

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