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Their Measure E Introduction We begin our study of trigonometry by discussing angles and two methods of measuring them: degrees and radians. As we will see in Section 3.1, it is the radian measure of an angle that enables us to define trigonometric functions on sets of real numbers. Terminal side Vertex ial side (a) Two half-rays V Terminal side Initial side (b) Standard position FIGURE 2.1.1 Initia and termina sides of an ange D AngLes An angle is formed by two half-rays, or half-lines, which have a common endpoint, called the vertex. We designate one ray the initial side of the angle and the other the terminal side. It is useful to consider the angle as having been formed by a rota tion from the initial side to the terminal side as shown in FIGURE 2.1.1(a). An angle is said to be in standard position if its vertex is placed at the origin of a rectangular coordinate 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 of 360 degrees (written 3600) to the angle formed by one complete counterclockwise rota tion, as shown in FIGURE 2.1.2. Other angles are then measured in terms of a 360° angle, of a complete rotation. If the rotation is counter with a 1° angle being formed by clockwise, the measure will be positive; if clockwise, the measure is negative. For exam ple, the angle in FIGURE 2.1.3(a) obtained by one-fourth of a complete counterclockwise rotation will be (360°) 90°. = Shown in Figure 2.1.3(b) is the angle formed by three-fourths of a complete clockwise rotation. This angle has measure (—360°) = —270°. V 4 360° (a) 900 angle FIGURE 2.1.2 FIGURE 2.1.3 360 degrees + 3600 O—360, Three coterminal anges FIGURE 2.1.4 90 (b) —270° angle Positive measure in (a); negative measure in (b) Angle of O Coterminat AngLes Comparison of Figure 2,1.3(a) with Figure 2.1.3(b) shows that 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 are coterminal. 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 angle results in a coterminal angle. Conversely, any two coterminal angles have degree meas ures 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°. CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY olution (a) We first determine how many full rotations are made in forming this gle. Dividing 960 by 360 we obtain a quotient of 2 and a remainder of 240. Equivalently e can write 960 = 2(360) + 240. hus, this angle is formed by making two counterclockwise rotations before complet of another rotation. As illustrated in FIGURE 2.1.5(a), the terminal side of 960° ig = 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° / (a) FIGURE 2.1.5 220: (c) (b) 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 of grees by decimals, such as 42.23°. Traditionally, however, fractions of degrees were pressed 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 decimal grees to Degrees, Minutes, and Seconds (DMS notation), and vice versa. The fol wing examples show how to perform these conversions by hand. EXAMPLE 2 Usinq (1) and (2 onvert: 86.23° to degrees, minutes, and seconds, ) 17°47’13” to decimal notation. i) iilution In each case we will use (1) and (2). i) Since 0.23° represents j of 1° and 1° 86.23° = = = 60 we have 86° + 0.23° 86° + (0.23)(60’) 86° + 13.8 ow 13.8’ = 13’ + 0.8’, so we must convert 0.8’ to seconds. Since 0.8’ represents and l’= 60”, we have of 86° + l3’+ 0.8’= 86° + 13’+ (0.8)(60”) = 86° + 13’+ 48”. ence, 86.23° 86°13’48”. lie use of the number 60 as a base dates back to the Babylonians. Another example of the e of this base in our culture in the measurement of time (1 hour = 60 minutes and minute = 60 seconds). 2.1 Angles and Their Measure 91 (b) Since 1° have = 60’,itfollowsthat 1’= ()°.Simi1arly, I” = ()‘= 17°47’13” Thuswe = 17° + 47’ + 13” = 17° + 47()° + l3()° 17° + 07833° + 00036° = 17.7869°. D Radian Measure Another measure for angles is radian measure, which is gen erally used in almost all applications of trigonometry that involve calculus. The radian measure of an angle 0 is based on the length of an arc on a circle. If we place the vertex of the angle U at the center of a circle of radius r, then 0 is called a central angle. As we know, an angle 0 in standard position can be viewed as having been formed by the ini tial side rotating from the positive x-axis to the terminal side. The region formed by the initial and terminal sides with a central angle 0 is called a sector of a circle. As shown ifl FIGURE 2.1.6(a), if the initial side of 0 traverses a distances along the circumference of the 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 cir cumference of the circle equal to the radius r of the circle, then we see from (3) that the measure of the angle 0 is 1 radian. See Figure 2.1.6(b). Terminal side Terminal r/ side /o S r / Initial _____// \ r side r / Initial __,,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, all we need do is to draw another circle centered at the vertex of 0 of radius r’and sub tended arc length s See FIGURE 2.1.7. Because the two circular sectors are similar the ratios s/r and s7r’ are equal. Therefore, regardless of which circle we use, we obtain the same radian measure for 0. In equation (3) any convenient unit of length maybe used for s and r, but the same unit must be used for both s and r. Thus, FIGURE 2.1.7 circLes Concentric 0(in radians) = s(units of length) r(units of length) appears to be a “dimensionless” quantity. For example, ifs = 6 in. and r = 2 in., then the radian measure of the angle is 0 = 4 in. —---- 2 In. 2, where 2 is simply a real number. This is the reason why sometimes the word radians is omit ted when an angle is measured in radians. We will come back to this idea in Section 3.1. 92 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY One complete rotation of the initial side of 0 will traverse an arc equal in length to the circumference of the circle 2irr. It follows from (3) that one rotation s = — r 27rr = — r 2 radians. We have the same convention as before: An angle formed by a counterclockwise rota tion is considered positive, whereas an angle formed by a clockwise rotation is nega and 3 tive. In FIGURE 2.1.8 we illustrate angles in standard position ofr/2, —r/2, radians, respectively. Note that the angle of ir/2 radians shown in (a) is obtained by one-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 rota tion, is —ir/2 radians. The angle shown in Figure 2.1.8(c) is coterminal with the angle shown in Figure 2.1.8(d). In general, the addition of any integer multiple of 2ir radians to an angle measured in radians results in a coterminal angle. Conversely, any two coter minal angles measured in radians will differ by an integer multiple of 2r. y y+ (a) FIGURE 2.1.8 (b) AngLes measured in radians EXAMPLE 3 y yt (d) (c) A Coterminat AngLe Find an angle between 0 and 2- radians that is coterminal with 0 Sketch the angle. = 1 17r14 radians. Solution Since 2r < 11 ‘w/4 < 3ir, we subtract the equivalent of one rotation, or 2 radians, to obtain llr 1lr 8w 32r 3r 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 in FIGURE 2.1.9. CoterminaL ang’es in ExampLe 3 FIGURE 2.1.9 0 Conversion FormuLas Although many scientific calculators have keys that con vert between degree and radian measure, there is an easy way to remember the rela tionship between the two measures. Since the circumference of a unit circle is 27r, one complete rotation has measure 2 radians as well as 360°. It follows that 360° = 2r radians or 180° rradians. (4) If we interpret (4) as 180(10) = ir (1 radian), then we obtain the following two formulas for converting between degree and radian measure. 2.1 AngLes and Their Measure 93 CONVERSION BETWEEN DEGREES AND RADIANS 10 (5) —fl—radian 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 :EXAMPLE4 and 57.29578°. 1 radian 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°) = F’ 20( —radian 1=—radian. \\180 9 J (b) To convert from radians to degrees we use (6): 1T 7 —radians 6 71T = — 6 . (1 radian) = 7irul8ON° —I j — 6\rJ (c) We again use (6): 2radians TABLE 2.1.1 Degrees 0 Radians 0 = 2(lradian) ul8ON° 2( 1 1360N° =1—) 1 210°. approximate answer rounded to two decimai places 114.59°. Table 2.1.1 provides the radian and degree measure of the most commonly used angles. 30 45 60 90 180 D TerminoLogy You may recall from geometry that a 90° angle is called a right angle and a 180° angle is called a straight angle. In radian measure, r/2 is a right angle and IT is a straight angle. An acute angle has measure between 0° and 90° (or between 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 if their sum is 90° (or ii-/2 radians). Two positive angles are supplementary if their sum is 180° (or IT radians). The angle 180° (or radians) is a straight angle. An angle whose terminal side coincides with a coordinate axis is called a quadrantal angle. For example, 90° (or IT/2 radians) is a quadrantal angle. A triangle that contains a right angle is called a right triangle. The lengths a, b, and c of the sides of a right triangle satisfy the Pythagorean relationship a 2 + b 2 = c , where c is the length of the side 2 opposite the right angle (the hypotenuse). 94 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY EXAMPLE 5 Complementary and Supplementary Angles Find the angle that is complementary to 0 ) Find the angle that is supplementary to q i) = 74.23°. rr/3 radians. olution (a) Since two angles are complementary if their sum is 90°, we find the angle tat is complementary to 0 74.23° is 90° — 0 = 90° — 74.23° 15.77°. = ) Since two angles are supplementary if their sum is tat is supplementary to = /3 radians is i. 3 7T 2r ir radians, we find the angle radians. I Arc Length In many applications it is necessary to find the lengths of the arc sub nded by a central angle 0 in a circle of radius r See FIGURE 2.1.10. From the definition f radian measure given in (3), 0 (in radians) 5 = —. r y multiplying both sides of the last equation by r we obtain the arc length formula rO. We summarize the result. HEOREM 2.1.1 FIGURE 2.1.10 Length of arcs determined by a centra ange 0 Arc Length Formula or a circle of radius r, a central angle of 0 radians subtends an arc of length (7) srO. EXAMPLE 6 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. 6 inches, we olution (a) From the arc length formula (7) with 0 = 2 radians and r ave s = rO 2 6 = 12. So the arc length is 12 inches. rI6 radians. Then from the ) We must first express 30° in radians. Recall that 30° rc length formula (7) we have s = r0 = (12)(/6) = 2r. So the arc length is 6.28 feet. Exercises I Studenis ofien apply ihe arc length formula incorrectly by using degree measure. Remember .c = rO is valid only if 0 is measured in radians. Answers to selected odd-numbered problems begin on page ANS-7. Problems 1—16, draw the given angle in standard position. Bear in mind that the ick of a degree symbol (°) in an angular measurement indicates that the angle is teasured in radians. i 1 60° 5. 1140° 2 —120° 6. —315° 3 135° 7. —240° 4 150° 8. —210° 2.1 Angtes and Their Measure 95 9. 13. — 10. 11. 12. 14. —3- 15. 3 16. 4 — In Problems 17—20, express the given angle in decimal notation. 17. 10°39’l7” 19. 5°10’ 18. l43°7’2” 20. 10°25’ In Problems 2 1—24, express the given angle in terms of degrees, minutes, and seconds. 21. 210.78° 23. 30.810 22. 15.45° 24. 110.5° In Problems 25—32, convert from degrees to radians. 25 10° 29. 270° 27 45° 31. —230° 26 15° 30. —120° 28 215° 32. 540° In Problems 33-40, convert from radians to degrees. 33. 2r 9 — 34. 117r 35. 6 39. 3.1 38. 7r 37. 27r 3 — 36. 512 40. 12 In Problems 41—44, find the angle between 0° and 360° that is coterminal with the given angle. 41. 875° 43. —610° 42. 400° 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 50. 48. 17r 49. 5.3’w 52. 7.5 51. —4 — 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. 96 55. 48.25° 56. 93° 57. 98.4° 58. 63.08° 59. 60. 61. 62. CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY 63. Find both the degree and the radian measures of the angle formed by (a) threefifths of a counterclockwise rotation and (b) five and one-eighth clockwise rotations. 64. Find both the degree and the radian measures of the obtuse angle formed by the hands 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 the hour 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 2 and (b) radius 4. 71. Find the measure of a central angle 0 in a circle of radius 5 if 0 subtends an arc of length 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 of length 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 circle 0. See Figure 2.1.10. [Hint: Use the proportional 2 of radius ris given byA = r ity property from geometry that the ratio of the area A of a circular sector to the 2 of the circle equals the ratio of the central angle 0 to one complete total area rr revolution 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.] Planet Mercury in ProbLem 68 FIGURE 2.1.11 Circular band in Problem 74 MisceLLaneous AppLications 75. Angular and Linear Speed If we divide (7) by time t we get the relationship v = rw, where v = sit is called the linear speed of a point on the circumference of a circle and w = 0,/t is called the angular speed of the point. A comrnunica tions satellite is placed in a circular geosynchronous orbit 35,786 km above the surface of the Earth. The time it takes the satellite to make one full revolution around the Earth is 23 hours, 56 minutes, 4 seconds and the radius of the Earth is Satellite 6378 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 forth along a 15-cm arc. Find (a) the central angle and (b) the area of the sector through which the pendulum sweeps in one swing. [Hint: To answer part (b), use the result of Problem 73.) 77. Sailing at Sea A nautical mile is defined as the arc length subtended on the sur face of the Earth by an angle of measure 1 minute. If the diameter of the Earth is 7927 miles, find how many statute (land) miles there are in a nautical mile. 78. Circumference of the Earth Around 230 BCE Eratosthenes calculated the cir cumference of the Earth from the following observations. At noon on the longest day of the year, the Sun was directly overhead in Syene, while it was inclined 7.2° from the vertical in Alexandria. He believed the two cities to be on the same longitudinal line and assumed that the rays of the Sun are parallel. Thus he con2.1 AngLes and Their Measure FIGURE 2.1.12 Satellite in Problem 75 97 Rays .- - from 00 stades -( - 2 Sun Syene ‘exandria FIGURE 2.1.13 — - Earth in 79. Problem 78 80. 81. Yo-yo in Problems 79 and 80 82. 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 from 559 feet, find Syene to Alexandria was measured as 5000 stades. If one stade that Show miles. (b) the circumference of the Earth in (a) stades and Eratosthenes’ 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). 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. 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) the linear speed. 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. Diameter of the Moon The average distance from the Earth to the Moon as given 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. The curved red arc represents the approximate diameter of the Moon FIGURE 2.1.14 Right TriangLe Trigonometry 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 trigonometric functions—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 as sin, cos, tan, cot. sec, and csc, respectively. We begin with some terminology and a definition. Hypotenuse C b Side opposite Defining the trigono function of the angLe U metric 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 by 2 + b . The side opposite the right 2 2 = c 0. From the Pythagorean theorem we know that a angle is called the hypotenuse; the remaining sides are referred to as the legs of the tri angle. The legs labeled a and b are, in turn, said to be the side adjacent to the angle 0 98 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY a Side adjacent FIGURE 2.2.1 I- id the side opposite the angle 0. We will also use the abbreviations hyp, adj, and opp denote 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 sinO=—=— C hyp opp b tan0=—-=— a adj hyp c sec0=—— a adj . adj a cosO=—=— C hyp adj a cot0=—=— opp b hyp c csc0=—=— opp b (1) 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 than ute angles. Then in Chapter 3 we will see how the trigonometric functions can be fined with domains consisting of real numbers rather than angles. The values of the six trigonometric functions depend only on the size of the angle and not on the size of the right triangle. To see this, consider the two right triangles own in FIGURE 2.2.2. Because the right triangles have the same acute angle 0 they are nilar and thus the ratios of the corresponding sides are equal. For example, from the d triangle in Figure 2.2.2(a) we have • sin 0 opp = hyp C b a (a) lb b = —, C [iereas in the smaller blue triangle in Figure 2.2.2(b) we find that • sin 0 opp hyp (b) FIGURE 2.2.2 Similar triangles b’ c —,. it since the red triangle is similar to the blue triangle we must have b’ b C’ C• other words, we get the same value for sin 0 regardless of which right triangle we use compute it. A similar argument can be made for the remaining five trigonometric nctions. EXAMPLE I Values of the Trigonometric Functions nd the exact values of the six trigonometric functions of the acute angle 0 in the right angle shown in FIGURE 2.2.3. ilution Matching Figure 2.2.3 with Figure 2.2.1 we see that the side opposite the gleO 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. 2.2 Right Triangle Trigonometry hyp 8 15 FIGURE 2.2.3 Right triangle in Example 1 99 Thus from (1) the values of the six trigonometric functions are 15 adj a cosO=—=—=—, C 17 hyp adj 15 b cotO=—=—=—, opp a 8 17 hyp c cscQ=—=—=—. opp 8 b b opp 8 sinO=—=—=—, C 17 hyp a 8 opp tanO__=—=—-— 15 b adj 17 c hyp secO=—=—=— a 15 adj D Quotient and ReciprocaL Identities There are many important relationships among the trigonometric functions. The basic ones listed next are referred to as the fundamental identities and should be memorized. Quotient Identities: tanO sinQ cosO = cotO , = cosQ sinO Reciprocal Identities: secO = 1 cosO 1 sinO cscO , 1 tanO cotO —--—, Identities (2) and (3) can be obtained from Definition 2.2.1. For example, the first of the quotient identities is verified as follows: sinQ cosO = opp/hyp adj/hyp = opp adj = tan ü The others can be verified in a similar manner. Using these identities, we can find the values of all six trigonometric functions once we know the values of sinO and cos 0. EXAMPLE Using (2) and (3) 2 and cos 0 Given sin 0 , find the values of the remaining four trigonometric functions. Solution From the fundamental identities, we have tan0 = secO = cscO = cot0 = sinO cosO 1 cos0 1 sinO 1 tan0 — - 4 = = = = 4 3 15 3 15 = 4 13 = 4 = — from (2) — — 4—from(3) — — —. Although we computed cot 0 using the reciprocal identity in (3), we could have also com puted cot 0 using the quotient identity in (2). EXAMPLE 3 Given cosO = Using a Right TriangLe andtan0 = 2V’. Find sin 0. Solution We can obtain sin 0 by multiplying the first identity in (2) by cos 0: sin0 100 = cos0tan0 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY 1 = —. 2V’ 2V = The next example illustrates that ifjust one trigonometric function value of an acute angle is known, it is possible to find the other five function values by drawing an appropriate 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 making pp = 2 and hyp = 7 as shown in FIGURE 2.2.4. From the Pythagorean theorem we have , 22 + (adj) 2 = 72 so that 2 (adj) 72 = — 22 = 45. adj = 3\/5 fhus, adj = The values of the remaining five trigonometric functions are obtained from the iefinitions in (1): cosO adj = hyp opp 3\/ 7 2 tan0=—= adj 3\/ hyp 7 secO=—= 3\/ adj 3\/ adj cotO = = opp 2 hyp 7 = opp 2 2V’ = FIGURE 2.2.4 Example 3 7v = 15 — 15 csc0 = —. ] 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, if he two acute angles of a right triangle ABC are labeled a and /3 and a is the length of he 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 sin a Right triangle in cos [3, — cos a b = — = B a A b C Acute angles ci’and/3in a righttriangle sin [3, FIGURE 2.2.5 a tana = seca = - I, = cot/3, cota = = csc/3, csca = — = tan/’3 = secl3. 3ecause the sum of the angles in any triangle is 1800 (or ir radians), the acute angles a md /3 in a right triangle are complementary. Thus the cosine of an acute angle equals he 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 of co,nplementaiy angles are equal. ] Cofunction hen (4) Identities If a and /3 are the acute angles in the triangle in Figure 2.2.5, or 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( sinO cosQ- — o) tan( cotO tanO — = cot( — — cscO = sec( secO = csc( — — e) o) or equivalently cosO sinO = sin(90° cos(90° — — cotO tanO 0) 0) tan(90° cot (90° = — — cscO secO 0) 0) = sec(90° csc(90° — — 0) 0). (6) In (5) and (6) it is understood that 0 is measured in radians and degrees, respectively. Using (5) and (6) EXAMPLE 5 From (5): complementary angles ,, (a) cot = tan( — (b) cos sinQ — ) ) , = tan = sin. From (6): (c) csc27° (d) cot 15° = sec(90° tan(90° — — 27°) 15°) = = sec63° 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 2 + b a 2 = . 2 C /c b /c + 2 a , we obtain 2 2 If we divide this last result by c = I or 2 (bN (aN 2 +1—I =1 \\cJ \\cJ (7) I—I 2 2 + b Similarly, if we divide a = 2 we obtain, in turn, 2 and b 2 by a c (bV 1+1— / (C \7 = - (8) and 2 (aN (CN +1=1— \b \\b) 102 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY (9) ig the appropriate definition in (1) in the results (7), (8), and (9) yield another set nportant identities. aagorean Identities: O = 1 2 sin + cos O 2 O = sec 2 1 + tan O 2 O + 1 = csc 2 cot O. 2 (10) (11) (12) the square of the trigonometric functions are written (sin 0)2 = ), (10), and (11), 7 = cos0, (tanO) = IXAMPLE 6 sin O 2 , tan 2 0, and so on. Using (11) is an acute angle and tan 0 = \/3 find the value cos 0. There are several ways of solving this problem. One way is to use the agorean identity (11): [tion 0 sec 0 2 + 1=5+1=6 +l=(V) =tan so secO = \/. Because secO = 1/V = V/6. 1/cosO we have cosO = 1/sec 0. Therefore NOTES FROM THE CLASSROOM we will see in Section 3.4, all the identities given in this section hold for any angle not just acute angles). Exercises Answers to selected odd-numbered problems begin on page ANS-7. oblems 1—10, find the values of the six trigonometric functions of the angle Gin iven triangle. 5 12 FIGURE 2.2.7 3 FIGURE 2.2.6 TriangLe for TriangLe for ProbLem 2 ProbLem 1 FIGURE 2.2.8 ProbLem 3 TnangLe for FIGURE 2.2.9 TnangLe for ProbLem 4 2.2 Right Triangle Trigonometry 103 >12 Triangle for FIGURE 2.2.10 FIGURE 2.2.11 Problem 5 Triangle for Problem 6 7. 1.2 FIGURE 2.2.12 TriangLe for FIGURE 2.2.13 ProbLem 7 Triangle for ProbLem 8 10. 9. y x FIGURE 2.2.14 S TriangLe for FIGURE 2.2.15 Problem 9 Triangle for Problem 10 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. 2 11. sinO = 13. sin 0 = 15. sinO = 17. csc0 = 19. cos0 = , 2 cos0 cos 0 —, 1 , 5 —, 1 —, 3 3\/ = tanO sec0 cscO 3 = 1 8 = — 5 = — 3 2V 12. sinO = cos 0 = 14. sin0 = cos 0 = cotO = , cot 0 = 7 , cot 0 = 7 16. cos0 = 18. sin 0 = 20. sin 0 = v 2 In Problems 2 1—28, find the value of the remaining trigonometric functions by drawing an appropriate triangle. 21. sin0 23. sec 0 = 25. tan 0 = 27. sec 0 29. 30. 31. 32. 104 = 12 13 — 5 7 = — Ifcos75° Ifcos75° = If tan(ir/8) = If tan(ir/8) = — — — — 22. cos0 = 24. csc 0 = 26. cot 0 = 28. tan 0 = ViO — 7 3 v’), find the exact value of sinl5°. V’),findtheexactvalueofsec75°. 1, find the exactvalue of cot(3ir/8). 1,findtheexactvalueoftan(3’ff/8). CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY I. In Problems 33-46, use the identities of this section to find the exact value of the given trigonometric expression. Do not use a calculator. 33. 3sin- + 3cos — 2 34. + sin35° sin 5 2 5° 35. 1 + 2 sin 18° cos l 8° + 2 33° 2 36. 1 + tan 37. 38. 2 —4csc 1 13° 2 3° + 4cot tan — sec2 — sec 3 2 3° sin 100 sin 100 sin 80° cos 100 41. 5cot4l°cot49° 42. cosl1°sec11° 43. sin 28° cot 28° csc 62° 44. lOsin—cot—sec— 45. sin 10°cos 80° + cos 10°sin80° 46. tan 30°cot60° 40. sec20° — csc70° — .T IT 3 3 IT — 3 sec30°csc60° [n Problems 47—54, given that cos 300 = V/2. Use the identities of this section to lind the exact value of the given trigonometric function. Do not use a calculator. 47. sin 30° 48. cos 60° 50. cot 30° 52. csc 30° 54. tan 30° + cot 60° 49. tan 60° 51. sec 30° 53. cos 30°tan 30° Trigonometric Functions of SpeciaL 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 trigonom try 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 these ialues using some results from plane geometry. J VaLues of sin 45° and cos 45° To find the values of the sine and cosine functions a 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 triangle ire equal; therefore, each acute angle measures 45°. We can find the length of the lypotenuse by using the Pythagorean theorem: 2 (hyp) = (1)2 + (1)2 = 2 gives hyp= hyp Ehus from(1) of Section 2.2 we obtain Isosceles right triangle FIGURE 2.3.1 sin45 opp hyp 1 \/ V 2 cos45 adj hyp 1 \/ 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 AC LAOB LAOC = AB (2) = = 1 i(60°) = and = 30°, LACO = 90c. A A 2 2 2 2 B B (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: 1 PP 2 hyp adj = cos30° = hyp 2 sin30 0 D VaLues of sin 60° and cos 60° Now using the 60° angle in the red right triangle 2. ACO in Figure 2.3.2(b) we identify opp = V’, adj = 1, and hyp Therefore sin 60° = cos60 0 hyp adj hyp = - 2 1 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° 106 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY sin 30° —— —. 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 reciprocal lentities in Section 2.2 along with the results in (3) and (4) we get = an 6 iT cot—= 6 iT sec-= sin(r/6) cos(/6) 1 tan(ir/6) 1 cos(ir/6) 1/2 — — = 1 — /2 — 3 — — =-‘; 1/v’ 1 2 2V”i 3 V/2 iT csc—= 6 sin(ir/6) 1/2 We leave finding the values tan 0, cot 0, sec 0, and csc 0 for 0 = 7T/4 and 0 = ir/3 s 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 we avejust determined for the special angles 30°, 45°, and 60°. As mentioned in the intro uction to this section, these function values are used so frequently that we feel that ey should be committed to memory. Knowing these values and the fundamental Ientities we discussed earlier will enable you to determine any of the trigonometric inctions for these special angles. TABLE2.3.1 0 0 (degrees) (radians) sin 0 cosU 1 2 \/ 2 iT 30° — 6 — \/ — 3 v/ iT 45° tan0 1 — 4 2 2 1 IT 60° 3 EXAMPLE 2 Find the Exact Values md the exact value of the given trigonometric expression. i) sin2 — cos (b) cos30°tan60° iT (c) 2 + 4sin— 3 iT — 6cos— 6 lution In each case we will use the information in Table 2.3.1. sin— — -r cos— ) cos30° tan60° = — — ±_LJ_ I 2422 = 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 and 5j for computing the values of these functions. To obtain keys labeled keys with the reciprocal or the values of csc, sec, or cot, we can use the The following example illustrates the process. key , . EXAMPLE 3 Using a CaLcuLator Use a calculator to approximate each of the following. (a) sin45° (c) sec 0.23 (b) cos8°l5’ (d) cot Solution (a) First we make sure that the calculator is set in degree mode. Then we key to obtain enter 45 and use the sin 45° 0.707 1068, 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 = 8.25°. Now with the calculator set in degree mode, we form: 8°15’= 80 + enter 8.25 and use the j key to obtain () cos 8°15’= cos 8.25° 0.98965 14. (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 key, and then take the reciprocal calculator set in radian mode, we enter 0.23, use the key. Thus we have of the result by pressing the 1 cos 0.23 sec0.23 1.0270458. (d) We observe that this angle is measured in radians and set the calculator accordingly. key to obtain key, and then the We first enter r, divide by 7, use the cot— 7 I Exercises = — tan 2.0765214. — 7 Answers to seected odd-numbered probLems 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° 108 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY 2. ir/3 In Problems 3—22, find the exact value of the given trigonometric expression. Do not use a calculator. IT cos — 3. 2 3 5. sec45°csc45° •IT 7. sin — cot 4. 2 tan — 6 6. sin60°cos30° IT 8. 6 sec IT — csc IT — 10. tan 600 cot 30° 9. 9 sec 450 csc 450 11. sincos + cosjsin 12. coscos 13. 6 tan 30° + 7 tan 60° 14. 3 sin 15. tan45° 16. sec2 + 4csc2 4 3 2 \/sin(ir/4) 18 17 — cot45° 8sin(Ir/4) tan(ir/4) sinsin 5 cos — sec(ir/3) cos(r/4) 19. sin 3O° + cos 2 45° 2 21 — — — tan(Ir/6) 1 + tan(w/4)tan(n-/6) 20. 2 + cot 3O° 2 22 — 3O° 2 lOcsc tan(7T/3) + tan(ir/4) 1 — tan(ir/3)tan(w/4) In Problems 23—32, use a calculator to find the approximate values of the six trigonometric functions of the given angle. Round your answer to four decimal places. 23 17° 25 143° 27. 71°30’15” 24 82° 26 3475° 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 tan— tan— tan— 180 180 180 89w tan—. 180 Trigonometric Functions of GeneraL Angles Introduction Until now we have defined the trigonometric functions only for acute angles. However, many applications of trigonometry involve angles that are not acute. Consequently, it is necessary to extend the definition of the six trigonometric functions in (1) of Section 2.2 to general angles. Naturally, we want the extended def inition to agree with the earlier definition whenever the angle is acute. To accomplish this we proceed in the following manner. 2.4 Trigonometric Functions of Generat Anges 109 y P(x, Y) r Let 0 be an acute angle in standard position_and choose a point P(x, y) on the ter Vx2 + y ,weseeinFIGIJPE2.4.1 thatx,y,and 2 minalsideofO.Ifweletr = d(O,P) hyp, r are the lengths of the sides of a right triangle. With y = opp, x = adj, and r 2.2.1, andwehavefromDefinition sinO FIGURE 2.4.1 An acute ange and cosO = tan0 (I) . The expressions in (1) provide us with a model on which to base our extended defini tion for any angle 0 in standard position, such as those illustrated in FIGURE 2.4.2. y y P(x, y) x x r x y P(x, y) P(x, y) (b) (a) ‘*. (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) 2 is the distance between (0, 0) and P(x, y), on the terminal side of 0. If r = V’x2 + y then the trigonometric functions are defined to be y sinO=— x cosO=— tan0= cot0= sec0=— x csc0=— y (2) 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 The angles are odd multiples of ir/2. 110 2 is never zero. Thus the (0, 0), then r = Vx2 + y inator is zero. Since P(x, y) 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 those having radian measure ±r/2, ±3ir/2, ±5’rr/2, and soon. Using set notation, CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY nd the fact that an odd integer can be written as 2n + 1, n an integer, the domains of ie tangent and the secant functions are (2n + {o(o (2n + l)r/2,n {ojo = 1)900,11 0, ±1, ±2,... } 0, ±1, ±2,.. ‘he cotangent and the cosecant functions are not defined for angles with their terminal ides on the x-axis because then y = 0. Thus the domains of cot 0 and csc 0 consist of 11 angles 0 except those having radian measure 0, ± r, ± 2, ± 3, and so on; that is, n7T,n = 0, ±1, ±2,.. .}or{00 00 180°n,n = 0, ±1, ±2,...}. Since r x/rI 1 and i/ri it follows that Ix r and 1. Therefore, as before lsinOI imilarly, because I r/x I 1 IcosOl and 1 and r/y cscoi 1 II The angles are integer multiples of if. r, or equivalently, 1. (3) 1 we have and IsecOl 1. (4) ‘he inequalities in (3) and (4) hold for every 0 in the domain of each of these functions. EXAMPLE 1 Values of the Trigonometric Functions md the exact values of the six trigonometric functions of the angle 0 if 0 is in standard osition and the terminal side of 0 contains the point P( —3, 1). olution The terminal side of the obtuse angle 0 is sketched in lentificationsx —3,y = 1, and FIGURE 2.4.3. With the yi = P(-3, I) ‘e have from (2), x Vi = sec0 = tanO = sin0 cos0 -3 = - = —3 3 —3 = —j-— = csc0 —, 3 3Vi 10 FIGURE 2.4.3 Example 1 —_____ = = cot0 —-— v i 1 d = —3 EXAMPLE 2 Angle 0 in —3, = Values of the Trigonometric Functions md the values of the six trigonometric functions of 0 if 0 —7r/2. = lution First we place 0 in standard position as shown in FIGURE 2.4.4. According to efinition 2.4.1, we can choose any point P(x, the terminal on of 0. For con y) side v2 + y 4 mience,weselectP(0, —1) so thatx = O.y = —1,andr = \ 2 = l.Thus, .Ir = coti = sini —1 1 — —l —1 cos 0 = — owever, the expressions tan0 ncex = 0. csc v/x and sec0 ( ( x 0 0 2) 1T\ = P(O, —1) rtv are undefined forD —/2 FIGURE 2.4.4 Angle 0 in Example 2 Algebraic Signs Depending on the quadrant in which the terminal_side of 0 lies, e or both coordinates of P(x, v) may be negative. Since r = Vx2 + y 2 is always 2.4 Trigonometric Functions of GeneraL Angles 111 positive, each of the six trigonometric functions of 0 has negative as well as positive val y/r is positive if the terminal side of 0 lies in quadrants I or ues. For example, sinO 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. I II cos8<O sin9>O cos6>O sinO>O tanO<O cot8<O tan6>O cot8>O secO<O csc8>O secO>O csc>O cos6<O sin9<O cosO>O sin8<O tanO>O cot8>O tanO<O cotO<O secB<O cscO<O sec8>O cscO<O IV III FIGURE 2.4.5 Agebraic signs of the six trigonometric functions EXAMPLE 3 Using Figure 2.4.5 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 P(, ) r FIGURE 2.4.6 An arbitrary The reciprocal, quotient, and Pythagorean 2.2 also hold for general angles. For exam Section given in acute angles identities for be ple, to derive the Pythagorean identities let 0 any angle in standard position. As showr in FIGURE 2.4.6, we let P(x, y) be any point other than the origin on the terminal side ol .Dividingbotl 2 2 = r 0.Ifweagainletr = d(O, P) = Vy2,thenwehavex2 + y gives 2 r by sides of the last equation Pythagorean Identities—Revisited U —+‘--= r Recognizing that x/r = r 1 cos 0 and y/r \ fy\ /X\ I—I +1—I =1. \rJ J or = / \ / ange 0 sin 0, we obtain the basic Pythagorean identit3 2 0 + cosO 2 sin = 1. (5 2 0 be written first. If both sides o In (5) we have followed the convention that sin obtain 0 we 2 sin and 0 2 cos turn, by in divided, (5) are 1 + tan 0 2 and 0 + 1 2 cot = O 2 sec (6 = . csc 6 2 (7 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. 112 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY EXAMPLE 4 Using (5) liven that cos 0 and that 0 is a quadrant IV angle, find the exact values of the maining five trigonometric functions of 0. olution Substituting cos 0 into (5) gives = (1)2 + sinO = 1 — — = 99 = —. ince the terminal side of 0 is in quadrant IV sin 0 is negative. Therefore, we must ]ect the negative square root of: sinO = [ — (See Figure 2.4.5. 2\/ —1• = low using sin 0 tan0 secO = cotO cos 0 1 cos 0 1 = csc0 tanO = sin0 e find that the values of the remaining four functions are tanO = = secO = —2, cot0 3, 1 = csc0 EXAMPLE 5 = -2V 1 —, 4 3V —2v’/3 4 Using (6) iven tanG = —2 and sinO inctions of 0. lution Letting tanG 0, find the exact values of the remaining five trigonometric > —2 in the identity I + tan 20 O 2 sec = 1 + (_2)2 = 2 0, we find sec = 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 sec0 :om sec 0 = = —\/.. 1/cos 0, it follows that 1 1 cos0=—= —\/ secO sing tanG = sin 0/cos 0, we obtain sinG ien 5 = cosG tanG 1 csc0 = cotO = 2\/ = 1 = I — tanG —-----)(_2) = 2V3)5 1 —2 = = 1 — 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. 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. FIGURE 2.4.7 illustrates this definition for angles with the terminal sides in each of the four quadrants. -V V 8 8=0’ >Jç e5 (a) (d) (c) (b) 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. (a) 0 = (b) 0 40° 271 (c) 0 = 210° 971 (d) 0 = Solution (a) From FIGURE 2.4.8(a) we see that 0’ 40°. 271/3 = 71/3. 0 = (b) From Figure 2.4.8(b), 0’ 180° 30°. 180° = 210° (c) FromFigure2.4.8(c),0’= 0 —971/4 is coterminal with (d) Since 0 — — — — 971 —-a- + 2ir 71 = we find that 0’ = 71/4. See Figure 2.4.8(d). y V 9, (a) FIGURE 2.4.8 114 1’ (b) Reference angLes in ExampLe 6 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY (c) (d) x I 1 Property of Reference AngLes The usefulness of reference angles in evaluat ing trigonometric functions is a result of the following property: 4 The absolute value of any trigonometric function of an angle 0 equals the value ofthatfimnction 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) r y=IyII x .. (a) FIGURE 2.4.9 (b) (c) Reference angles We can now describe a step-by-step procedure for determining a trigonometric unction value of any angle 0. FINDING THE VALUE OF A TRIGONOMETRIC FUNCTION Suppose 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 considering in which quadrant the terminal side of the angle 0 lies. EXAMPLE 7 Finding VaLues Using Reference AngLes md the exact values of sin 0, cos 0. and tan 0 for each of the following angles. 27r 97T (b) 0 = 210° (c) 0 = ) 0 = 4 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 and ie tangent are negative, we conclude \/ 2r s1n—— = 2 cos—— = —i-,1 and 2ir tan— 3 = —\/. 2.4 Trigonometric Functions of General Angles 115 (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° tan 210° tan 30° = ——, see Figure 2.4.5 for the correct algebraic signs — 2 = (c) From part (d) of Example 6 we know that the reference angle 0’ = 0 = —9ir/4 is a quadrant IV angle, it follows that 9irN 1 .f sini I \ 4/ —-——— cos_T) —sin— 4 = — 2 ‘r = cos = —tan-a- = —1. Finding AngLes Find all angles 0 satisfying 0° y Since 1T = iT EXAMPLE 8 iT/4. 0 < 3600 such that sin 0 = Solution From what we know about the special angles 30°, 60°, and 90°, we know that = 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°. o FIGURE 2.4.10 Solutions in Example 8 .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. 3,r 4 (b) (a) FiGURE 2.4.11 116 Solutions in Example 9 CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY NOTES FROM THE CLASSROOM In this section we have purposely avoided using calculators. For a complete understanding of trigonometry, it is essential that you master the concepts and be able to perform without the aid of a calculator the types of calculations and simplifications we have dis cussed. The exercise set that follows should be worked without the use of a calculator. Exercises P1 Answers to se’ected odd-numbered probems begin on page ANS-7. e recommend that you do not use a calculator in solving any of the problems thatfohlow. 7 4 n Problems 1—10, evaluate the six trigonometric functions of the angle 0 if 0 is in stanlard position and the terminal side of 0 contains the given point. 1. 3. 5. 7. 9. (6, 8) (5, —12) (0,2) (—2, 3) (—v’,—l) 2. 4. 6. 8. 10. (—1,2) (—8, —15) (—3,0) (5,—I) (V/) n Problems 11—18, find the quadrant in which the terminal side of an angle 0 lies if 0 atisfies the given conditions. Li. .3. .5. .7. sin0 <Oandtan0 >0 tanO <0 and sec0 <0 cotO > 0 and sinO > 0 sinO > Oandcos0 <0 12. 14. 16. 18. I cos9 > Oandsin0 <0 secO <Oandcsc0 <0 csc0 > Oandcot0 <0 tan0 <Oandcsc0 >0 n Problems 19—28, the value of one of the trigonometric functions of an angle 0 is iven. From the given value and the additional information, determine the values of the ive remaining trigonometric functions of 0. .9. 1. 3. 5. 7. sinO tan 0 csc 0 sinO tan0 9. 0. 1. 2. 3. 4. If cos 0 = find all possible values of sin 9. If sin 0 = find all possible values of cos 0. If 2 sinO cos 0 = 0, find all possible values of sin 0 and cos 0. Ifcot0 = find all possible values ofcsc0. If sec 0 = —5, find all possible values of sinO and cos 0. If 3 cos 0 = sin 0, find all possible values of tan 0, cot 0, sec 0, and csc 0. = = = = = 0 is in quadrant II 3, 0 is in quadrant III —10, 0 is in quadrant IV cosO > 0 8, sec0 > 0 , —, 20. 22. 24. 26. 28. cos 0 cot 0 sec 0 cos0 sec0 = = = 0 is in quadrant II 2, 0 is in quadrant III 3, 0 is in quadrant IV sin0 <0 —4, csc0 <0 —, —, = , —, — , 2.4 Trigonometric Functions of Genera Anges 117 35. Complete the following table. 0 (degrees) 0 (radians) sin 0 cos 6 0° 0 0 1 300 IT/6 1/2 450 ir/4 \//2 \//2 60° /3 ‘.//2 1/2 90° ir/2 1 0 120° 2/3 \//2 —1/2 135° 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 csc 6 sec 0 tan 0 — —v’s 36. Complete the following table. 0 (degrees) 0 (radians) 0° 0 30° ir/6 45° T/4 60° Tr/3 2V’/3 90° /2 I 120° 2ir/3 135° 3r/4 150° 5ir/6 180° 118 210° 7/6 225° Sir/4 240° 4ir/3 270° 3rr/2 300° 5ir/3 315° 7-/4 330° 1l7r/6 360° 27r CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY — 2 1 cot 0 — 2\//3 2 — v/3 0 In Problems 37—52, find the exact value of the given expression. 37. cos5r 38. 13r 39. cot— 40. tan— 41. sin(_) 23ir 42. cos—-— 43. 45. 47. 49. 51. 9r csc (—i) & t 23ir 44. tan—T sec(—120°) sin 1500 tan 405° cot (—720°) 46. 48. 50. 52. In Problems 53—58, find all angles 0, where 0 condition. 0 53. tanO = 54. sin0 = 55. coso = 56. sec6 = 57. csc0 = 58. cotO = 2 —1 In Problems 59—64, find all angles 0, where 0 59. sinOO 61. sec0 = —\/ 63. cot0 = —V h csc495° cos(—45°) sin3l5° sec(—300°) 0 < 60.cos0 62. csc0 64. tan0 < 360°, satisfying the given —— 2\/ 3 1 2r, satisfying the given condition. = = = —1 2 1 MisceLLaneous AppLications 65. Free Throw Under certain conditions the maximum height y attained by a basketball released from a height h at an angle a measured from the horizontal with an initial velocity v 0 is given by y = h + (v sin a)/2g, 2 where g is the acceleration due to gravity. Compute the maximum height reached 0 = 8 m/s, a = 64.47°, and g = 9.81 rn/s by a free throw if h = 2.15 m, v . 2 66. Putting the Shot The range of a shot put released from a height h above the ground with an initial velocity v 0 at an angle 0 to the horizontal can be approximated by cosS 0 v R = g 0 sin + (v Free throw 2 + 2gh), \/vö sin where g is the acceleration due to gravity. , compare the ranges 2 (a) If v 0 = 13.7 m/s, 4 = 40°, and g = 9.81 m/s 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? 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 1 g 978.0309 + (a) Find g 1 at the equator (0 latitude. y Line through origin in Probem 71 FIGURE 2.4.12 CONCEPTS REVIEW 0 2 5.18552 sin = 0. sin 2 0.00570 2 0°), (b) at the north pole, and (c) at 45° north For Discussion 68. Is there an angle 0 such that cos 0 = ? Explain. 1? Explain. 69. Is there an angle 0 such that 2 csc 0 without the aid of a calculator that both to determine possible it how is 70. Discuss 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 of each ofthefollowing concepts. Straight angle Quadrantal angle Supplementary angles Right angle Arc length Conversion: degrees to radians radians to degrees Reference angle Right triangles: side adjacent side opposite Initial side of an angle Tenninal side of an angle Standard position of an angle Coterminal angles Minutes Seconds Degree measure of an angle Central angle Radian measure of an angle Acute angle Complementary angles Obtuse angle CHAPTER 2 hypotenuse Trigonometric functions: of acute angles of general angles Quotient identities Reciprocal identities Pythagorean identities Cofunctions Cofunction identities Review Exercises A. True/False In Problems 1—10, answer true or false. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 120 — sin(-rr/6) cos(/3) sin(r/2) = sin(5ir/2) sin = 30°__ tan7r = 0 1 cscOl (90° —0) = 1 2 O + sin 2 sin The angles 120° and —240° are coterminal. 2 and cos 0 = 5. If tan 0 = then sin 0 If sec (3 = \/7, then cos 0 = 30’ is equivalent to 0.5°. = , CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY Answers to selected odd-numbered problems begin on page ANS-8. 15. 27. sinO 29. 3(\/ rathans 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 75. 77. 79. 81. 19. 5.17° 10.6547° 23. 30048360 210°46’48” 27. r/4 ir/18 31. —23/18 3ir/2 400 35. 1200 39. 177.62° 225° 43. 110° 155° 47. 7ir/4 —205° 4 2.28 51. 2 1.3r —-/4 (b) 131.75° (a) 41.75° (a) The given angle is greater than 900. (b) 81.60 (b) 3r/4 (a) /4 than greater is Tr/2. given angle (a) The (b) ir/3 (b) —1845°; —10.25r (a) 216°; 1.2ir because the hour hand moves counterclockwise: —60°, —7r/3 (b) 2h (a) 16h (b) 15 (a) 9 (b) 85.94° (a) 1.5 (b) 3.074641 km/s (a) 0.000072921 rad/s miles 1.15 statute (a) 3irrad/s IT cm/s 0 (b) 30 (a) 711.1 rev/mm (b) 4468 rad/min — 103 *696 1. sinO =3,coso 3. sinO = 3,tanO ‘3,cot9 = 3,seco =4,csco = , tanO = 3, cotO cosO secO = — 33.3 37. —1 41. 5 45.1 49. \/ 53. =3 3. °, 3, Page 108 Exercises 2.3 V’, csc45° = 1, sec45° 7. 5.2 3.3 ii. 3(’v’ + 9. 18 17. 2\/ 15. 0 13. 9V’ 21. 2—V’ 19.3 23. sinl7° = 0.2924,cosl7° = 0.9563,tanl7° = 0.3057, 3.4203 cotl7° = 3.2709,secl7° = l.0457,cscl7° 25. sin 14.3° = 0.2470, cos 14.3° = 0.9690, tan 14.30 = 0.2549, 4.0486 3.9232. sec 14.3° = 1.0320, csc 14.3° cot 14.30 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 0.6229, cos0.6725 = 0.7823, 31. sinO,6725 1.2558, 0.7963.cot0.6725 tan0.6725 1.6053 secO.6725 = l.2783,csc0.6725 1. tan45° 1, cot45° = 11. tan6 = 4, cotO 4, secO = cot& = , 13. tanO 5 15. cosO 17. sinO 19. sinO 21. cos6 = = =3 ,secB =-,csc = cotO ,cscO = = 8, secO = , csc8 = Exercises 2.4 rPage 117 = , = tanO tanO 2V, cotO = , cosO = = —-f, cotG 9. sinG = —*,cosG = secG = _E6,cotO = cotO = = fs, -. cotO secO = = 4. secO —vT 21. sinG = —°,cosG cotO = 3 23. sinG = = cotG = 3 27. sinG = = 34 = cscb , cscO tanG = = \/,cscO secO = , = 19. cosG cotG cotG = —4 = tanG = csc6 = 13. II 17. II 25. cosO = = 3 3, 3, 4, 4,coto = 1. sinG = cosO = tanG = cscO = secO 7 3 7 5 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 cosO = —‘,tanG = —4, cscO = 7. sinG = V’ 3, cosO 3, tanO 3, cotO 3 = 23. sinO =4,coso =o2,tanO =,cotO 25. sinO oo, ,cscO secO=—- = 11. III 15. I ,tanO =,cotO _,cosQ = cscO 1 — 3 secO 5. sinO =4,coso =‘,tanO =cot0 secO = 7] cscO = 4 = , tanO = , cotO = 2V, 7. sinO = cosO 3 cscO secO = 9. sinO = coto — = V’Tö, cscO = 31. v’ 35.2 39. 0 43. 47. 3, Exercises 2.2 4, tano cos6 = —, = —,cscG —-f3,cscG - tanG 1 j 4 34 , cosG = = = 4, secO = ,secG —, secG = —‘s/iô = —3vTi tanG —-j, = cscO —5, secO cscO =,secG = , cos6 5 = = V,cotG = 4 29. ± 2 = 31. sinG 33. cosO = = ±, cosG —3, sinG ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS = ± ±6 ANS-7