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Summer 2012 L.E.A.D. Ambassador Team 3 (Advanced Algebra and Trigonometry Course Content) “Algebra and Trigonometry” (Third Edition) Beecher, Penna, Bittinger Addision Wesley (February 2, 2007) BBEPMC05_0321279115.QXP 432 Chapter 5 1/10/05 1:00 PM Page 432 • The Trigonometric Functions 2.1 5.1 Trigonometric Polynomial Functions Functionsand of Acute Modeling Angles Determine the six trigonometric ratios for a given acute angle of a right triangle. Determine the trigonometric function values of 30, 45, and 60. Using a calculator, find function values for any acute angle, and given a function value of an acute angle, find the angle. Given the function values of an acute angle, find the function values of its complement. The Trigonometric Ratios We begin our study of trigonometry by considering right triangles and acute angles measured in degrees. An acute angle is an angle with measure greater than 0 and less than 90. Greek letters such as (alpha), (beta), (gamma), (theta), and (phi) are often used to denote an angle. Consider a right triangle with one of its acute angles labeled . The side opposite the right angle is called the hypotenuse. The other sides of the triangle are referenced by their position relative to the acute angle . One side is opposite and one is adjacent to . Hypotenuse Side opposite u u Side adjacent to u The lengths of the sides of the triangle are used to define the six trigonometric ratios: sine (sin), cosine (cos), tangent (tan), Hypotenuse Opposite u u Adjacent to u Hypotenuse u Adjacent to u Figure 2 The sine of is the length of the side opposite divided by the length of the hypotenuse (see Fig. 1): sin Figure 1 Opposite u cosecant (csc), secant (sec), cotangent (cot). length of side opposite . length of hypotenuse The ratio depends on the measure of angle and thus is a function of . The notation sin actually means sin , where sin, or sine, is the name of the function. The cosine of is the length of the side adjacent to divided by the length of the hypotenuse (see Fig. 2): cos length of side adjacent to . length of hypotenuse The six trigonometric ratios, or trigonometric functions, are defined as follows. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 433 Section 5.1 • Trigonometric Functions of Acute Angles 433 Trigonometric Function Values of an Acute Angle Let be an acute angle of a right triangle. Then the six trigonometric functions of are as follows: side opposite , hypotenuse side adjacent to cos , hypotenuse side opposite tan , side adjacent to sin Hypotenuse Opposite u u Adjacent to u hypotenuse , side opposite hypotenuse sec , side adjacent to side adjacent to cot . side opposite csc EXAMPLE 1 In the right triangle shown at left, find the six trigonometric function values of (a) and (b) . a 13 12 u 5 Solution We use the definitions. opp 12 hyp 13 csc , , a) sin hyp 13 opp 12 adj hyp 13 5 sec cos , , hyp 13 adj 5 opp 12 adj 5 cot tan , adj 5 opp 12 opp 5 , hyp 13 adj 12 cos , hyp 13 opp 5 tan , adj 12 b) sin hyp 13 , opp 5 hyp 13 sec , adj 12 adj 12 cot opp 5 The references to opposite, adjacent, and hypotenuse are relative to . csc The references to opposite, adjacent, and hypotenuse are relative to . 12 In Example 1(a), we note that the value of sin , 13 , is the reciprocal of the value of csc . Likewise, we see the same reciprocal relationship between the values of cos and sec and between the values of tan and cot . For any angle, the cosecant, secant, and cotangent values are the reciprocals of the sine, cosine, and tangent function values, respectively. 13 12 , Reciprocal Functions 1 1 csc sec , , cos sin cot 1 tan Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 434 Chapter 5 1/10/05 1:00 PM Page 434 • The Trigonometric Functions If we know the values of the sine, cosine, and tangent functions of an angle, we can use these reciprocal relationships to find the values of the cosecant, secant, and cotangent functions of that angle. Study Tip Success on the next exam can be planned. Include study time (even if only 30 minutes a day) in your daily schedule and commit to making this time a priority. Choose a time when you are most alert and a setting in which you can concentrate. You will be surprised how much more you can learn and retain if study time is included each day rather than in one long session before the exam. 4 3 4 EXAMPLE 2 Given that sin 5 , cos 5 , and tan 3 , find csc , sec , and cot . Solution Using the reciprocal relationships, we have csc 1 1 5 , sin 4 4 5 cot 1 1 3 . tan 4 4 3 sec 1 1 5 , cos 3 3 5 and Triangles are said to be similar if their corresponding angles have the same measure. In similar triangles, the lengths of corresponding sides are in the same ratio. The right triangles shown below are similar. Note that the corresponding angles are equal and the length of each side of the second triangle is four times the length of the corresponding side of the first triangle. a 20 12 3 a 5 b 4 b 16 Let’s observe the sine, cosine, and tangent values of in each triangle. Can we expect corresponding function values to be the same? FIRST TRIANGLE SECOND TRIANGLE 3 5 4 cos 5 3 tan 4 sin sin 12 20 16 cos 20 12 tan 16 3 5 4 5 3 4 For the two triangles, the corresponding values of sin , cos , and tan are the same. The lengths of the sides are proportional — thus the Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 435 Section 5.1 • 435 Trigonometric Functions of Acute Angles ratios are the same. This must be the case because in order for the sine, cosine, and tangent to be functions, there must be only one output (the ratio) for each input (the angle ). The trigonometric function values of depend only on the measure of the angle, not on the size of the triangle. The Six Functions Related We can find the other five trigonometric function values of an acute angle when one of the function-value ratios is known. EXAMPLE 3 If sin 67 and is an acute angle, find the other five trigonometric function values of . Solution 6 7 pythagorean theorem review section 1.1. We know from the definition of the sine function that the ratio opp . hyp is Using this information, let’s consider a right triangle in which the hypotenuse has length 7 and the side opposite has length 6. To find the length of the side adjacent to , we recall the Pythagorean theorem: 7 6 b a2 b 2 c 2 a 2 62 72 a2 36 49 a2 49 36 13 a 13. a We now use the lengths of the three sides to find the other five ratios: 6 , 7 13 cos , 7 6 tan , 13 sin 7 b 13 6 csc sec or 613 , 13 7 , 6 7 , 13 13 cot . 6 or 713 , 13 Function Values of 30, 45, and 60 In Examples 1 and 3, we found the trigonometric function values of an acute angle of a right triangle when the lengths of the three sides were known. In most situations, we are asked to find the function values when the measure of the acute angle is given. For certain special angles such as Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 436 Chapter 5 1/10/05 1:00 PM Page 436 • The Trigonometric Functions 30, 45, and 60, which are frequently seen in applications, we can use geometry to determine the function values. A right triangle with a 45 angle actually has two 45 angles. Thus the triangle is isosceles, and the legs are the same length. Let’s consider such a triangle whose legs have length 1. Then we can find the length of its hypotenuse, c, using the Pythagorean theorem as follows: 12 12 c 2, or c 2 2, or c 2. Such a triangle is shown below. From this diagram, we can easily determine the trigonometric function values of 45. 1 opp 2 0.7071, hyp 2 2 adj 1 2 cos 45 0.7071, hyp 2 2 opp 1 1 tan 45 adj 1 sin 45 45 2 1 45 1 It is sufficient to find only the function values of the sine, cosine, and tangent, since the others are their reciprocals. It is also possible to determine the function values of 30 and 60. A right triangle with 30 and 60 acute angles is half of an equilateral triangle, as shown in the following figure. Thus if we choose an equilateral triangle whose sides have length 2 and take half of it, we obtain a right triangle that has a hypotenuse of length 2 and a leg of length 1. The other leg has length a, which can be found as follows: 2 a 1 30 2 60 1 a2 12 22 a2 1 4 a2 3 a 3. We can now determine the function values of 30 and 60: 1 0.5, 2 3 cos 30 0.8660, 2 1 3 tan 30 0.5774, 3 3 sin 30 3 30 2 60 1 3 0.8660, 2 1 cos 60 0.5, 2 3 3 1.7321. tan 60 1 sin 60 Since we will often use the function values of 30, 45, and 60, either the triangles that yield them or the values themselves should be memorized. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 437 Section 5.1 • Trigonometric Functions of Acute Angles 30 45 60 sin 12 22 32 cos 32 22 12 tan 33 1 3 437 Let’s now use what we have learned about trigonometric functions of special angles to solve problems. We will consider such applications in greater detail in Section 5.2. EXAMPLE 4 Height of a Hot-air Balloon. As a hot-air balloon began to rise, the ground crew drove 1.2 mi to an observation station. The initial observation from the station estimated the angle between the ground and the line of sight to the balloon to be 30. Approximately how high was the balloon at that point? (We are assuming that the wind velocity was low and that the balloon rose vertically for the first few minutes.) Solution We begin with a drawing of the situation. We know the measure of an acute angle and the length of its adjacent side. h 30° 1.2 mi Since we want to determine the length of the opposite side, we can use the tangent ratio, or the cotangent ratio. Here we use the tangent ratio: opp h adj 1.2 1.2 tan 30 h 3 3 1.2 h Substituting; tan 30 3 3 0.7 h. tan 30 The balloon is approximately 0.7 mi, or 3696 ft, high. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 438 Chapter 5 1/10/05 1:00 PM Page 438 • The Trigonometric Functions Function Values of Any Acute Angle Historically, the measure of an angle has been expressed in degrees, minutes, and seconds. One minute, denoted 1, is such that 60 1, 1 or 1 60 1. One second, denoted 1 , is such that 60 1, or 1 1 60 1. Then 61 degrees, 27 minutes, 4 seconds could be written as 61274 . This DMS form was common before the widespread use of scientific calculators. Now the preferred notation is to express fractional parts of degrees in decimal degree form. Although the DMS notation is still widely used in navigation, we will most often use the decimal form in this text. Most calculators can convert DMS notation to decimal degree notation and vice versa. Procedures among calculators vary. Normal Sci Eng Float 0123456789 Radian Degree Func Par Pol Seq Connected Dot Sequential Simul Real abi reˆθ i Full Horiz G –T GCM EXAMPLE 5 Convert 54230 to decimal degree notation. Solution We can use a graphing calculator set in DEGREE mode to convert between DMS form and decimal degree form. (See window at left.) To convert DMS form to decimal degree form, we enter 54230 using the ANGLE menu for the degree and minute symbols and ALPHA for the symbol representing seconds. Pressing ENTER gives us 54230 5.71, rounded to the nearest hundredth of a degree. 5°42'30" 5.708333333 Without a calculator, we can convert as follows: 54230 5 42 30 30 5 42 60 42.5 60 1 EXAMPLE 6 1 42.5 ; 42.5 60 60 42.5 0.71 60 5.71. GCM 1 30 ; 30 60 60 30 0.5 60 5 42.5 5 1 Convert 72.18 to DMS notation. Solution To convert decimal degree form to DMS form, we enter 72.18 and access the DMS feature in the ANGLE menu. The result is 72.18 721048 . 72.18 DMS 72°10'48" Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 439 Section 5.1 • Trigonometric Functions of Acute Angles 439 Without a calculator, we can convert as follows: 72.18 72 0.18 1 72 0.18 60 72 10.8 72 10 0.8 1 72 10 0.8 60 72 10 48 721048 . 1 60 1 60 So far we have measured angles using degrees. Another useful unit for angle measure is the radian, which we will study in Section 5.4. Calculators work with either degrees or radians. Be sure to use whichever mode is appropriate. In this section, we use the degree mode. Keep in mind the difference between an exact answer and an approximation. For example, sin 60 3 . 2 This is exact! But using a calculator, you get an answer like sin 60 0.8660254038. This is an approximation! Calculators generally provide values only of the sine, cosine, and tangent functions. You can find values of the cosecant, secant, and cotangent by taking reciprocals of the sine, cosine, and tangent functions, respectively. GCM EXAMPLE 7 Find the trigonometric function value, rounded to four decimal places, of each of the following. a) tan 29.7 b) sec 48 c) sin 841039 Solution a) We check to be sure that the calculator is in DEGREE mode. The function value is tan 29.7 0.5703899297 Rounded to four decimal places 0.5704. b) The secant function value can be found by taking the reciprocal of the cosine function value: sec 48 1 1.49447655 1.4945. cos 48 c) We enter sin 841039 . The result is sin 841039 0.9948409474 0.9948. We can use the TABLE feature on a graphing calculator to find an angle for which we know a trigonometric function value. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 440 Chapter 5 1/10/05 1:00 PM Page 440 • The Trigonometric Functions GCM EXAMPLE 8 Find the acute angle, to the nearest tenth of a degree, whose sine value is approximately 0.20113. Solution With a graphing calculator set in DEGREE mode, we first enter the equation y sin x . With a minimum value of 0 and a step-value of 0.1, we scroll through the table of values looking for the y-value closest to 0.20113. X 11.1 11.2 11.3 11.4 11.5 11.6 11.7 X 11.6 Y1 .19252 .19423 .19595 .19766 .19937 .20108 .20279 sin 11.6 0.20108 We find that 11.6 is the angle whose sine value is about 0.20113. The quickest way to find the angle with a calculator is to use an inverse function key. (We first studied inverse functions in Section 4.1 and will consider inverse trigonometric functions in Section 6.4.) First check to be sure that your calculator is in DEGREE mode. Usually two keys must be pressed in sequence. For this example, if we press 2ND SIN .20113 ENTER , we find that the acute angle whose sine is 0.20113 is approximately 11.60304613, or 11.6. EXAMPLE 9 Ladder Safety. A paint crew has purchased new 30-ft extension ladders. The manufacturer states that the safest placement on a wall is to extend the ladder to 25 ft and to position the base 6.5 ft from the wall. (Source : R. D. Werner Co., Inc.) What angle does the ladder make with the ground in this position? 25 ft 6.5 ft Solution We make a drawing and then use the most convenient trigonometric function. Because we know the length of the side adjacent to and the length of the hypotenuse, we choose the cosine function. From the definition of the cosine function, we have cos adj 6.5 ft 0.26. hyp 25 ft Using a calculator, we find the acute angle whose cosine is 0.26: 74.92993786. Pressing 2ND COS 0.26 ENTER Thus when the ladder is in its safest position, it makes an angle of about 75 with the ground. Cofunctions and Complements We recall that two angles are complementary whenever the sum of their measures is 90. Each is the complement of the other. In a right triangle, Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 441 Section 5.1 • Trigonometric Functions of Acute Angles 441 the acute angles are complementary, since the sum of all three angle measures is 180 and the right angle accounts for 90 of this total. Thus if one acute angle of a right triangle is , the other is 90 . The six trigonometric function values of each of the acute angles in the triangle below are listed at the right. Note that 53 and 37 are complementary angles since 53 37 90. 53 37 sin 37 0.6018 cos 37 0.7986 tan 37 0.7536 csc 37 1.6616 sec 37 1.2521 cot 37 1.3270 sin 53 0.7986 cos 53 0.6018 tan 53 1.3270 csc 53 1.2521 sec 53 1.6616 cot 53 0.7536 Try this with the acute, complementary angles 20.3 and 69.7 as well. What pattern do you observe? Look for this same pattern in Example 1 earlier in this section. Note that the sine of an angle is also the cosine of the angle’s complement. Similarly, the tangent of an angle is the cotangent of the angle’s complement, and the secant of an angle is the cosecant of the angle’s complement. These pairs of functions are called cofunctions. A list of cofunction identities follows. Cofunction Identities 90 u sin cos 90 , tan cot 90 , sec csc 90 , u cos sin 90 , cot tan 90 , csc sec 90 EXAMPLE 10 Given that sin 18 0.3090 , cos 18 0.9511 , and tan 18 0.3249, find the six trigonometric function values of 72. Solution Using reciprocal relationships, we know that 1 3.2361, sin 18 1 sec 18 1.0515, cos 18 1 cot 18 3.0777. tan 18 csc 18 and Since 72 and 18 are complementary, we have sin 72 cos 18 0.9511, tan 72 cot 18 3.0777, sec 72 csc 18 3.2361, cos 72 sin 18 0.3090, cot 72 tan 18 0.3249, csc 72 sec 18 1.0515. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 442 Chapter 5 5.1 1/10/05 1:00 PM Page 442 • The Trigonometric Functions Exercise Set In Exercises 1–6, find the six trigonometric function values of the specified angle. 1. 22 1 , cos , and 3 3 tan 22, find csc , sec , and cot . 8. Given that sin 2. 0.4 0.3 17 8 3. Given a function value of an acute angle, find the other five trigonometric function values. 24 9. sin 25 10. cos 0.7 b 15 0.5 11. tan 2 f 4. e 6 6 3 u u 33 5. 7 φ 4 15. cos 12. cot 13 13. csc 1.5 a 14. sec 17 5 5 16. sin 10 11 Find the exact function value. 3 17. cos 45 2 18. tan 30 3 2 2 19. sec 60 2 20. sin 45 2 3 21. cot 60 22. csc 45 2 3 23. sin 30 12 24. cos 60 12 23 25. tan 45 1 26. sec 30 3 3 27. csc 30 2 28. cot 60 3 29. Distance Across a River. Find the distance a across the river. 62.4 m B 6. Grill c 30° A a 36 m 9 C 8.2 7. Given that sin tan 2 5 , cos , and 3 3 5 , find csc , sec , and cot . 2 30. Distance Between Bases. A baseball diamond is actually a square 90 ft on a side. If a line is drawn from third base to first base, then a right triangle Answers to Exercises 1–16 can be found on p. IA-31. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 443 Section 5.1 QPH is formed, where QPH is 45. Using a trigonometric function, find the distance from third base to first base. 127.3 ft • 443 Trigonometric Functions of Acute Angles 67. csc 89.5 1.0000 68. sec 35.28 1.2250 69. cot 30256 1.7032 70. sin 59.2 0.8590 Find the acute angle , to the nearest tenth of a degree, for the given function value. 71. sin 0.5125 30.8 72. tan 2.032 63.8 P Q (third) 45 (first) h 90 ft 90 ft GCM H Convert to decimal degree notation. Round to two decimal places. 31. 943 9.72 32. 5215 52.25 73. tan 0.2226 12.5 74. cos 0.3842 67.4 75. sin 0.9022 64.4 76. tan 3.056 71.9 77. cos 0.6879 46.5 78. sin 0.4005 23.6 79. cot 2.127 25.2 1 Hint: tan . cot 80. csc 1.147 60.7 81. sec 1.279 38.6 82. cot 1.351 36.5 33. 3550 35.01 34. 6453 64.88 35. 32 3.03 36. 194723 19.79 37. 493846 49.65 38. 761134 76.19 GCM 39. 155 0.25 40. 682 68.00 GCM 41. 553 5.01 42. 4410 0.74 Find the exact acute angle for the given function value. 2 3 83. sin 45 84. cot 60 2 3 85. cos Convert to degrees, minutes, and seconds. Round to the nearest second. 43. 17.6 1736 44. 20.14 20824 45. 83.025 83130 46. 67.84 675024 47. 11.75 1145 48. 29.8 2948 49. 47.8268 474936 50. 0.253 01511 51. 0.9 54 52. 30.2505 30152 53. 39.45 3927 54. 2.4 224 Find the function value. Round to four decimal places. 55. cos 51 0.6293 56. cot 17 3.2709 57. tan 413 0.0737 58. sin 26.1 0.4399 59. sec 38.43 1.2765 60. cos 741040 61. cos 40.35 0.7621 62. csc 45.2 1.4093 63. sin 69 0.9336 64. tan 6348 2.0323 65. tan 85.4 12.4288 66. cos 4 0.9976 1 2 87. tan 1 89. csc 60 45 23 3 86. sin 1 2 88. cos 3 30 2 30 90. tan 3 60 91. cot 3 30 60 92. sec 2 45 Use the cofunction and reciprocal identities to complete each of the following. 1 93. cos 20 sin; sec 70 _____ 20 26 1 cos; csc _____ 64 95. tan 52 cot 1 38; cot _____ 52 96. sec 13 csc 1 77; cos _____ 13 94. sin 64 0.2727 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 444 Chapter 5 1/10/05 1:00 PM Page 444 • The Trigonometric Functions 97. Given that sin 65 0.9063, tan 65 2.1445, sec 65 2.3662, cos 65 0.4226, cot 65 0.4663, csc 65 1.1034, find the six function values of 25. 109. 5x 625 [4.5] 4 110. log 3x 1 log x 1 2 [4.5] 101 97 111. log7 x 3 [4.5] 343 98. Given that sin 8 0.1392, tan 8 0.1405, sec 8 1.0098, Solve. 108. e t 10,000 [4.5] 9.21 cos 8 0.9903, cot 8 7.1154, csc 8 7.1853, find the six function values of 82. Synthesis 112. Given that cos 0.9651, find csc 90° . 1.0362 113. Given that sec 1.5304, find sin 90 . 0.6534 99. Given that sin 71105 0.9465, cos 71105 0.3228, and tan 71105 2.9321, find the six function values of 184955 . 100. Given that sin 38.7 0.6252, cos 38.7 0.7804, and tan 38.7 0.8012, find the six function values of 51.3. 114. Find the six trigonometric function values of . q 1 q a 115. Show that the area of this right triangle is 101. Given that sin 82 p, cos 82 q, and tan 82 r , find the six function values of 8 in terms of p, q, and r. 1 2 bc sin A. A c Collaborative Discussion and Writing 102. Explain the difference between reciprocal functions and cofunctions. 103. Explain why it is not necessary to memorize the function values for both 30 and 60. B b a C 116. Show that the area of this triangle is 1 2 ab sin . Skill Maintenance a Make a hand-drawn graph of the function. Then check your work using a graphing calculator. 104. fx 2x 105. fx e x/2 106. gx log2 x 107. hx ln x u b Let h the height of the triangle. Then Area where sin 1 bh, 2 1 h , or h a sin , so Area ab sin . a 2 Answers to Exercises 97–101, 104–107, 114, and 115 can be found on p. IA-31. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 445 Section 5.2 5.2 Applications of Right Triangles • Applications of Right Triangles 445 Solve right triangles. Solve applied problems involving right triangles and trigonometric functions. Solving Right Triangles Now that we can find function values for any acute angle, it is possible to solve right triangles. To solve a triangle means to find the lengths of all sides and the measures of all angles. EXAMPLE 1 In ABC (shown at left), find a, b, and B, where a and b represent lengths of sides and B represents the measure of B. Here we use standard lettering for naming the sides and angles of a right triangle: Side a is opposite angle A, side b is opposite angle B, where a and b are the legs, and side c, the hypotenuse, is opposite angle C, the right angle. B 106.2 A Solution In ABC , we know three of the measures: A 61.7, B ?, C 90, a a ?, b ?, c 106.2. Since the sum of the angle measures of any triangle is 180 and C 90, the sum of A and B is 90. Thus, 61.7 b C B 90 A 90 61.7 28.3. We are given an acute angle and the hypotenuse. This suggests that we can use the sine and cosine ratios to find a and b, respectively: sin 61.7 opp a hyp 106.2 and cos 61.7 adj b . hyp 106.2 Solving for a and b, we get a 106.2 sin 61.7 and b 106.2 cos 61.7 a 93.5 b 50.3. Thus, A 61.7, B 28.3, C 90, a 93.5, b 50.3, c 106.2. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 446 Chapter 5 1/10/05 Page 446 • The Trigonometric Functions d E 1:00 PM F EXAMPLE 2 Solution 13 In DEF (shown at left), find D and F. Then find d. In DEF , we know three of the measures: D ?, E 90, F ?, 23 D d ?, e 23, f 13. We know the side adjacent to D and the hypotenuse. This suggests the use of the cosine ratio: cos D 13 adj . hyp 23 We now find the angle whose cosine is of a degree, D 55.58. Pressing 2ND 13 23 . To the nearest hundredth COS 1323 ENTER Since the sum of D and F is 90, we can find F by subtracting: F 90 D 90 55.58 34.42. We could use the Pythagorean theorem to find d, but we will use a trigonometric function here. We could use cos F, sin D, or the tangent or cotangent ratios for either D or F. Let’s use tan D: tan D opp d , adj 13 or tan 55.58 d . 13 Then d 13 tan 55.58 19. The six measures are D 55.58, E 90, F 34.42, d 19, e 23, f 13. Applications Right triangles can be used to model and solve many applied problems in the real world. North Rim 6.2 mi b South Rim c 50° EXAMPLE 3 Hiking at the Grand Canyon. A backpacker hiking east along the North Rim of the Grand Canyon notices an unusual rock formation directly across the canyon. She decides to continue watching the landmark while hiking along the rim. In 2 hr, she has gone 6.2 mi due east and the landmark is still visible but at approximately a 50 angle to the North Rim. (See the figure at left.) a) How many miles is she from the rock formation? b) How far is it across the canyon from her starting point? Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 447 Section 5.2 • Applications of Right Triangles 447 Solution a) We know the side adjacent to the 50 angle and want to find the hypotenuse. We can use the cosine function: 6.2 mi c 6.2 mi 9.6 mi. c cos 50 cos 50 After hiking 6.2 mi, she is approximately 9.6 mi from the rock formation. b) We know the side adjacent to the 50 angle and want to find the opposite side. We can use the tangent function: b 6.2 mi b 6.2 mi tan 50 7.4 mi. tan 50 Thus it is approximately 7.4 mi across the canyon from her starting point. Many applications with right triangles involve an angle of elevation or an angle of depression. The angle between the horizontal and a line of sight above the horizontal is called an angle of elevation. The angle between the horizontal and a line of sight below the horizontal is called an angle of depression. For example, suppose that you are looking straight ahead and then you move your eyes up to look at an approaching airplane. The angle that your eyes pass through is an angle of elevation. If the pilot of the plane is looking forward and then looks down, the pilot’s eyes pass through an angle of depression. Horizontal Angle of depression Angle of elevation Horizontal A Angle of depression 475 ft 850 ft B Angle of elevation C EXAMPLE 4 Aerial Photography. An aerial photographer who photographs farm properties for a real estate company has determined from experience that the best photo is taken at a height of approximately 475 ft and a distance of 850 ft from the farmhouse. What is the angle of depression from the plane to the house? Solution When parallel lines are cut by a transversal, alternate interior angles are equal. Thus the angle of depression from the plane to the house, , is equal to the angle of elevation from the house to the plane, Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 448 Chapter 5 1/10/05 1:00 PM Page 448 • The Trigonometric Functions so we can use the right triangle shown in the figure. Since we know the side opposite B and the hypotenuse, we can find by using the sine function. We first find sin : sin sin B 475 ft 0.5588. 850 ft Using a calculator in DEGREE mode, we find the acute angle whose sine is approximately 0.5588: 34. Pressing 2nd SIN 0.5588 ENTER Thus the angle of depression is approximately 34. EXAMPLE 5 Cloud Height. To measure cloud height at night, a vertical beam of light is directed on a spot on the cloud. From a point 135 ft away from the light source, the angle of elevation to the spot is found to be 67.35. Find the height of the cloud. Solution From the figure, we have h h 135 ft h 135 ft tan 67.35 324 ft. tan 67.35 67.35° 135 ft The height of the cloud is about 324 ft. Some applications of trigonometry involve the concept of direction, or bearing. In this text we present two ways of giving direction, the first below and the second in Exercise Set 5.3. Bearing: First-Type One method of giving direction, or bearing, involves reference to a north – south line using an acute angle. For example, N55W means 55 west of north and S67E means 67 east of south. N N N N55W N N60E W E W E W E W S67E S S S S35W S Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley E BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 449 Section 5.2 • Applications of Right Triangles 449 EXAMPLE 6 Distance to a Forest Fire. A forest ranger at point A sights a fire directly south. A second ranger at point B, 7.5 mi east, sights the same fire at a bearing of S2723W. How far from A is the fire? N A W 7.5 mi B E 6237 d 2723 Fire S F S2723W Solution We first find the complement of 2723: B 90 2723 6237 62.62. Angle B is opposite side d in the right triangle. From the figure shown above, we see that the desired distance d is part of a right triangle. We have d tan 62.62 7.5 mi d 7.5 mi tan 62.62 14.5 mi. The forest ranger at point A is about 14.5 mi from the fire. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 450 Chapter 5 1/10/05 1:00 PM Page 450 • The Trigonometric Functions EXAMPLE 7 Comiskey Park. In the new Comiskey Park, the home of the Chicago White Sox baseball team, the first row of seats in the upper deck is farther away from home plate than the last row of seats in the old Comiskey Park. Although there is no obstructed view in the new park, some of the fans still complain about the present distance from home plate to the upper deck of seats. (Source : Chicago Tribune, September 19, 1993) From a seat in the last row of the upper deck directly behind the batter, the angle of depression to home plate is 29.9, and the angle of depression to the pitcher’s mound is 24.2. Find (a) the viewing distance to home plate and (b) the viewing distance to the pitcher’s mound. 24.2 29.9 d2 Pitcher Batter u2 h d1 u1 x 60.5 ft Study Tip Tutoring is available to students using this text. The AddisonWesley Math Tutor Center, staffed by mathematics instructors, can be reached by telephone, fax, or e-mail. When you are having difficulty with an exercise, this live tutoring can be a valuable resource. These instructors have a copy of your text and are familiar with the content objectives in this course. Solution From geometry we know that 1 29.9 and 2 24.2. The standard distance from home plate to the pitcher’s mound is 60.5 ft. In the drawing, we let d1 be the viewing distance to home plate, d2 the viewing distance to the pitcher’s mound, h the elevation of the last row, and x the horizontal distance from the batter to a point directly below the seat in the last row of the upper deck. We begin by determining the distance x. We use the tangent function with 1 29.9 and 2 24.2: tan 29.9 h x and tan 24.2 h x 60.5 or h x tan 29.9 and h x 60.5 tan 24.2. Then substituting x tan 29.9 for h in the second equation, we obtain x tan 29.9 x 60.5 tan 24.2. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 451 Section 5.2 • Applications of Right Triangles 451 Solving for x, we get x tan 29.9 x tan 24.2 60.5 tan 24.2 x tan 29.9 x tan 24.2 x tan 24.2 60.5 tan 24.2 x tan 24.2 xtan 29.9 tan 24.2 60.5 tan 24.2 60.5 tan 24.2 x tan 29.9 tan 24.2 x 216.5. We can then find d1 and d2 using the cosine function: cos 29.9 216.5 d1 and cos 24.2 216.5 60.5 d2 or 216.5 and cos 29.9 d1 249.7 d1 277 cos 24.2 d2 303.7. d2 The distance to home plate is about 250 ft,* and the distance to the pitcher’s mound is about 304 ft. *In the old Comiskey Park, the distance to home plate was only 150 ft. , , T 63.3, s 0.38, t 0.34 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:00 PM Page 451 Section 5.2 5.2 • Applications of Right Triangles Exercise Set In Exercises 1–6, solve the right triangle. 2. 1. F 5. 6. P P 4738, n 34.4, p 25.4 A 6 d 30 E f 10 b F 60, d 3, f 5.2 M 45 C a 126 c B s 67.3 B A 22.7, a 52.7, c 136.6 f N p G H 6126, f 36.2, h 31.8 H 4222 R a A 17.3 23.2 4. C 2834 In Exercises 7–16, solve the right triangle. (Standard lettering has been used.) B A 45, a 7.1, b 7.1 3. h F n D 451 c 26.7 a T t 0.17 S T 63.3, s 0.38, t 0.34 A b C 7. A 8743, a 9.73 B 217, b 0.39, c 9.74 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 452 Chapter 5 1/10/05 1:00 PM Page 452 • The Trigonometric Functions 8. a 12.5, b 18.3 A 34.3, B 55.7, c 22.2 9. b 100, c 450 A 77.2, B 12.8, a 439 10. B 56.5, c 0.0447 A 33.5, a 0.0247, b 0.0373 11. A 47.58, c 48.3 B 42.42, a 35.7, b 32.6 20. Height of a Tree. A supervisor must train a new team of loggers to estimate the heights of trees. As an example, she walks off 40 ft from the base of a tree and estimates the angle of elevation to the tree’s peak to be 70. Approximately how tall is the tree? About 110 ft 12. B 20.6, a 7.5 A 69.4, b 2.8, c 8.0 13. A 35, b 40 B 55, a 28.0, c 48.8 14. B 69.3, b 93.4 A 20.7, a 35.3, c 99.8 15. b 1.86, c 4.02 A 62.4, B 27.6, a 3.56 16. a 10.2, c 20.4 A 30, B 60, b 17.7 17. Safety Line to Raft. Each spring Bryan uses his vacation time to ready his lake property for the summer. He wants to run a new safety line from point B on the shore to the corner of the anchored diving raft. The current safety line, which runs perpendicular to the shore line to point A, is 40 ft long. He estimates the angle from B to the corner of the raft to be 50. Approximately how much rope does he need for the new safety line if he allows 5 ft of rope at each end to fasten the rope? About 62.2 ft 70° 40 ft 21. Sand Dunes National Park. While visiting the Sand Dunes National Park in Colorado, Cole approximated the angle of elevation to the top of a sand dune to be 20. After walking 800 ft closer, he guessed that the angle of elevation had increased by 15. Approximately how tall is the dune he was observing? About 606 ft Diving raft 40 ft B 50° Shoreline A 18. Enclosing an Area. Alicia is enclosing a triangular area in a corner of her fenced rectangular backyard for her Labrador retriever. In order for a certain tree to be included in this pen, one side needs to be 14.5 ft and make a 53 angle with the new side. How long is the new side? About 24.1 ft 20° 800 ft 22. Tee Shirt Design. A new tee shirt design is to have a regular octagon inscribed in a circle, as shown in the figure. Each side of the octagon is to be 3.5 in. long. Find the radius of the circumscribed circle. About 4.6 in. 3.5 in. 19. Easel Display. A marketing group is designing an easel to display posters advertising their newest products. They want the easel to be 6 ft tall and the back of it to fit flush against a wall. For optimal eye contact, the best angle between the front and back legs of the easel is 23. How far from the wall should the front legs be placed in order to obtain this angle? About 2.5 ft Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley r BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 453 Section 5.2 23. Inscribed Pentagon. A regular pentagon is inscribed in a circle of radius 15.8 cm. Find the perimeter of the pentagon. About 92.9 cm 24. Height of a Weather Balloon. A weather balloon is directly west of two observing stations that are 10 mi apart. The angles of elevation of the balloon from the two stations are 17.6 and 78.2. How high is the balloon? About 3.4 mi • Applications of Right Triangles 453 29. Distance from a Lighthouse. From the top of a lighthouse 55 ft above sea level, the angle of depression to a small boat is 11.3. How far from the foot of the lighthouse is the boat? About 275 ft 25. Height of a Kite. For a science fair project, a group of students tested different materials used to construct kites. Their instructor provided an instrument that accurately measures the angle of elevation. In one of the tests, the angle of elevation was 63.4 with 670 ft of string out. Assuming the string was taut, how high was the kite? About 599 ft 30. Lightning Detection. In extremely large forests, it is not cost-effective to position forest rangers in towers or to use small aircraft to continually watch for fires. Since lightning is a frequent cause of fire, lightning detectors are now commonly used instead. These devices not only give a bearing on the location but also measure the intensity of the lightning. A detector at point Q is situated 15 mi west of a central fire station at point R. The bearing from Q to where lightning hits due south of R is S37.6E. How far is the hit from point R? 26. Height of a Building. A window washer on a ladder looks at a nearby building 100 ft away, noting that the angle of elevation of the top of the building is 18.7 and the angle of depression of the bottom of the building is 6.5. How tall is the nearby building? 31. Lobster Boat. A lobster boat is situated due west of a lighthouse. A barge is 12 km south of the lobster boat. From the barge, the bearing to the lighthouse is N6320E. How far is the lobster boat from the lighthouse? About 24 km About 19.5 mi About 45 ft 18.7° 100 ft 6.5° 12 km North 63° 20' 27. Distance Between Towns. From a hot-air balloon 2 km high, the angles of depression to two towns in line with the balloon are 81.2 and 13.5. How far apart are the towns? About 8 km 28. Angle of Elevation. What is the angle of elevation of the sun when a 35-ft mast casts a 20-ft shadow? About 60.3 32. Length of an Antenna. A vertical antenna is mounted atop a 50-ft pole. From a point on level ground 75 ft from the base of the pole, the antenna subtends an angle of 10.5. Find the length of the antenna. About 23 ft 35 ft 10.5° 50 ft 20 ft 75 ft Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 454 Chapter 5 1/10/05 1:01 PM Page 454 • The Trigonometric Functions Collaborative Discussion and Writing 33. In this section, the trigonometric functions have been defined as functions of acute angles. Thus the set of angles whose measures are greater than 0 and less than 90 is the domain for each function. What appear to be the ranges for the sine, the cosine, and the tangent functions given this domain? 41. Construction of Picnic Pavilions. A construction company is mass-producing picnic pavilions for national parks, as shown in the figure. The rafter ends are to be sawed in such a way that they will be vertical when in place. The front is 8 ft high, the back is 6 12 ft high, and the distance between the front and back is 8 ft. At what angle should the rafters be cut? Cut so that 79.38 34. Explain in your own words five ways in which length c can be determined in this triangle. Which way seems the most efficient? u u c 14 8 ft 6 6 q ft Skill Maintenance Find the distance between the points. 35. 8, 2 and 6, 4 [1.1] 102, or about 14.142 36. 9, 3 and 0, 0 [1.1] 310, or about 9.487 37. Convert to an exponential equation: log 0.001 3. [4.3] 10 3 0.001 38. Convert to a logarithmic equation: e 4 t . [4.3] ln t 4 Synthesis 39. Find h, to the nearest tenth. 3.3 8 ft 42. Diameter of a Pipe. A V-gauge is used to find the diameter of a pipe. The advantage of such a device is that it is rugged, it is accurate, and it has no moving parts to break down. In the figure, the measure of angle AVB is 54. A pipe is placed in the V-shaped slot and the distance VP is used to estimate the diameter. The line VP is calibrated by listing as its units the corresponding diameters. This, in effect, establishes a function between VP and d. A Q h d V 36 P B 7 40. Find a, to the nearest tenth. 5.9 5 72 a a) Suppose that the diameter of a pipe is 2 cm. What is the distance VP? About 1.96 cm b) Suppose that the distance VP is 3.93 cm. What is the diameter of the pipe? About 4.00 cm c) Find a formula for d in terms of VP. d 1.02 VP d) Find a formula for VP in terms of d. VP 0.98d 43. Sound of an Airplane. It is a common experience to hear the sound of a low-flying airplane and look at the wrong place in the sky to see the plane. Suppose that a plane is traveling directly at you at a speed of Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 455 Section 5.2 200 mph and an altitude of 3000 ft, and you hear the sound at what seems to be an angle of inclination of 20. At what angle should you actually look in order to see the plane? Consider the speed of sound to be 1100 ftsec. 27 Perceived location of plane P 3000 ft Actual location of plane when heard • Applications of Right Triangles mountain and a line drawn from the top of the mountain to the horizon, as shown in the figure. The height of Mt. Shasta in California is 14,162 ft. From the top of Mt. Shasta, one can see the horizon on the Pacific Ocean. The angle formed between a line to the horizon and the vertical is found to be 8753. Use this information to estimate the radius of the earth, in miles. R 3928 mi A 455 R 14,162 u 8753 20° V R 44. Measuring the Radius of the Earth. One way to measure the radius of the earth is to climb to the top of a mountain whose height above sea level is known and measure the angle between a vertical line to the center of the earth from the top of the Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 455 Section 5.3 5.3 Trigonometric Functions of Any Angle • Trigonometric Functions of Any Angle Find angles that are coterminal with a given angle and find the complement and the supplement of a given angle. Determine the six trigonometric function values for any angle in standard position when the coordinates of a point on the terminal side are given. Find the function values for any angle whose terminal side lies on an axis. Find the function values for an angle whose terminal side makes an angle of 30, 45, or 60 with the x-axis. Use a calculator to find function values and angles. Angles, Rotations, and Degree Measure An angle is a familiar figure in the world around us. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 455 BBEPMC05_0321279115.QXP 456 Chapter 5 1/10/05 1:01 PM Page 456 • The Trigonometric Functions An angle is the union of two rays with a common endpoint called the vertex. In trigonometry, we often think of an angle as a rotation. To do so, think of locating a ray along the positive x-axis with its endpoint at the origin. This ray is called the initial side of the angle. Though we leave that ray fixed, think of making a copy of it and rotating it. A rotation counterclockwise is a positive rotation, and a rotation clockwise is a negative rotation. The ray at the end of the rotation is called the terminal side of the angle. The angle formed is said to be in standard position. y Terminal side y Vertex y A positive rotation (or angle) x x x Initial side A negative rotation (or angle) The measure of an angle or rotation may be given in degrees. The Babylonians developed the idea of dividing the circumference of a circle into 360 equal parts, or degrees. If we let the measure of one of these parts be 1, then one complete positive revolution or rotation has a measure of 360. One half of a revolution has a measure of 180, one fourth of a revolution has a measure of 90, and so on. We can also speak of an angle of measure 60, 135, 330, or 420. The terminal sides of these angles lie in quadrants I, II, IV, and I, respectively. The negative rotations 30, 110, and 225 represent angles with terminal sides in quadrants IV, III, and II, respectively. y y y 90 180 x x 360 y y 270 x 270 330 30 x x y 60 x 90 y 135 225 y 420 610 x Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley x 110 BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 457 Section 5.3 • Trigonometric Functions of Any Angle 457 If two or more angles have the same terminal side, the angles are said to be coterminal. To find angles coterminal with a given angle, we add or subtract multiples of 360. For example, 420, shown above, has the same terminal side as 60, since 420 360 60. Thus we say that angles of measure 60 and 420 are coterminal. The negative rotation that measures 300 is also coterminal with 60 because 60 360 300. The set of all angles coterminal with 60 can be expressed as 60 n 360, where n is an integer. Other examples of coterminal angles shown above are 90 and 270, 90 and 270, 135 and 225, 30 and 330, and 110 and 610. EXAMPLE 1 Find two positive and two negative angles that are coterminal with (a) 51 and (b) 7. Solution a) We add and subtract multiples of 360. Many answers are possible. y y 51 51 x x 411 1131 51 360 411 51 3(360) 1131 y 309 y 51 669 51 x 51 360 309 x 51 2(360) 669 Thus angles of measure 411, 1131, 309, and 669 are coterminal with 51. b) We have the following: 7 360 353, 7 360 367, 7 2360 713, 7 10360 3607. Thus angles of measure 353, 713, 367, and 3607 are coterminal with 7. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 458 Chapter 5 1/10/05 1:01 PM Page 458 • The Trigonometric Functions Angles can be classified by their measures, as seen in the following figure. 161 180 Obtuse: 90 u 180 Straight : u 180 90 24 Right: u 90 Acute: 0 u 90 Recall that two acute angles are complementary if their sum is 90. For example, angles that measure 10 and 80 are complementary because 10 80 90. Two positive angles are supplementary if their sum is 180. For example, angles that measure 45 and 135 are supplementary because 45 135 180. 10 80 Complementary angles EXAMPLE 2 Solution 45 135 Supplementary angles Find the complement and the supplement of 71.46. We have 90 71.46 18.54, 180 71.46 108.54. Thus the complement of 71.46 is 18.54 and the supplement is 108.54. y Trigonometric Functions of Angles or Rotations x 2 y 2 r2 r y u x P(x, y) x Many applied problems in trigonometry involve the use of angles that are not acute. Thus we need to extend the domains of the trigonometric functions defined in Section 5.1 to angles, or rotations, of any size. To do this, we first consider a right triangle with one vertex at the origin of a coordinate system and one vertex on the positive x-axis. (See the figure at left.) The other vertex is at P, a point on the circle whose center is at the origin and whose radius r is the length of the hypotenuse of the triangle. This triangle is a reference triangle for angle , which is in standard position. Note that y is the length of the side opposite and x is the length of the side adjacent to . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 459 Section 5.3 • Trigonometric Functions of Any Angle 459 Recalling the definitions in Section 5.1, we note that three of the trigonometric functions of angle are defined as follows: sin opp y , hyp r cos adj x , hyp r tan opp y . adj x Since x and y are the coordinates of the point P and the length of the radius is the length of the hypotenuse, we can also define these functions as follows: y-coordinate , radius x-coordinate cos , radius y-coordinate . tan x-coordinate sin We will use these definitions for functions of angles of any measure. The following figures show angles whose terminal sides lie in quadrants II, III, and IV. y P(x, y) y r x y u y u x x y P(x, y) x x r u r x y P(x, y) A reference triangle can be drawn for angles in any quadrant, as shown. Note that the angle is in standard position; that is, it is always measured from the positive half of the x-axis. The point Px, y is a point, other than the vertex, on the terminal side of the angle. Each of its two coordinates may be positive, negative, or zero, depending on the location of the terminal side. The length of the radius, which is also the length of the hypotenuse of the reference triangle, is always considered positive. Note that x 2 y 2 r 2, or r x 2 y 2. Regardless of the location of P, we have the following definitions. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 460 Chapter 5 1/10/05 x Page 460 • The Trigonometric Functions y P(x, y) r y 1:01 PM u x Trigonometric Functions of Any Angle Suppose that Px, y is any point other than the vertex on the terminal side of any angle in standard position, and r is the radius, or distance from the origin to Px, y. Then the trigonometric functions are defined as follows: y-coordinate radius x-coordinate cos radius y-coordinate tan x-coordinate y , r x , r y , x sin radius y-coordinate radius sec x-coordinate x-coordinate cot y-coordinate csc r , y r , x x . y Values of the trigonometric functions can be positive, negative, or zero, depending on where the terminal side of the angle lies. The length of the radius is always positive. Thus the signs of the function values depend only on the coordinates of the point P on the terminal side of the angle. In the first quadrant, all function values are positive because both coordinates are positive. In the second quadrant, first coordinates are negative and second coordinates are positive; thus only the sine and the cosecant values are positive. Similarly, we can determine the signs of the function values in the third and fourth quadrants. Because of the reciprocal relationships, we need to learn only the signs for the sine, cosine, and tangent functions. Positive: sin Negative: cos, tan y II (, ) (, ) III Positive: tan Negative: sin, cos Positive: All Negative: None I (, ) x (, ) IV Positive: cos Negative: sin, tan Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 461 Section 5.3 b) y Trigonometric Functions of Any Angle 461 Find the six trigonometric function values for each angle EXAMPLE 3 shown. a) • c) y y (1, 3) u 3 4 3 (4, 3) r u 1 x r u r 1 x 1 x (1, 1) Solution a) We first determine r, the distance from the origin 0, 0 to the point 4, 3. The distance between 0, 0 and any point x, y on the terminal side of the angle is r x 02 y 02 x 2 y 2. Substituting 4 for x and 3 for y, we find r 42 32 16 9 25 5. Using the definitions of the trigonometric functions, we can now find the function values of . We substitute 4 for x, 3 for y, and 5 for r : y 3 3 , r 5 5 x 4 4 cos , r 5 5 3 3 y tan , x 4 4 sin r 5 5 , y 3 3 r 5 5 sec , x 4 4 4 4 x . cot y 3 3 csc As expected, the tangent and the cotangent values are positive and the other four are negative. This is true for all angles in quadrant III. b) We first determine r, the distance from the origin to the point 1, 1: r 12 12 1 1 2. Substituting 1 for x, 1 for y, and 2 for r, we find y 1 2 , r 2 2 x 1 2 cos , r 2 2 y 1 tan 1, x 1 sin r 2 2, y 1 r 2 sec 2, x 1 x 1 1. cot y 1 csc Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 462 Chapter 5 1/10/05 1:01 PM Page 462 • The Trigonometric Functions c) We determine r, the distance from the origin to the point 1, 3 : r 12 3 2 1 3 4 2. Substituting 1 for x, 3 for y, and 2 for r, we find the trigonometric function values of are 3 , 2 1 1 cos , 2 2 3 tan 3, 1 sin y (8, 4) (4, 2) 108642 u 4 6 8 10 x csc 2 23 , 3 3 2 sec 2, 1 1 3 cot . 3 3 Any point other than the origin on the terminal side of an angle in standard position can be used to determine the trigonometric function values of that angle. The function values are the same regardless of which point is used. To illustrate this, let’s consider an angle in 1 standard position whose terminal side lies on the line y 2 x . We can determine two second-quadrant solutions of the equation, find the length r for each point, and then compare the sine, cosine, and tangent function values using each point. 1 If x 4, then y 2 4 2. y qx 1 If x 8, then y 2 8 4. For 4, 2, r 42 22 20 25. For 8, 4, r 82 42 80 45. Using 4, 2 and r 25, we find that 2 1 5 , 25 5 5 2 1 tan . 4 2 sin and cos 4 2 25 , 25 5 5 cos 8 2 25 , 45 5 5 Using 8, 4 and r 45, we find that 4 1 5 , 45 5 5 4 1 tan . 8 2 sin and We see that the function values are the same using either point. Any point other than the origin on the terminal side of an angle can be used to determine the trigonometric function values. The trigonometric function values of depend only on the angle, not on the choice of the point on the terminal side that is used to compute them. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 463 Section 5.3 • Trigonometric Functions of Any Angle 463 The Six Functions Related When we know one of the function values of an angle, we can find the other five if we know the quadrant in which the terminal side lies. The procedure is to sketch a reference triangle in the appropriate quadrant, use the Pythagorean theorem as needed to find the lengths of its sides, and then find the ratios of the sides. 2 EXAMPLE 4 Given that tan 3 and is in the second quadrant, find the other function values. Solution We first sketch a second-quadrant angle. Since tan Expressing 2 2 as since is in 3 3 quadrant II we make the legs lengths 2 and 3. The hypotenuse must then have length 22 32, or 13. Now we read off the appropriate ratios: y (3, 2) 2 y 2 2 , x 3 3 13 sin u 3 2 , or 13 3 cos , 13 2 tan , 3 x 13 , 2 13 sec , 3 213 , 13 313 or , 13 csc cot 3 . 2 Terminal Side on an Axis An angle whose terminal side falls on one of the axes is a quadrantal angle. One of the coordinates of any point on that side is 0. The definitions of the trigonometric functions still apply, but in some cases, function values will not be defined because a denominator will be 0. EXAMPLE 5 Find the sine, cosine, and tangent values for 90, 180, 270, and 360. Solution We first make a drawing of each angle in standard position and label a point on the terminal side. Since the function values are the same for all points on the terminal side, we choose 0, 1, 1, 0, 0, 1, and 1, 0 for convenience. Note that r 1 for each choice. y y y y (0, 1) 90 180 x (1, 0) (1, 0) 270 x x x 360 (0, 1) Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 464 Chapter 5 1/10/05 1:01 PM Page 464 • The Trigonometric Functions Then by the definitions we get 1 1, 1 0 cos 90 0, 1 1 tan 90 , Not defined 0 sin 90 0 0, 1 1 cos 180 1, 1 0 tan 180 0, 1 sin 180 1 1, 1 0 cos 270 0, 1 1 , Not defined tan 270 0 0 0, 1 1 cos 360 1, 1 0 tan 360 0. 1 sin 360 sin 270 In Example 5, all the values can be found using a calculator, but you will find that it is convenient to be able to compute them mentally. It is also helpful to note that coterminal angles have the same function values. For example, 0 and 360 are coterminal; thus, sin 0 0, cos 0 1, and tan 0 0. EXAMPLE 6 Find each of the following. a) sin 90 b) csc 540 Solution a) We note that 90 is coterminal with 270. Thus, sin 90 sin 270 1 1. 1 b) Since 540 180 360, 540 and 180 are coterminal. Thus, csc 540 csc 180 1 1 , sin 180 0 which is not defined. Trigonometric values can always be checked using a calculator. When the value is undefined, the calculator will display an ERROR message. ERR: DIVIDE BY 0 1: Quit 2: Goto Reference Angles: 30, 45, and 60 We can also mentally determine trigonometric function values whenever the terminal side makes a 30, 45, or 60 angle with the x-axis. Consider, for example, an angle of 150. The terminal side makes a 30 angle with the x-axis, since 180 150 30. y 150 P(3, 1) Reference 2 150 angle 1 30 180 O N 3 P(3, 1) 2 30 3 1 N x Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 465 Section 5.3 • Trigonometric Functions of Any Angle 465 As the figure shows, ONP is congruent to ONP; therefore, the ratios of the sides of the two triangles are the same. Thus the trigonometric function values are the same except perhaps for the sign. We could determine the function values directly from ONP , but this is not necessary. If we remember that in quadrant II, the sine is positive and the cosine and the tangent are negative, we can simply use the function values of 30 that we already know and prefix the appropriate sign. Thus, sin 150 sin 30 1 , 2 3 , 2 1 tan 150 tan 30 , 3 cos 150 cos 30 and or 3 . 3 Triangle ONP is the reference triangle and the acute angle NOP is called a reference angle. Reference Angle The reference angle for an angle is the acute angle formed by the terminal side of the angle and the x-axis. EXAMPLE 7 Find the sine, cosine, and tangent function values for each of the following. b) 780 a) 225 Solution a) We draw a figure showing the terminal side of a 225 angle. The reference angle is 225 180, or 45. y y 2 45 1 Reference angle 45 225 x 45 1 x 225 Recall from Section 5.1 that sin 45 22, cos 45 22, and tan 45 1. Also note that in the third quadrant, the sine and the cosine are negative and the tangent is positive. Thus we have sin 225 2 , 2 cos 225 2 , 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley and tan 225 1. BBEPMC05_0321279115.QXP 466 Chapter 5 1/10/05 1:01 PM Page 466 • The Trigonometric Functions b) We draw a figure showing the terminal side of a 780 angle. Since 780 2360 60, we know that 780 and 60 are coterminal. y y 2 60 60 3 1 x x 60 Reference angle 780 780 The reference angle for 60 is the acute angle formed by the terminal side of the angle and the x-axis. Thus the reference angle for 60 is 60. We know that since 780 is a fourth-quadrant angle, the cosine is positive and the sine and the tangent are negative. Recalling that sin 60 32, cos 60 12, and tan 60 3, we have 3 , 2 tan 780 3. sin 780 and cos 780 1 , 2 Function Values for Any Angle When the terminal side of an angle falls on one of the axes or makes a 30, 45, or 60 angle with the x-axis, we can find exact function values without the use of a calculator. But this group is only a small subset of all angles. Using a calculator, we can approximate the trigonometric function values of any angle. In fact, we can approximate or find exact function values of all angles without using a reference angle. EXAMPLE 8 Find each of the following function values using a calculator and round the answer to four decimal places, where appropriate. a) c) e) g) cos(112) 1/cos(500) tan(83.4) cos 112 tan 83.4 cos 2400 cot 135 Solution .3746065934 1.305407289 8.64274761 b) sec 500 d) csc 351.75 f ) sin 175409 Using a calculator set in DEGREE mode, we find the values. a) cos 112 0.3746 1 1.3054 b) sec 500 cos 500 c) tan 83.4 8.6427 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 467 Section 5.3 • Trigonometric Functions of Any Angle 467 1 6.9690 sin 351.75 e) cos 2400 0.5 f ) sin 175409 0.0755 1 1 g) cot 135 tan 135 d) csc 351.75 1/sin(351.75) 6.968999424 cos(2400) .5 sin(175409 ) .0755153443 In many applications, we have a trigonometric function value and want to find the measure of a corresponding angle. When only acute angles are considered, there is only one angle for each trigonometric function value. This is not the case when we extend the domain of the trigonometric functions to the set of all angles. For a given function value, there is an infinite number of angles that have that function value. There can be two such angles for each value in the range from 0 to 360. To determine a unique answer in the interval 0, 360, the quadrant in which the terminal side lies must be specified. The calculator gives the reference angle as an output for each function value that is entered as an input. Knowing the reference angle and the quadrant in which the terminal side lies, we can find the specified angle. EXAMPLE 9 find . Given the function value and the quadrant restriction, a) sin 0.2812, 90 180 b) cot 0.1611, 270 360 Solution a) We first sketch the angle in the second quadrant. We use the calculator to find the acute angle (reference angle) whose sine is 0.2812. The reference angle is approximately 16.33. We find the angle by subtracting 16.33 from 180: y u 16.33 x Thus, 163.67. b) We begin by sketching the angle in the fourth quadrant. Because the tangent and cotangent values are reciprocals, we know that y u 180 16.33 163.67. x 80.85 tan 1 6.2073. 0.1611 We use the calculator to find the acute angle (reference angle) whose tangent is 6.2073, ignoring the fact that tan is negative. The reference angle is approximately 80.85. We find angle by subtracting 80.85 from 360: 360 80.85 279.15. Thus, 279.15. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 468 Chapter 5 5.3 1/10/05 1:01 PM Page 468 • The Trigonometric Functions Exercise Set The terminal side of angle in standard position lies on the given line in the given quadrant. Find sin , cos , and tan . 29. 2x 3y 0; quadrant IV For angles of the following measures, state in which quadrant the terminal side lies. It helps to sketch the angle in standard position. 1. 187 III 2. 14.3 IV 3. 24515 III 4. 120 III 30. 4x y 0; quadrant II 5. 800 I 6. 1075 IV 31. 5x 4y 0; quadrant I 7. 460.5 III 8. 315 IV 32. y 0.8x ; quadrant III 9. 912 II 10. 131560 I 12. 345.14 I 11. 537 II Find two positive angles and two negative angles that are coterminal with the given angle. Answers may vary. 13. 74 14. 81 A function value and a quadrant are given. Find the other five function values. Give exact answers. 1 33. sin , quadrant III 3 34. tan 5, quadrant I 15. 115.3 16. 27510 35. cot 2, quadrant IV 17. 180 18. 310 36. cos Find the complement and the supplement. 19. 17.11 72.89, 162.89 20. 4738 4222, 13222 21. 12314 22. 9.038 80.962, 170.962° 23. 45.2 44.8, 134.8 24. 67.31 22.69, 112.69 Find the six trigonometric function values for the angle shown. 25. 26. y y (12, 5) r b x u x r (7, 3) 27. 28. y y x (23, 4) r 3 , quadrant IV 5 38. sin 5 , quadrant III 13 Find the reference angle and the exact function value if it exists. 45; 2 3 39. cos 150 30°; 40. sec 225 2 2 2 Not defined 41. tan 135 45; 1 42. sin 45 45°; 43. sin 7560 0 44. tan 270 2 45. cos 495 45°; 2 47. csc 210 30; 2 48. sin 300 60; 49. cot 570 30; 3 50. cos 120 55. cos 180 1 56. csc 90 1 57. tan 240 60; 3 58. cot 180 46. tan 675 45; 1 3 2 60; 21 3 51. tan 330 30; 52. cot 855 45; 1 3 53. sec 90 Not defined 54. sin 90 1 (9, 1) f 37. cos 4 , quadrant II 5 r a x Answers to Exercises 13–18, 21, and 25–38 can be found on pp. IA-31 and IA-32. Not defined Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 469 Section 5.3 59. sin 495 45°; 2 2 60. sin 1050 30; 12 61. csc 225 45; 2 62. sin 450 1 63. cos 0 1 64. tan 480 60; 3 65. cot 90 0 66. sec 315 45; 2 67. cos 90 0 68. sin 135 45; 69. cos 270 0 • Trigonometric Functions of Any Angle In aerial navigation, directions are given in degrees clockwise from north. Thus, east is 90 , south is 180 , and west is 270. Several aerial directions or bearings are given below. Aerial Navigation. 0 N 2 2 70. tan 0 0 74. 620 75. 215 76. 290 77. 272 78. 91 Use a calculator in Exercises 79 – 82, but do not use the trigonometric function keys. 79. Given that sin 41 0.6561, cos 41 0.7547, tan 41 0.8693, find the trigonometric function values for 319. 0 N 50 W 270 Find the signs of the six trigonometric function values for the given angles. 71. 319 72. 57 73. 194 469 E 90 S 180 0 N 0 N W 270 W 350 270 E 90 S 180 S 180 83. An airplane flies 150 km from an airport in a direction of 120. How far east of the airport is the plane then? How far south? East: about 130 km; south: 75 km 0° N 82. Given that sin 35 0.5736, cos 35 0.8192, tan 35 0.7002, find the trigonometric function values for 215. 120° W 270° E 90° 150 km S 180° 81. Given that find the trigonometric function values for 115. E 90 223 find the trigonometric function values for 333. sin 65 0.9063, cos 65 0.4226, tan 65 2.1445, E 90 S 180 80. Given that sin 27 0.4540, cos 27 0.8910, tan 27 0.5095, 115 W 270 84. An airplane leaves an airport and travels for 100 mi in a direction of 300. How far north of the airport is the plane then? How far west? North: 50 mi; west: about 87 mi 0° N 100 mi W 270° E 90° 300° S 180° Answers to Exercises 71– 82 can be found on p. IA-32. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 470 Chapter 5 1/10/05 1:01 PM Page 470 • The Trigonometric Functions 85. An airplane travels at 150 kmh for 2 hr in a direction of 138 from Omaha. At the end of this time, how far south of Omaha is the plane? Determine the domain and the range of the function. x4 111. fx x2 86. An airplane travels at 120 kmh for 2 hr in a direction of 319 from Chicago. At the end of this time, how far north of Chicago is the plane? 112. gx About 223 km About 181 km Find the function value. Round to four decimal places. 87. tan 310.8 1.1585 88. cos 205.5 0.9026 89. cot 146.15 1.4910 90. sin 16.4 0.2823 91. sin 11842 0.8771 92. cos 27345 0.0654 x2 9 2x 7x 15 2 Find the zeros of the function. [2.3] 2, 3 113. fx 12 x [2.1] 12 114. gx x 2 x 6 Find the x-intercepts of the graph of the function. 115. fx 12 x 116. gx x 2 x 6 [2.1] 12, 0 [2.3] 2, 0, 3, 0 93. cos 295.8 0.4352 94. tan 1086.2 0.1086 Synthesis 95. cos 5417 0.9563 96. sec 24055 2.0573 97. csc 520 2.9238 98. sin 3824 0.6947 117. Valve Cap on a Bicycle. The valve cap on a bicycle wheel is 12.5 in. from the center of the wheel. From the position shown, the wheel starts to roll. After the wheel has turned 390, how far above the ground is the valve cap? Assume that the outer radius of the tire is 13.375 in. 19.625 in. Given the function value and the quadrant restriction, find . INTERVAL 270, 360 275.4 100. tan 0.2460 180, 270 193.8 101. cos 0.9388 180, 270 200.1 102. sec 1.0485 90, 180 162.5 103. tan 3.0545 270, 360 288.1 104. sin 0.4313 180, 270 205.6 105. csc 1.0480 0, 90 72.6 106. cos 0.0990 90, 180 95.7 FUNCTION VALUE 99. sin 0.9956 13.375 in. 12.5 in. 118. Seats of a Ferris Wheel. The seats of a ferris wheel are 35 ft from the center of the wheel. When you board the wheel, you are 5 ft above the ground. After you have rotated through an angle of 765, how far above the ground are you? About 15.3 ft Collaborative Discussion and Writing 107. Why do the function values of depend only on the angle and not on the choice of a point on the terminal side? 108. Why is the domain of the tangent function different from the domains of the sine and the cosine functions? 35 ft Skill Maintenance Graph the function. Sketch and label any vertical asymptotes. 1 109. fx 2 110. gx x 3 2x 1 x 25 5 ft Answers to Exercises 109 – 112 can be found on p. IA-32. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 471 Section 5. 4 • 471 Radians, Arc Length, and Angular Speed Find points on the unit circle determined by real numbers. Convert between radian and degree measure; find coterminal, complementary, and supplementary angles. Find the length of an arc of a circle; find the measure of a central angle of a circle. Convert between linear speed and angular speed. 5.4 Radians, Arc Length, and Angular Speed Another useful unit of angle measure is called a radian. To introduce radian measure, we use a circle centered at the origin with a radius of length 1. Such a circle is called a unit circle. Its equation is x 2 y 2 1. y (x, y) 1 circles review section 1.1. x x2 y2 1 Distances on the Unit Circle The circumference of a circle of radius r is 2 r . Thus for the unit circle, where r 1, the circumference is 2. If a point starts at A and travels around the circle (Fig. 1), it will travel a distance of 2. If it travels 1 halfway around the circle (Fig. 2), it will travel a distance of 2 2, or . y y y y p C A B A x x u D d B A x A x 2p Figure 1 Figure 2 Figure 3 Figure 4 If a point C travels 18 of the way around the circle (Fig. 3), it will 1 1 travel a distance of 8 2, or 4. Note that C is 4 of the way from A to B. 1 If a point D travels 6 of the way around the circle (Fig. 4), it will travel 1 1 a distance of 6 2, or 3. Note that D is 3 of the way from A to B. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 472 Chapter 5 1/10/05 1:01 PM Page 472 • The Trigonometric Functions EXAMPLE 1 How far will a point travel if it goes (a) 41 , (b) 121 , (c) 83 , and (d) 65 of the way around the unit circle? Solution 1 1 1 a) 4 of the total distance around the circle is 4 2, which is 2 , or 2. 1 1 b) The distance will be 12 2, which is 6 , or 6. 3 3 c) The distance will be 8 2, which is 4 , or 34. 5 5 d) The distance will be 6 2, which is 3 , or 53. Think of 53 as 2 3 . These distances are illustrated in the following figures. y y y y f q p A x x x x A point may travel completely around the circle and then continue. 1 For example, if it goes around once and then continues 4 of the way 1 around, it will have traveled a distance of 2 4 2, or 52 (Fig. 5). Every real number determines a point on the unit circle. For the positive number 10, for example, we start at A and travel counterclockwise a distance of 10. The point at which we stop is the point “determined” by the number 10. Note that 2 6.28 and that 10 1.62 . Thus 3 the point for 10 travels around the unit circle about 1 5 times (Fig. 6). y y r 10 A x Figure 5 x Figure 6 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 473 Section 5. 4 • 473 Radians, Arc Length, and Angular Speed For a negative number, we move clockwise around the circle. Points for 4 and 32 are shown in the figure below. The number 0 determines the point A. y y w A A x x d EXAMPLE 2 On the unit circle, mark the point determined by each of the following real numbers. a) 9 4 b) 7 6 Solution 1 a) Think of 94 as 2 4 . (See the figure on the left below.) Since 94 0, the point moves counterclockwise. The point goes com1 pletely around once and then continues 4 of the way from A to B. y y k B A x F A B D A x b) The number 76 is negative, so the point moves clockwise. From 6 A to B, the distance is , or 6 , so we need to go beyond B another distance of 6, clockwise. (See the figure on the right above.) Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 474 Chapter 5 1/10/05 1:01 PM Page 474 • The Trigonometric Functions Radian Measure Degree measure is a common unit of angle measure in many everyday applications. But in many scientific fields and in mathematics (calculus, in particular), there is another commonly used unit of measure called the radian. Consider the unit circle. Recall that this circle has radius 1. Suppose we measure, moving counterclockwise, an arc of length 1, and mark a point T on the circle. y T Arc length is 1 u 1 radian r1 x 1 If we draw a ray from the origin through T, we have formed an angle. The measure of that angle is 1 radian. The word radian comes from the word radius. Thus measuring 1 “radius” along the circumference of the circle determines an angle whose measure is 1 radian. One radian is about 57.3. Angles that measure 2 radians, 3 radians, and 6 radians are shown below. 1 radian 57.3 y y u 2 radians 1 y u 3 radians 1 u 6 radians 1 1 1 1 1 1 1 1 x 1 x x 1 1 1 When we make a complete (counterclockwise) revolution, the terminal side coincides with the initial side on the positive x-axis. We then have an angle whose measure is 2 radians, or about 6.28 radians, which is the circumference of the circle: 2 r 2 1 2. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 475 Section 5. 4 • Radians, Arc Length, and Angular Speed 475 Thus a rotation of 360 (1 revolution) has a measure of 2 radians. A half revolution is a rotation of 180, or radians. A quarter revolution is a rotation of 90, or 2 radians, and so on. y u 90 q radians 1.57 radians 1 y u 180 p radians 3.14 radians 1 x y u 270 w radians 4.71 radians 1 x y u 360 2p radians 6.28 radians x 1 To convert between degrees and radians, we first note that 360 2 radians. It follows that 180 radians. To make conversions, we multiply by 1, noting that: Converting Between Degree and Radian Measure radians 180 1. 180 radians radians To convert from degree to radian measure, multiply by . 180 180 To convert from radian to degree measure, multiply by . radians GCM EXAMPLE 3 Convert each of the following to radians. a) 120 b) 297.25 Solution a) 120 120 radians 180 Multiplying by 1 120 radians 180 2 radians, or about 2.09 radians 3 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley x BBEPMC05_0321279115.QXP 476 Chapter 5 1/10/05 1:01 PM Page 476 • The Trigonometric Functions b) 297.25 297.25 radians 180 297.25 radians 180 297.25 radians 180 5.19 radians 120° We also can use a calculator set in RADIAN mode to convert the angle measures. We enter the angle measure followed by ° (degrees) from the ANGLE menu. 2.094395102 297.25° 5.187991202 GCM Convert each of the following to degrees. EXAMPLE 4 a) 3 radians 4 b) 8.5 radians Solution 3 3 180 radians Multiplying by 1 a) radians 4 4 radians 3 3 180 180 135 4 4 180 b) 8.5 radians 8.5 radians radians 8.5180 487.01 (3/4) r 135 8.5 r 487.0141259 With a calculator set in DEGREE mode, we can enter the angle measure followed by r(radians) from the ANGLE menu. The radian – degree equivalents of the most commonly used angle measures are illustrated in the following figures. y y 135 f 180 p h 225 90 60 45 qu 30 d A 0 w 2p 360 x j 270 315 270 315 w j h 2p 360 180 p 0 x A d f 30 q u 45 135 90 60 225 When a rotation is given in radians, the word “radians” is optional and is most often omitted. Thus if no unit is given for a rotation, the rotation is understood to be in radians. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 477 Section 5. 4 • Radians, Arc Length, and Angular Speed 477 We can also find coterminal, complementary, and supplementary angles in radian measure just as we did for degree measure in Section 5.3. EXAMPLE 5 Find a positive angle and a negative angle that are coterminal with 23. Many answers are possible. Solution To find angles coterminal with a given angle, we add or subtract multiples of 2 : 2 2 6 8 2 , 3 3 3 3 16 2 2 18 32 . 3 3 3 3 y y 8p 3 i i 16p 3 x x Thus, 83 and 163 are two of the many angles coterminal with 23. EXAMPLE 6 Solution Find the complement and the supplement of 6. Since 90 equals 2 radians, the complement of 6 is 3 2 , or 2 6 6 6 6 . 3 Since 180 equals radians, the supplement of 6 is 6 5 . 6 6 6 6 Thus the complement of 6 is 3 and the supplement is 56. Arc Length and Central Angles y s s1 x 1 r Radian measure can be determined using a circle other than a unit circle. In the figure at left, a unit circle (with radius 1) is shown along with another circle (with radius r, r 1). The angle shown is a central angle of both circles. From geometry, we know that the arcs that the angle subtends have their lengths in the same ratio as the radii of the circles. The radii of the circles are r and 1. The corresponding arc lengths are s and s1. Thus we have the proportion r s , s1 1 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 478 Chapter 5 1/10/05 1:01 PM Page 478 • The Trigonometric Functions which also can be written as s1 s . 1 r Now s1 is the radian measure of the rotation in question. It is common to use a Greek letter, such as , for the measure of an angle or rotation and the letter s for arc length. Adopting this convention, we rewrite the proportion above as s . r In any circle, the measure (in radians) of a central angle, the arc length the angle subtends, and the length of the radius are related in this fashion. Or, in general, the following is true. Radian Measure The radian measure of a rotation is the ratio of the distance s traveled by a point at a radius r from the center of rotation, to the length of the radius r: s . r y r u s x When using the formula sr , must be in radians and s and r must be expressed in the same unit. EXAMPLE 7 Find the measure of a rotation in radians when a point 2 m from the center of rotation travels 4 m. Solution We have s r 4m 2. 2m The unit is understood to be radians. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 479 Section 5. 4 y Radians, Arc Length, and Angular Speed 479 EXAMPLE 8 Find the length of an arc of a circle of radius 5 cm associated with an angle of 3 radians. s r = 5 cm • Solution θ = 3 x We have s , r or s r . Thus s 5 cm 3, or about 5.24 cm. Linear Speed and Angular Speed Linear speed is defined to be distance traveled per unit of time. If we use v for linear speed, s for distance, and t for time, then v s . t Similarly, angular speed is defined to be amount of rotation per unit of time. For example, we might speak of the angular speed of a bicycle wheel as 150 revolutions per minute or the angular speed of the earth as 2 radians per day. The Greek letter (omega) is generally used for angular speed. Thus for a rotation and time t, angular speed is defined as . t As an example of how these definitions can be applied, let’s consider the refurbished carousel at the Children’s Museum in Indianapolis, Indiana. It consists of three circular rows of animals. All animals, regardless of the row, travel at the same angular speed. But the animals in the outer row travel at a greater linear speed than those in the inner rows. What is the relationship between the linear speed v and the angular speed ? To develop the relationship we seek, recall that, for rotations measured in radians, sr . This is equivalent to s r . We divide by time, t, to obtain s r Dividing by t t t s r t t v Now st is linear speed v and t is angular speed . Thus we have the relationship we seek, v r. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 480 Chapter 5 1/10/05 1:01 PM Page 480 • The Trigonometric Functions Linear Speed in Terms of Angular Speed The linear speed v of a point a distance r from the center of rotation is given by v r, where is the angular speed in radians per unit of time. For the formula v r, the units of distance for v and r must be the same, must be in radians per unit of time, and the units of time for v and must be the same. EXAMPLE 9 Linear Speed of an Earth Satellite. An earth satellite in circular orbit 1200 km high makes one complete revolution every 90 min. What is its linear speed? Use 6400 km for the length of a radius of the earth. Solution 6400 km To use the formula v r, we need to know r and : r 6400 km 1200 km 1200 km 7600 km, 2 . t 90 min 45 min Radius of earth plus height of satellite We have, as usual, omitted the word radians. Now, using v r, we have v 7600 km 7600 km km 531 . 45 min 45 min min Thus the linear speed of the satellite is approximately 531 kmmin. EXAMPLE 10 Angular Speed of a Capstan. An anchor is hoisted at a rate of 2 ftsec as the chain is wound around a capstan with a 1.8-yd diameter. What is the angular speed of the capstan? 1.8 yd Capstan Chain Anchor Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 481 Section 5. 4 • Radians, Arc Length, and Angular Speed 481 Solution We will use the formula v r in the form vr , taking care to use the proper units. Since v is given in feet per second, we need r in feet: 1.8 3 ft d yd 2.7 ft. 2 2 1 yd r Then will be in radians per second: 2 ftsec 2 ft 1 v 0.741sec. r 2.7 ft sec 2.7 ft Thus the angular speed is approximately 0.741 radiansec. The formulas t and v r can be used in combination to find distances and angles in various situations involving rotational motion. EXAMPLE 11 Angle of Revolution. A 2004 Tundra V8 is traveling at a speed of 65 mph. Its tires have an outside diameter of 30.56 in. Find the angle through which a tire turns in 10 sec. 30.56 in. Solution Recall that t , or t . Thus we can find if we know and t. To find , we use the formula v r. The linear speed v of a point on the outside of the tire is the speed of the Tundra, 65 mph. For convenience, we first convert 65 mph to feet per second: mi 1 hr 1 min 5280 ft hr 60 min 60 sec 1 mi ft 95.333 . sec v 65 The radius of the tire is half the diameter. Now r d2 30.56 in.2 15.28 in. We will convert to feet, since v is in feet per second: r 15.28 in. 1 ft 12 in. 15.28 ft 1.27 ft. 12 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 482 Chapter 5 1/10/05 1:01 PM Page 482 • The Trigonometric Functions Using v r, we have 95.333 ft 1.27 ft , sec so Study Tip The Student’s Solutions Manual is an excellent resource if you need additional help with an exercise in the exercise sets. It contains worked-out solutions to the odd-numbered exercises in the exercise sets. 95.333 ftsec 75.07 . 1.27 ft sec Then in 10 sec, t 75.07 10 sec 751. sec Thus the angle, in radians, through which a tire turns in 10 sec is 751. Answers to Exercises 1–4 can be found on p. IA-32. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 482 Chapter 5 5.4 1/10/05 1:01 PM Page 482 • The Trigonometric Functions Exercise Set For each of Exercises 1– 4, sketch a unit circle and mark the points determined by the given real numbers. 3 3 1. a) b) c) 4 2 4 17 11 d) e) f) 4 4 2. a) 2 9 d) 4 6 10 d) 6 3. a) 4. a) 2 5 d) 2 5 b) 4 13 e) 4 2 3 14 e) 6 b) 3 b) 4 17 e) 6 Find two real numbers between 2 and 2 that determine each of the points on the unit circle. y 5. M 1 x c) 2 23 f) 4 7 6 23 f) 4 Q P N y 6. c) 5 c) 6 9 f) 4 N M x Q 2 4 , ; 3 3 3 N: , ; 2 2 5 3 P: , ; 4 4 11 Q: , 6 6 M: M: , ; 7 N: , ; 4 4 4 2 P: , ; 3 3 7 5 Q: , 6 6 P Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley Answers to Exercises 1– 4 can be found on p. IA-32. BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 483 Section 5. 4 For Exercises 7 and 8, sketch a unit circle and mark the approximate location of the point determined by the given real number. 7. a) 2.4 b) 7.5 c) 32 d) 320 8. a) 0.25 c) 47 b) 1.8 d) 500 Find a positive angle and a negative angle that are coterminal with the given angle. Answers may vary. 5 11 9 7 9. , 10. , 4 3 4 4 3 3 7 11. 6 13. 19 5 , 6 6 2 3 4 8 , 3 3 12. 14. 3 , 3 4 5 11 , 4 4 Find the complement and the supplement. 5 15. 16. 3 12 17. 3 8 18. 4 19. 12 20. 6 • 41. 345 6.02 42. 75 1.31 43. 95 1.66 44. 24.8 0.43 483 Convert to degree measure. Round the answer to two decimal places. 3 7 45. 135 46. 210 4 6 3 47. 8 1440 48. 49. 1 57.30 50. 17.6 1008.41 51. 2.347 134.47 52. 25 1432.39 53. 5 225 4 57. 60 54. 6 1080 55. 90 5156.62 56. 37.12 2126.82 2 51.43 7 58. 9 20 59. Certain positive angles are marked here in degrees. Find the corresponding radian measures. y 90 60 45 30 135 0 360 x 180 Convert to radian measure. Leave the answer in terms of . 21. 75 512 22. 30 6 23. 200 109 Radians, Arc Length, and Angular Speed 315 225 270 24. 135 34 25. 214.6 214.6180 26. 37.71 37.71180 27. 180 28. 90 2 29. 12.5 572 30. 6.3 7200 31. 340 179 32. 60 3 60. Certain negative angles are marked here in degrees. Find the corresponding radian measures. Convert to radian measure. Round the answer to two decimal places. 33. 240 4.19 34. 15 0.26 35. 60 1.05 y 270 225 315 360 0 x 180 p 36. 145 2.53 37. 117.8 2.06 38. 231.2 4.04 39. 1.354 0.02 40. 584 10.19 135 90 30 45 60 Answers to Exercises 7, 8, 15–20, 59, and 60 can be found on pp. IA-32 and IA-33. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 484 Chapter 5 1/10/05 1:01 PM Page 484 • The Trigonometric Functions Complete the table. Round the answers to two decimal places. Arc Length and Central Angles. DISTANCE, S (ARC LENGTH) ANGLE, RADIUS, R 1 ft 2 61. 8 ft 62. 200 cm 254.65 cm 45 63. 16 yd 3.2 yd 5 64. 5.50 in. 4.2 in. 5 12 3 2.29 65. In a circle with a 120-cm radius, an arc 132 cm long subtends an angle of how many radians? how many degrees, to the nearest degree? 1.1; 63 71. Linear Speed. A flywheel with a 15-cm diameter is rotating at a rate of 7 radianssec. What is the linear speed of a point on its rim, in centimeters per minute? 3150 cmmin 72. Linear Speed. A wheel with a 30-cm radius is rotating at a rate of 3 radianssec. What is the linear speed of a point on its rim, in meters per minute? 54 mmin 73. Angular Speed of a Printing Press. This text was printed on a four-color web heatset offset press. A cylinder on this press has a 13.37-in. diameter. The linear speed of a point on the cylinder’s surface is 18.33 feet per second. What is the angular speed of the cylinder, in revolutions per hour? Printers often refer to the angular speed as impressions per hour (IPH). (Source: Scott Coulter, Quebecor World, Taunton, MA) About 18,852 revolutions per hour 66. In a circle with a 10-ft diameter, an arc 20 ft long subtends an angle of how many radians? how many degrees, to the nearest degree? 4; 229 67. In a circle with a 2-yd radius, how long is an arc associated with an angle of 1.6 radians? 3.2 yd 68. In a circle with a 5-m radius, how long is an arc associated with an angle of 2.1 radians? 10.5 m 69. Angle of Revolution. Through how many radians does the minute hand of a clock rotate from 12:40 P.M. to 1:30 P.M.? 5, or about 5.24 3 11 12 1 10 2 8 4 9 3 7 6 5 70. Angle of Revolution. A tire on a 2004 Saturn Ion has an outside diameter of 24.877 in. Through what angle (in radians) does the tire turn while traveling 1 mi? 5094 24.877 in. 74. Linear Speeds on a Carousel. When Alicia and Zoe ride the carousel described earlier in this section, Alicia always selects a horse on the outside row, whereas Zoe prefers the row closest to the center. These rows are 19 ft 3 in. and 13 ft 11 in. from the center, respectively. The angular speed of the carousel is 2.4 revolutions per minute. What is the difference, in miles per hour, in the linear speeds of Alicia and Zoe? (Source: The Children’s Museum, Indianapolis, IN) 0.92 mph 75. Linear Speed at the Equator. The earth has a 4000-mi radius and rotates one revolution every 24 hr. What is the linear speed of a point on the equator, in miles per hour? 1047 mph Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 485 Section 5. 4 76. Linear Speed of the Earth. The earth is about 93,000,000 mi from the sun and traverses its orbit, which is nearly circular, every 365.25 days. What is the linear velocity of the earth in its orbit, in miles per hour? 66,659 mph 77. Determining the Speed of a River. A water wheel has a 10-ft radius. To get a good approximation of the speed of the river, you count the revolutions of the wheel and find that it makes 14 revolutions per minute (rpm). What is the speed of the river, in miles per hour? 10 mph • Radians, Arc Length, and Angular Speed 485 79. John Deere Tractor. A rear wheel and tire on a John Deere 8520 farm tractor has a 39-in. radius. Find the angle (in radians) through which a wheel rotates in 12 sec if the tractor is traveling at a speed of 22 mph. About 119 39 in. 10 ft Collaborative Discussion and Writing 80. Explain in your own words why it is preferable to omit the word, or unit, radians in radian measures. 81. In circular motion with a fixed angular speed, the length of the radius is directly proportional to the linear speed. Explain why with an example. 78. The Tour de France. Lance Armstrong won the 2004 Tour de France bicycle race. The wheel of his bicycle had a 67-cm diameter. His overall average linear speed during the race was 40.560 kmh. What was the angular speed of the wheel, in revolutions per hour? (Source: tourdefrancenews.com) About 19,270 revolutionshr 82. Two new cars are each driven at an average speed of 60 mph for an extended highway test drive of 2000 mi. The diameters of the wheels of the two cars are 15 in. and 16 in., respectively. If the cars use tires of equal durability and profile, differing only by the diameter, which car will probably need new tires first? Explain your answer. Skill Maintenance In each of Exercises 83–90, fill in the blanks with the correct terms. Some of the given choices will not be used. inverse a relation a horizontal line vertical asymptote a vertical line horizontal asymptote exponential function even function logarithmic function odd function natural sine of common cosine of logarithm tangent of one-to-one Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 486 Chapter 5 1/10/05 1:01 PM Page 486 • The Trigonometric Functions 83. The domain of a(n) the range of the inverse f 1. function f is [4.1] one-to-one 84. The is the length of the side adjacent to divided by the length of the hypotenuse. [5.1] cosine of 95. Angular Speed of a Gear Wheel. One gear wheel turns another, the teeth being on the rims. The wheels have 40-cm and 50-cm radii, and the smaller wheel rotates at 20 rpm. Find the angular speed of the larger wheel, in radians per second. 1.676 radianssec 85. The function fx a , where x is a real number, , a 0 and a 1, is called the base a. [4.2] exponential function x 86. The graph of a rational function may or may not cross a(n) . [3.5] horizontal asymptote 87. If the graph of a function f is symmetric with respect to the origin, we say that it is a(n) . [1.7] odd function 88. Logarithms, base e, are called logarithms. [4.3] natural 89. If it is possible for a(n) to intersect the graph of a function more than once, then the function is not one-to-one and its is not a function. [4.1] horizontal line; inverse 90. A(n) [4.3] logarithm 50 cm 40 cm 96. Angular Speed of a Pulley. Two pulleys, 50 cm and 30 cm in diameter, respectively, are connected by a belt. The larger pulley makes 12 revolutions per minute. Find the angular speed of the smaller pulley, in radians per second. 2.094 radianssec is an exponent. Synthesis 91. On the earth, one degree of latitude is how many kilometers? how many miles? (Assume that the radius of the earth is 6400 km, or 4000 mi, approximately.) 111.7 km; 69.8 mi 92. A point on the unit circle has y-coordinate 215. What is its x-coordinate? Check using a calculator. 25 93. A mil is a unit of angle measure. A right angle has a measure of 1600 mils. Convert each of the following to degrees, minutes, and seconds. a) 100 mils 53730 b) 350 mils 194115 94. A grad is a unit of angle measure similar to a degree. A right angle has a measure of 100 grads. Convert each of the following to grads. a) 48 53.33 grads 5 b) 142.86 grads 7 30 cm 50 cm 97. Distance Between Points on the Earth. To find the distance between two points on the earth when their latitude and longitude are known, we can use a right triangle for an excellent approximation if the points are not too far apart. Point A is at latitude 382730 N, longitude 825715 W; and point B is at latitude 382845 N, longitude 825630 W. Find the distance from A to B in nautical miles. (One minute of latitude is one nautical mile.) 1.46 nautical miles 98. Hands of a Clock. At what times between noon and 1:00 P.M. are the hands of a clock perpendicular? 16.3636 min after 12:00 noon, or about 12:16:22 P.M. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 487 Section 5.5 5.5 Circular Functions: Graphs and Properties Study Tip Take advantage of the numerous detailed art pieces in this text. They provide a visual image of the concept being discussed. Taking the time to study each figure is an efficient way to learn and retain the concepts. • Circular Functions: Graphs and Properties 487 Given the coordinates of a point on the unit circle, find its reflections across the x-axis, the y-axis, and the origin. Determine the six trigonometric function values for a real number when the coordinates of the point on the unit circle determined by that real number are given. Find function values for any real number using a calculator. Graph the six circular functions and state their properties. The domains of the trigonometric functions, defined in Sections 5.1 and 5.3, have been sets of angles or rotations measured in a real number of degree units. We can also consider the domains to be sets of real numbers, or radians, introduced in Section 5.4. Many applications in calculus that use the trigonometric functions refer only to radians. Let’s again consider radian measure and the unit circle. We defined radian measure for as s . r y (x, y) When r 1, s , 1 1 or s . su u x The arc length s on the unit circle is the same as the radian measure of the angle . In the figure above, the point x, y is the point where the terminal side of the angle with radian measure s intersects the unit circle. We can now extend our definitions of the trigonometric functions using domains composed of real numbers, or radians. In the definitions, s can be considered the radian measure of an angle or the measure of an arc length on the unit circle. Either way, s is a real number. To each real number s, there corresponds an arc length s on the unit circle. Trigonometric functions with domains composed of real numbers are called circular functions. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 488 Chapter 5 1/10/05 1:01 PM Page 488 • The Trigonometric Functions Basic Circular Functions For a real number s that determines a point x, y on the unit circle: y (x, y) s sin s second coordinate y , cos s first coordinate x , second coordinate y tan s x 0, first coordinate x 1 1 csc s y 0, second coordinate y 1 1 sec s x 0, first coordinate x first coordinate x cot s y 0. second coordinate y 1 x We can consider the domains of trigonometric functions to be real numbers rather than angles. We can determine these values for a specific real number if we know the coordinates of the point on the unit circle determined by that number. As with degree measure, we can also find these function values directly using a calculator. y Reflections on the Unit Circle (35 , 45 ) 1 x Let’s consider the unit circle and a few of its points. For any point x, y on the unit circle, x 2 y 2 1, we know that 1 x 1 and 1 y 1. If we know the x- or y-coordinate of a point on the unit circle, we can find 3 the other coordinate. If x 5 , then 35 2 y 2 1 (35 , 45 ) y 2 1 259 16 25 y 45 . 3 4 3 4 Thus, 5 , 5 and 5 , 5 are points on the unit circle. There are two 3 points with an x-coordinate of 5 . Now let’s consider the radian measure 3 and determine the coordinates of the point on the unit circle determined by 3. We construct a right triangle by dropping a perpendicular segment from the point to the x-axis. y y 30 (x, y) 1 uu (x, y) su x 1 y 60 q Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley x BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 489 Section 5.5 • 489 Circular Functions: Graphs and Properties Since 3 60, we have a 30– 60 right triangle in which the side opposite the 30 angle is one half of the hypotenuse. The hypotenuse, 1 1 or radius, is 1, so the side opposite the 30 angle is 2 1, or 2 . Using the Pythagorean theorem, we can find the other side: 1 2 2 y y2 1 1 3 4 4 3 3 . 4 2 u y2 1 y q, 3 2 x We know that y is positive since the point is in the first quadrant. Thus the coordinates of the point determined by 3 are x 12 and y 32, or 12, 32 . We can always check to see if a point is on the unit circle by substituting into the equation x 2 y 2 1: 1 2 2 3 2 2 1 3 1. 4 4 Because a unit circle is symmetric with respect to the x-axis, the y-axis, and the origin, we can use the coordinates of one point on the unit circle to find coordinates of its reflections. EXAMPLE 1 Each of the following points lies on the unit circle. Find their reflections across the x-axis, the y-axis, and the origin. a) c) a) 3 4 , 5 5 , 2 2 2 2 E, R x E, R E, R b) 2 2 , 2 2 1 3 , 2 2 Solution b) y E, R c) y 2 , 2 2 2 y q, 3 2 q, 3 2 x x 2 , 2 2 2 2 , 2 2 2 q, 3 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley q, 3 2 BBEPMC05_0321279115.QXP 490 Chapter 5 1/10/05 1:01 PM Page 490 • The Trigonometric Functions Finding Function Values Knowing the coordinates of only a few points on the unit circle along with their reflections allows us to find trigonometric function values of the most frequently used real numbers, or radians. y (0, 1) q, q u (1, 0) p 3 2 d 2 , 2 2 2 3 2 , q 0 2p (1, 0) x A EXAMPLE 2 a) tan 3 3 4 4 d) cos 3 b) cos c) sin (0, 1) Find each of the following function values. 6 e) cot f ) csc 7 2 Solution We locate the point on the unit circle determined by the rotation, and then find its coordinates using reflection if necessary. a) The coordinates of the point determined by 3 are 12, 32 . b) The reflection of 22, 22 across the y-axis is 22, 22 . y y q, 3 2 , 2 2 2 2 u 2 , 2 2 2 d f x Thus, tan y 32 3. 3 x 12 x 3 2 x . 4 2 d) The reflection of 12, 32 across the origin is 12, 32 . Thus, cos c) The reflection of 32, 12 across the x-axis is 32, 12 . y y 3 ,q 2 q, 3 2 u A x A x o , q 3 2 Thus, sin 6 y 1 . 2 q, 3 2 Thus, cos 1 4 x . 3 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 491 Section 5.5 e) The coordinates of the point determined by are 1, 0. • Circular Functions: Graphs and Properties 491 f ) The coordinates of the point determined by 72 are 0, 1. y y (0, 1) t (1, 0) p x x x 1 , which is not defined. y 0 We can also think of cot as the reciprocal of tan . Since tan yx 01 0 and the reciprocal of 0 is not defined, we know that cot is not defined. Thus, cot Normal Sci Eng Float 0123456789 Radian Degree Func Par Pol Seq Connected Dot Sequential Simul Real abi reˆθ i Full Horiz G –T Thus, csc 7 1 1 1. 2 y 1 Using a calculator, we can find trigonometric function values of any real number without knowing the coordinates of the point that it determines on the unit circle. Most calculators have both degree and radian modes. When finding function values of radian measures, or real numbers, we must set the calculator in RADIAN mode. (See the window at left.) EXAMPLE 3 Find each of the following function values of radian measures using a calculator. Round the answers to four decimal places. a) cos 2 5 b) tan 3 d) sec c) sin 24.9 Solution cos(2π/5) tan(3) .3090169944 a) cos Using a calculator set in RADIAN mode, we find the values. 2 0.3090 5 b) tan 3 0.1425 .1425465431 sin(24.9) .2306457059 7 c) sin 24.9 0.2306 7 1 1.1099 cos 7 Note in part (d) that the secant function value can be found by taking the reciprocal of the cosine value. Thus we can enter cos 7 and use the reciprocal key. d) sec Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 492 Chapter 5 1/10/05 1:01 PM Page 492 • The Trigonometric Functions We can graph the unit circle using a graphing calculator. We use PARAMETRIC mode with the following window and let X1T cos T and Y1T sin T. Here we use DEGREE mode. EXPLORING WITH TECHNOLOGY WINDOW Tmin 0 Tmax 360 Tstep 15 Xmin 1.5 Xmax 1.5 Xscl 1 Ymin 1 Ymax 1 Yscl 1 1 X1Tcos(T) Y1Tsin(T) 1.5 1.5 T 30 X .8660254 1 Y .5 Using the trace key and an arrow key to move the cursor around the unit circle, we see the T, X, and Y values appear on the screen. What do they represent? Repeat this exercise in RADIAN mode. What do the T, X, and Y values represent? (For more on parametric equations, see Section 9.7.) From the definitions on p. 488, we can relabel any point x, y on the unit circle as cos s, sin s, where s is any real number. y (x, y) (cos s, sin s) s (1, 0) x Graphs of the Sine and Cosine Functions y , 2 2 2 2 (0, 1) f q, 3 2 2 2 2, 2 qu 3 d , q A 2 0 (1, 0) p 2p x (1, 0) A 3 , q f d 2 q u 2 2 2 2 2 , 2 2 , 2 (0, 1) 3 q, 2 Properties of functions can be observed from their graphs. We begin by graphing the sine and cosine functions. We make a table of values, plot the points, and then connect those points with a smooth curve. It is helpful to first draw a unit circle and label a few points with coordinates. We can either use the coordinates as the function values or find approximate sine and cosine values directly with a calculator. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 493 Section 5.5 • 493 Circular Functions: Graphs and Properties s sin s cos s s sin s cos s 0 6 4 3 2 34 54 32 74 2 0 0.5 0.7071 0.8660 1 0.7071 0 0.7071 1 0.7071 0 1 0.8660 0.7071 0.5 0 0.7071 1 0.7071 0 0.7071 1 0 6 4 3 2 34 54 32 74 2 0 0.5 0.7071 0.8660 1 0.7071 0 0.7071 1 0.7071 0 1 0.8660 0.7071 0.5 0 0.7071 1 0.7071 0 0.7071 1 The graphs are as follows. y y sin s 2 1 2p w p q q p w 2p s 2p s 1 2 The sine function y 2 2p w p q y cos s q p w 1 2 The cosine function We can check these graphs using a graphing calculator. y cos x y sin x 2 2 2 π 2π 2 Xscl 2 π π 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2π 2 Xscl π 2 BBEPMC05_0321279115.QXP 494 Chapter 5 1/10/05 1:01 PM Page 494 • The Trigonometric Functions The sine and cosine functions are continuous functions. Note in the graph of the sine function that function values increase from 0 at s 0 to 1 at s 2, then decrease to 0 at s , decrease further to 1 at s 32, and increase to 0 at 2. The reverse pattern follows when s decreases from 0 to 2. Note in the graph of the cosine function that function values start at 1 when s 0, and decrease to 0 at s 2. They decrease further to 1 at s , then increase to 0 at s 32, and increase further to 1 at s 2. An identical pattern follows when s decreases from 0 to 2. From the unit circle and the graphs of the functions, we know that the domain of both the sine and cosine functions is the entire set of real numbers, , . The range of each function is the set of all real numbers from 1 to 1, 1, 1. Domain and Range of Sine and Cosine Functions The domain of the sine and cosine functions is , . The range of the sine and cosine functions is 1, 1. 3 2 2π 3 3 2 2π 3 EXPLORING WITH TECHNOLOGY Another way to construct the sine and cosine graphs is by considering the unit circle and transferring vertical distances for the sine function and horizontal distances for the cosine function. Using a graphing calculator, we can visualize the transfer of these distances. We use the calculator set in PARAMETRIC and RADIAN modes and let X1T cos T 1 and Y1T sin T for the unit circle centered at 1, 0 and X2T T and Y2T sin T for the sine curve. Use the following window settings. Tmin 0 Tmax 2 Tstep .1 Xmin 2 Xmax 2 Xscl 2 Ymin 3 Ymax 3 Yscl 1 With the calculator set in SIMULTANEOUS mode, we can actually watch the sine function (in red) “unwind” from the unit circle (in blue). In the two screens at left, we partially illustrate this animated procedure. Consult your calculator’s instruction manual for specific keystrokes and graph both the sine curve and the cosine curve in this manner. (For more on parametric equations, see Section 9.7.) A function with a repeating pattern is called periodic. The sine and cosine functions are examples of periodic functions. The values of these functions repeat themselves every 2 units. In other words, for any s, we have sin s 2 sin s and cos s 2 cos s . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 495 Section 5.5 • Circular Functions: Graphs and Properties 495 To see this another way, think of the part of the graph between 0 and 2 and note that the rest of the graph consists of copies of it. If we translate the graph of y sin x or y cos x to the left or right 2 units, we will obtain the original graph. We say that each of these functions has a period of 2. Periodic Function A function f is said to be periodic if there exists a positive constant p such that fs p fs for all s in the domain of f. The smallest such positive number p is called the period of the function. y s s 2p T x The period p can be thought of as the length of the shortest recurring interval. We can also use the unit circle to verify that the period of the sine and cosine functions is 2. Consider any real number s and the point T that it determines on a unit circle, as shown at left. If we increase s by 2, the point determined by s 2 is again the point T. Hence for any real number s, sin s 2 sin s cos s 2 cos s . and It is also true that sin s 4 sin s , sin s 6 sin s , and so on. In fact, for any integer k, the following equations are identities: sin s k2 sin s and cos s k2 cos s , or sin s sin s 2k and cos s cos s 2k . The amplitude of a periodic function is defined as one half of the distance between its maximum and minimum function values. It is always positive. Both the graphs and the unit circle verify that the maximum value of the sine and cosine functions is 1, whereas the minimum value of each is 1. Thus, 1 the amplitude of the sine function 2 1 1 1 y y sin x 2p 1 p Amplitude: 1 p 1 Period: 2p Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2p x BBEPMC05_0321279115.QXP 496 Chapter 5 1/10/05 1:01 PM Page 496 • The Trigonometric Functions and the amplitude of the cosine function is 12 1 1 1. y y cos x 1 Amplitude: 1 p 2p p 2p x 1 Period: 2p Using the TABLE feature on a graphing calculator, compare the y-values for y1 sin x and y2 sin x and for y3 cos x and y4 cos x. We set TblMin 0 and Tbl 12. EXPLORING WITH TECHNOLOGY X Y1 Y2 X Y3 Y4 0 .2618 .5236 .7854 1.0472 1.309 1.5708 0 .25882 .5 .70711 .86603 .96593 1 0 .2588 .5 .7071 .866 .9659 1 0 .2618 .5236 .7854 1.0472 1.309 1.5708 1 .96593 .86603 .70711 .5 .25882 0 1 .96593 .86603 .70711 .5 .25882 0 X0 X0 What appears to be the relationship between sin x and sin x and between cos x and cos x? Consider any real number s and its opposite, s. These numbers determine points T and T1 on a unit circle that are symmetric with respect to the x-axis. y y s s T(x, y) T(x, y) sin s cos s x sin (s) x T1(x, y) T1(x, y) cos (s) s s Because their second coordinates are opposites of each other, we know that for any number s, sin s sin s . Because their first coordinates are the same, we know that for any number s, cos s cos s . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 497 Section 5.5 • Circular Functions: Graphs and Properties 497 Thus we have shown that the sine function is odd and the cosine function is even. The following is a summary of the properties of the sine and cosine functions. even and odd functions review section 1.7. CONNECTING THE CONCEPTS COMPARING THE SINE AND COSINE FUNCTIONS SINE FUNCTION COSINE FUNCTION y sin x 2p 1. 2. 3. 4. 5. 6. p y 1 y y cos x 1 1 2p p 2p x Continuous Period: 2 Domain: All real numbers Range: 1, 1 Amplitude: 1 Odd: sin s sin s 1. 2. 3. 4. 5. 6. p 1 p 2p x Continuous Period: 2 Domain: All real numbers Range: 1, 1 Amplitude: 1 Even: cos s cos s Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions To graph the tangent function, we could make a table of values using a calculator, but in this case it is easier to begin with the definition of tangent and the coordinates of a few points on the unit circle. We recall that tan s sin s y . x cos s y y (x, y) (cos s, sin s) y sin s s tan s x cos s (1, 0) x (0, 1) 2 , 2 2 2 2 , 2 2 2 (1, 0) x (1, 0) 2 , 2 2 2 (0, 1) Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2 , 2 2 2 BBEPMC05_0321279115.QXP 498 Chapter 5 1/10/05 1:01 PM Page 498 • The Trigonometric Functions The tangent function is not defined when x, the first coordinate, is 0. That is, it is not defined for any number s whose cosine is 0: s 3 5 , , ,.... 2 2 2 We draw vertical asymptotes at these locations (see Fig. 1 below). y y 2 2 1 1 2pw p q q p w 2p s 2pw p q q p w 2p s 1 2 2 Figure 1 Figure 2 We also note that tan s 0 at s 0, , 2, 3, . . . , 7 3 5 9 tan s 1 at s . . . , , , ,..., , 4 4 4 4 4 9 5 3 7 tan s 1 at s . . . , , , ,.... , 4 4 4 4 4 We can add these ordered pairs to the graph (see Fig. 2 above) and investigate the values in 2, 2 using a calculator. Note that the function value is 0 when s 0, and the values increase without bound as s increases toward 2. The graph gets closer and closer to an asymptote as s gets closer to 2, but it never touches the line. As s decreases from 0 to 2, the values decrease without bound. Again the graph gets closer and closer to an asymptote, but it never touches it. We now complete the graph. y tan s y y tan x 2 2 1 2 π 2π 2p w p q q p w 1 DOT Mode 2 2 The tangent function Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2p s BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 499 Section 5.5 • 499 Circular Functions: Graphs and Properties From the graph, we see that the tangent function is continuous except where it is not defined. The period of the tangent function is . Note that although there is a period, there is no amplitude because there are no maximum and minimum values. When cos s 0, tan s is not defined tan s sin scos s. Thus the domain of the tangent function is the set of all real numbers except 2 k, where k is an integer. The range of the function is the set of all real numbers. The cotangent function cot s cos ssin s is not defined when y, the second coordinate, is 0 — that is, it is not defined for any number s whose sine is 0. Thus the cotangent is not defined for s 0, , 2, 3, . . . . The graph of the function is shown below. y cot s y y cot x 1/tan x 2 2 1 2 π 2π 2p w p q q p w 2p s 1 DOT Mode 2 The cotangent function The cosecant and sine functions are reciprocal functions, as are the secant and cosine functions. The graphs of the cosecant and secant functions can be constructed by finding the reciprocals of the values of the sine and cosine functions, respectively. Thus the functions will be positive together and negative together. The cosecant function is not defined for those numbers s whose sine is 0. The secant function is not defined for those numbers s whose cosine is 0. In the graphs below, the sine and cosine functions are shown by the gray curves for reference. y csc s y csc x 1/sin x y y sin s 2 2 1 2 π 2π 2p w p q q p 1 DOT Mode 2 The cosecant function Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley w 2p s BBEPMC05_0321279115.QXP 500 Chapter 5 1/10/05 1:01 PM Page 500 • The Trigonometric Functions y y sec s y sec x 1/cos x y cos s 2 2 2 π 2π 2p w p q q p w 2p s 1 DOT Mode 2 2 The secant function The following is a summary of the basic properties of the tangent, cotangent, cosecant, and secant functions. These functions are continuous except where they are not defined. CONNECTING THE CONCEPTS COMPARING THE TANGENT, COTANGENT, COSECANT, AND SECANT FUNCTIONS TANGENT FUNCTION COTANGENT FUNCTION 1. Period: 2. Domain: All real numbers except 2 k, where k is an integer 3. Range: All real numbers COSECANT FUNCTION 1. Period: 2. Domain: All real numbers except k, where k is an integer 3. Range: All real numbers SECANT FUNCTION 1. Period: 2 2. Domain: All real numbers except k, where k is an integer 3. Range: , 1 1, 1. Period: 2 2. Domain: All real numbers except 2 k, where k is an integer 3. Range: , 1 1, In this chapter, we have used the letter s for arc length and have avoided the letters x and y, which generally represent first and second coordinates. Nevertheless, we can represent the arc length on a unit circle by any variable, such as s, t, x, or . Each arc length determines a point that can be labeled with an ordered pair. The first coordinate of that ordered pair is the cosine of the arc length, and the second coordinate is the sine of the arc length. The identities we have developed hold no matter what symbols are used for variables — for example, cos s cos s , cos x cos x , cos cos , and cos t cos t . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 501 Section 5.5 5.5 3. 2 21 , 5 5 4. 1 3 , 2 2 5. The number 4 determines a point on the unit circle with coordinates 22, 22 . What are the coordinates of the point determined by 4? 6. A number determines a point on the unit circle with coordinates 23, 53 . What are the coordinates of the point determined by ? Find the function value using coordinates of points on the unit circle. Give exact answers. 1 7. sin 0 8. cos 3 2 9. cot 7 6 501 3 10. tan 14. tan 15. sec 2 Not defined 16. cos 10 17. cos 6 3 2 18. sin 2 3 19. sin 5 4 20. cos 11 6 1 2 4 1 1 3 2 3 2 21. sin 5 0 22. tan 3 2 Not defined 5 2 24. tan 5 3 3 0 1.3065 28. sin 11.7 0.7620 29. cot 342 2.1599 30. tan 1.3 31. cos 6 1 32. sin 0.3090 10 34. sec 10 7 33. csc 4.16 1.1747 35. tan 7 4 1 41. tan 5 6 2 2 27. sec 37 11 4 13. cos 11. sin 3 0 3 2 Find the function value using a calculator set in RADIAN mode. Round the answer to four decimal places, where appropriate. 2 25. tan 0.4816 26. cos 0.3090 7 5 37. sin 3 12. csc 4 23. cot Circular Functions: Graphs and Properties Exercise Set The following points are on the unit circle. Find the coordinates of their reflections across (a) the x-axis, (b) the y-axis, and (c) the origin. 3 7 2 5 1. , 2. , 4 4 3 3 • 39. sin 0 2 9 4 3.6021 4.4940 36. cos 2000 0.3675 0.7071 38. cot 7 Not defined 40. cos 29 0.7481 0 0.8391 42. sin 8 3 0.8660 In Exercises 43 – 48, make hand-drawn graphs. 43. a) Sketch a graph of y sin x . b) By reflecting the graph in part (a), sketch a graph of y sin x. c) By reflecting the graph in part (a), sketch a graph of y sin x . Same as (b) d) How do the graphs in parts (b) and (c) compare? The same 44. a) Sketch a graph of y cos x . b) By reflecting the graph in part (a), sketch a graph of y cos x. Same as (a) c) By reflecting the graph in part (a), sketch a graph of y cos x . d) How do the graphs in parts (a) and (b) compare? The same 45. a) Sketch a graph of y sin x . See Exercise 43(a). b) By translating, sketch a graph of y sin x . c) By reflecting the graph of part (a), sketch a graph of y sin x . Same as (b) d) How do the graphs of parts (b) and (c) compare? The same Answers to Exercises 1–6, 43(a), 43(b), 44(a), 44(c), and 45(b) can be found on p. IA-33. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 502 Chapter 5 1/10/05 1:01 PM Page 502 • The Trigonometric Functions 46. a) Sketch a graph of y sin x . See Exercise 47(a). b) By translating, sketch a graph of y sin x . c) By reflecting the graph of part (a), sketch a graph of y sin x . Same as (b) d) How do the graphs of parts (b) and (c) compare? Skill Maintenance 47. a) Sketch a graph of y cos x . b) By translating, sketch a graph of y cos x . c) By reflecting the graph of part (a), sketch a graph of y cos x . Same as (b) d) How do the graphs of parts (b) and (c) compare? 59. fx x, gx 12 x 4 1 The same The same 48. a) Sketch a graph of y cos x . See Exercise 47(a). b) By translating, sketch a graph of y cos x . c) By reflecting the graph of part (a), sketch a graph of y cos x . Same as (b) d) How do the graphs of parts (b) and (c) compare? The same Graph both functions in the same viewing window and describe how g is a transformation of f. 57. fx x 2, gx 2x 2 3 58. fx x 2, gx x 22 60. fx x , gx x 3 3 Write an equation for a function that has a graph with the given characteristics. Check using a graphing calculator. 61. The shape of y x 3, but reflected across the x-axis, shifted right 2 units, and shifted down 1 unit 3 [1.7] y x 2 1 62. The shape of y 1x , but shrunk vertically by a factor of 14 and shifted up 3 units [1.7] y Synthesis 1 3 4x 49. Of the six circular functions, which are even? Which are odd? Complete. (For example, sin x 2 sin x .) 63. cos x cos x 50. Of the six circular functions, which have period ? Which have period 2 ? 64. sin x 65. sin x 2k , k sin x Consider the coordinates on the unit circle for Exercises 51–54. 51. In which quadrants is the tangent function positive? negative? Positive: I, III; negative: II, IV 66. cos x 2k , k cos x sin x 67. sin x sin x 68. cos x cos x 52. In which quadrants is the sine function positive? negative? Positive: I, II; negative: III, IV 69. cos x cos x 70. cos x cos x 53. In which quadrants is the cosine function positive? negative? Positive: I, IV; negative: II, III 71. sin x sin x 72. sin x sin x 54. In which quadrants is the cosecant function positive? negative? Positive: I, II; negative: III, IV Collaborative Discussion and Writing 73. Find all numbers x that satisfy the following. Check using a graphing calculator. a) sin x 1 2 2k, k b) cos x 1 2k, k c) sin x 0 k, k 55. Describe how the graphs of the sine and cosine functions are related. 74. Find f g and g f , where fx x 2 2x and gx cos x . 56. Explain why both the sine and cosine functions are continuous, but the tangent function, defined as sinecosine, is not continuous. Use a graphing calculator to determine the domain, the range, the period, and the amplitude of the function. 75. y sin x2 76. y cos x 1 Answers to Exercises 46(b), 47(a), 47(b), 48(b), 49, 50, 57–60, and 74–76 can be found on p. IA-33. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:01 PM Page 503 Section 5.5 • Circular Functions: Graphs and Properties y Determine the domain of the function. 77. fx cos x 79. fx sin x cos x 78. gx C 1 sin x 83. y sin x cos x 1 D u 80. gx log sin x Graph. 81. y 3 sin x E P x x k, k an integer. 503 O A B x 82. y sin x 84. y cos x 85. One of the motivations for developing trigonometry with a unit circle is that you can actually “see” sin and cos on the circle. Note in the figure at right that AP sin and OA cos . It turns out that you can also “see” the other four trigonometric functions. Prove each of the following. a) BD tan b) OD sec c) OE csc d) CE cot 86. Using graphs, determine all numbers x that satisfy sin x cos x . 87. Using a calculator, consider sin xx , where x is between 0 and 2. As x approaches 0, this function approaches a limiting value. What is it? 1 Answers to Exercises 77 and 79 – 86 can be found on pp. IA-33 and IA-34. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 522 Chapter 5 1/10/05 1:02 PM Page 522 • The Trigonometric Functions Chapter 5 Summary and Review Important Properties and Formulas Trigonometric Function Values of an Acute Angle Let be an acute angle of a right triangle. The six trigonometric functions of are as follows: opp , hyp hyp csc , opp adj , hyp hyp sec , adj sin cos Hypotenuse Opposite u u Adjacent to u opp , adj adj cot . opp tan Reciprocal Functions csc 1 , sin sec 1 , cos cot 1 tan Function Values of Special Angles 0 30 45 60 90 sin 0 12 22 32 1 cos 1 32 22 12 0 tan 0 33 1 3 Not defined Cofunction Identities 90 u u sin cos 90 , tan cot 90 , sec csc 90 , cos sin 90 , cot tan 90 , csc sec 90 (continued) Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:02 PM Page 523 Chapter 5 • Summary and Review Trigonometric Functions of Any Angle If Px, y is any point on the terminal side of any angle in standard position, and r is the distance from the origin to Px, y, where r x 2 y 2, then y , r r csc , y x , r r sec , x sin cos Positive: sin Negative: cos, tan y u x x Basic Circular Functions For a real number s that determines a point x, y on the unit circle: sin s y , cos s x , y tan s . x y (x, y) Positive: All Negative: None s I x (, ) x (, ) (, ) III IV Positive: tan Negative: sin, cos Positive: cos Negative: sin, tan Radian–Degree Equivalents y 135 f 180 p h 225 P(x, y) r y y , x x cot . y 1 II (, ) y tan Signs of Function Values The signs of the function values depend only on the coordinates of the point P on the terminal side of an angle. 523 90 60 45 qu 30 d A 0 w 2p 360 x j 270 315 Sine is an odd function: sin s sin s . Cosine is an even function: cos s cos s . Transformations of Sine and Cosine Functions To graph y A sin Bx C D and y A cos Bx C D : 1. Stretch or shrink the graph horizontally 2 according to B. Period B 2. Stretch or shrink the graph vertically according to A. (Amplitude A) 3. Translate the graph horizontally according to C CB. Phase shift B 4. Translate the graph vertically according to D. Linear Speed in Terms of Angular Speed v r Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 524 Chapter 5 1/10/05 1:02 PM Page 524 • The Trigonometric Functions Review Exercises 1. Find the six trigonometric function values of . 8 u 3 27. a 30.5, B 51.17 [5.2] A 38.83, b 37.9, 73 c 48.6 91 , find the other five 10 trigonometric function values. 2. Given that sin Find the exact function value, if it exists. 3. cos 45 [5.1] 22 4. cot 60 [5.1] 33 5. cos 495 [5.3] 22 6. sin 150 [5.3] 12 7. sec 270 8. tan 600 [5.3] 3 [5.3] Not defined 9. csc 60 [5.1] 233 Solve each of the following right triangles. Standard lettering has been used. 26. a 7.3, c 8.6 [5.2] b 4.5, A 58.1, B 31.9 28. One leg of a right triangle bears east. The hypotenuse is 734 m long and bears N5723E. Find the perimeter of the triangle. [5.2] 1748 m 29. An observer’s eye is 6 ft above the floor. A mural is being viewed. The bottom of the mural is at floor level. The observer looks down 13 to see the bottom and up 17 to see the top. How tall is the mural? [5.2] 14 ft 10. cot 45 [5.1] 1 11. Convert 22.27 to degrees, minutes, and seconds. Round to the nearest second. [5.1] 221612 17° 13° 6 ft 12. Convert 473327 to decimal degree notation. Round to two decimal places. [5.1] 47.56 Find the function value. Round to four decimal places. 13. tan 2184 [5.3] 0.4452 14. sec 27.9 [5.3] 1.1315 15. cos 181342 16. sin 24524 17. cot 33.2 18. sin 556.13 [5.3] 0.9092 [5.3] 0.9498 [5.3] 1.5292 [5.3] 0.2778 Find in the interval indicated. Round the answer to the nearest tenth of a degree. 19. cos 0.9041, 180, 270 [5.3] 205.3 20. tan 1.0799, 0, 90 [5.3] 47.2 Find the exact acute angle , in degrees, given the function value. 3 21. sin [5.1] 60 22. tan 3 [5.1] 60 2 23. cos 2 [5.1] 45 2 24. sec For angles of the following measures, state in which quadrant the terminal side lies. 30. 142115 [5.3] II 31. 635.2 [5.3] I 32. 392 [5.3] IV Find a positive angle and a negative angle that are coterminal with the given angle. Answers may vary. 7 5 33. 65 [5.3] 425, 295 34. [5.4] , 3 3 3 Find the complement and the supplement. 35. 13.4 36. 6 37. Find the six trigonometric function values for the angle shown. 23 [5.1] 30 3 25. Given that sin 59.1 0.8581, cos 59.1 0.5135, and tan 59.1 1.6709, find the six function values for 30.9. y (2, 3) 3 r 2 u x Answers to Exercises 1, 2, 25, and 35–37 can be found on pp. IA-35 and IA-36. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:02 PM Page 525 Chapter 5 38. Given that tan 25 and that the terminal side is in quadrant III, find the other five function values. 54. sin 5 3 56. tan 6 [5.5] • 3 2 55. sin 3 3 525 Review Exercises 7 6 1 [5.5] 2 57. cos 13 [5.5] 1 39. An airplane travels at 530 mph for 3 12 hr in a direction of 160 from Minneapolis, Minnesota. At the end of that time, how far south of Minneapolis is the airplane? [5.3] About 1743 mi Find the function value. Round to four decimal places. 58. sin 24 [5.5] 0.9056 59. cos 75 [5.5] 0.9218 40. On a unit circle, mark and label the points determined by 76, 34, 3, and 94. 60. cot 16 For angles of the following measures, convert to radian measure in terms of , and convert to radian measure not in terms of . Round the answer to two decimal places. 41. 145.2 [5.4] 121 , 2.53 42. 30 [5.4] , 150 0.52 6 Convert to degree measure. Round the answer to two decimal places. 3 43. [5.4] 270 44. 3 [5.4] 171.89 2 45. 4.5 [5.4] 257.83 46. 11 [5.4] 1980 47. Find the length of an arc of a circle, given a central angle of 4 and a radius of 7 cm. [5.4] 74, or 5.5 cm 48. An arc 18 m long on a circle of radius 8 m subtends an angle of how many radians? how many degrees, to the nearest degree? [5.4] 2.25, 129 49. At one time, inside La Madeleine French Bakery and Cafe in Houston, Texas, there was one of the few remaining working watermills in the world. The 300-yr-old French-built waterwheel had a radius of 7 ft and made one complete revolution in 70 sec. What was the linear speed in feet per minute of a point on the rim? (Source: La Madeleine French Bakery and Cafe, Houston, TX) [5.4] About 37.9 ftmin 50. An automobile wheel has a diameter of 14 in. If the car travels at a speed of 55 mph, what is the angular velocity, in radians per hour, of a point on the edge of the wheel? [5.4] 497,829 radianshr 51. The point is on a unit circle. Find the coordinates of its reflections across the x-axis, the y-axis, and the origin. [5.5] 35 , 45 , 35 , 45 , 35 , 45 3 5, 45 Find the exact function value, if it exists. 5 52. cos [5.5] 1 53. tan [5.5] 1 4 [5.5] 61. tan [5.5] Not defined 3 7 [5.5] 6.1685 63. cos 62. sec 14.3 [5.5] 4.3813 5 [5.5] 0.8090 64. Graph by hand each of the six trigonometric functions from 2 to 2. 65. What is the period of each of the six trigonometric functions? [5.5] Period of sin, cos, sec, csc: 2 ; period of tan, cot: 66. Complete the following table. FUNCTION DOMAIN RANGE sine , 1, 1 cosine , 1, 1 , tangent Domain of tangent: x x 2 k, k 67. Complete the following table with the sign of the specified trigonometric function value in each of the four quadrants. FUNCTION I II III IV sine cosine tangent Determine the amplitude, the period, and the phase shift of the function, and sketch the graph of the function. Then check the graph using a graphing calculator. 68. y sin x 2 69. y 3 1 cos 2x 2 2 Answers to Exercises 38, 40, 64, 68, and 69 can be found on p. IA-36. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 526 Chapter 5 1/10/05 1:02 PM Page 526 • The Trigonometric Functions In Exercises 70–73, without a graphing calculator, match the function with one of the graphs (a)– (d), which follow. Then check your work using a graphing calculator. y y a) b) 2 2 2 1 1 2 x 2 1 2 2 77. Does 5 sin x 7 have a solution for x ? Why or why not? 78. Explain the disadvantage of a graphing calculator when graphing a function like 1 76. Describe the shape of the graph of the cosine function. How many maximum values are there of the cosine function? Where do they occur? 2 x fx sin x . x Synthesis y c) All values 80. Graph y 3 sin x2, and determine the domain, the range, and the period. 2 2 2 79. For what values of x in 0, 2 is sin x x true? y d) 1 2 x 2 2 4 1 6 2 81. In the graph below, y1 sin x is shown and y2 is shown in red. Express y2 as a transformation of the graph of y1. 2 x [5.6] y2 2 sin x y1 sin x, y2 ? 2 2 4 70. y cos 2x [5.6] (d) 71. y 1 sin x 1 2 [5.6] (a) 72. y 2 sin [5.6] (c) 1 x3 2 73. y cos x [5.6] (b) 2 π 2 74. Sketch a graph of y 3 cos x sin x for values of x between 0 and 2. Collaborative Discussion and Writing 75. Compare the terms radian and degree. 2π 4 82. Find the domain of y log cos x. 83. Given that sin x 0.6144 and that the terminal side is in quadrant II, find the other basic circular function values. [5.3] cos x 0.7890, tan x 0.7787, cot x 1.2842, sec x 1.2674, csc x 1.6276 Answers to Exercises 74–78, 80, and 82 can be found on p. IA-36. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 526 Chapter 5 1/10/05 1:02 PM Page 526 • The Trigonometric Functions Chapter 5 Test 1. Find the six trigonometric function values of . 65 Find the exact function value, if it exists. 2. sin 120 [5.3] 32 3. tan 45 [5.3] 1 4. cos 3 4 7 [5.4] 1 5. sec 5 4 [5.4] 2 6. Convert 382756 to decimal degree notation. Round to two decimal places. [5.1] 38.47 Answer to Exercise 1 can be found on p. IA-36. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC05_0321279115.QXP 1/10/05 1:02 PM Page 527 Chapter 5 Find the function values. Round to four decimal places. 7. tan 526.4 8. sin 12 [5.3] 0.2419 9. sec 5 9 [5.3] 0.2079 [5.4] 5.7588 23. Which is the graph of the function? y a) b) 10. cos 76.07 [5.4] 0.7827 11. Find the exact acute angle , in degrees, for which sin 12 . [5.1] 30 12. Given that sin 28.4 0.4756, cos 28.4 0.8796, and tan 28.4 0.5407, find the six trigonometric function values for 61.6. 2 [5.2] B 54.1, a 32.6, c 55.7 2 15. Find the supplement of 1 2 x 5 . [5.4] 6 6 2 1 1 2 2 2 1 1 2 x 2 2 x 2 x y d) 2 [5.3] Answers may vary; 472, 248 y 1 y c) [5.6] (c) 2 527 Test 2 13. Solve the right triangle with b 45.1 and A 35.9. Standard lettering has been used. 14. Find a positive angle and a negative angle coterminal with a 112 angle. • 1 1 2 2 16. Given that sin 441 and that the terminal side is in quadrant IV, find the other five trigonometric function values. 24. Height of a Kite. The angle of elevation of a kite is 65 with 490 ft of string out. Assuming the string is taut, how high is the kite? [5.2] About 444 ft 17. Convert 210 to radian measure in terms of . [5.4] 76 3 18. Convert to degree measure. [5.4] 135 4 25. Location. A pickup-truck camper travels at 50 mph for 6 hr in a direction of 115 from Buffalo, Wyoming. At the end of that time, how far east of Buffalo is the camper? [5.2] About 272 mi 19. Find the length of an arc of a circle given a central angle of 3 and a radius of 16 cm. [5.4] 163 16.755 cm 26. Linear Speed. A ferris wheel has a radius of 6 m and revolves at 1.5 rpm. What is the linear speed, in meters per minute? [5.4] 18 56.55 mmin Consider the function y sin x 2 1 for Exercises 20–23. 20. Find the amplitude. [5.5] 1 21. Find the period. [5.5] 2 Synthesis 27. Determine the domain of fx 22. Find the phase shift. [5.5] 2 Answers to Exercises 12, 16, and 27 can be found on p. IA-37. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 3 cos x . BBEPMC06_0321279115.QXP 1/10/05 1:29 PM Page 529 Trigonometric Identities, Inverse Functions, and Equations 6.1 6.2 6.3 6.4 6.5 Identities: Pythagorean and Sum and Difference Identities: Cofunction, Double-Angle, and Half-Angle Proving Trigonometric Identities Inverses of the Trigonometric Functions Solving Trigonometric Equations 6 SUMMARY AND REVIEW TEST A P P L I C A T I O N T he number of daylight hours in Fairbanks, Alaska, varies from about 3.9 hr to 20.6 hr (Source: Astronomical Applications Department, U.S. Naval Observatory, Washington, DC). The function Hd 8.3578 sin 0.0166d 1.2711 12.2153 can be used to approximate the number of daylight hours H on a certain day d in Fairbanks. We can use this function to determine on which day of the year there will be about 10.5 hr of daylight. This problem appears as Exercise 52 in Exercise Set 6.5. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 530 Chapter 6 1/10/05 1:30 PM Page 530 • Trigonometric Identities, Inverse Functions, and Equations 2.1 6.1 Polynomial Identities: Functions Pythagorean and andModeling Sum and Difference State the Pythagorean identities. Simplify and manipulate expressions containing trigonometric expressions. Use the sum and difference identities to find function values. An identity is an equation that is true for all possible replacements of the variables. The following is a list of the identities studied in Chapter 5. Basic Identities 1 1 , , csc x sin x csc x sin x 1 , sec x 1 tan x , cot x cos x 1 , cos x 1 cot x , tan x sec x sin x sin x , cos x cos x , tan x tan x , sin x , cos x cos x cot x sin x tan x In this section, we will develop some other important identities. Pythagorean Identities We now consider three other identities that are fundamental to a study of trigonometry. They are called the Pythagorean identities. Recall that the equation of a unit circle in the xy-plane is y (x, y), or (cos s, sin s) s x 2 y 2 1. (1, 0) x x2 y2 1 For any point on the unit circle, the coordinates x and y satisfy this equation. Suppose that a real number s determines a point on the unit circle with coordinates x, y, or cos s, sin s. Then x cos s and y sin s . Substituting cos s for x and sin s for y in the equation of the unit circle gives us the identity cos s2 sin s2 1, Substituting cos s for x and sin s for y which can be expressed as sin2 s cos2 s 1. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 531 Section 6.1 531 • Identities: Pythagorean and Sum and Difference It is conventional in trigonometry to use the notation sin2 s rather than sin s2. Note that sin2 s sin s 2. y 2 y y (sin x)2 (sin x)(sin x) 1 2 y sin x 2 sin (x ⋅ x) 2 1 2 x 2 1 2 x 1 The identity sin2 s cos2 s 1 gives a relationship between the sine and the cosine of any real number s. It is an important Pythagorean identity. EXPLORING WITH TECHNOLOGY Addition of y-values provides a unique way of developing the identity sin2 x cos2 x 1. First, graph y1 sin2 x and y2 cos2 x. By visually adding the y-values, sketch the graph of the sum, y3 sin2 x cos2 x. Then graph y3 using a graphing calculator and check your sketch. The resulting graph appears to be the line y4 1, which is the graph of sin2 x cos2 x. These graphs do not prove the identity, but they do provide a check in the interval shown. y1 sin2 x y2 cos2 x 2 2 2π 2π 2π 2π 2 2 y3 sin2 x cos2 x, y4 1 2 2π 2π 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 532 Chapter 6 1/10/05 1:30 PM Page 532 • Trigonometric Identities, Inverse Functions, and Equations We can divide by sin2 s on both sides of the preceding identity: sin2 s cos2 s 1 2 . sin2 s sin2 s sin s Dividing by sin2 s Simplifying gives us a second identity: 1 cot2 s csc2 s . This equation is true for any replacement of s with a real number for which sin2 s 0, since we divided by sin2 s . But the numbers for which sin2 s 0 (or sin s 0) are exactly the ones for which the cotangent and cosecant functions are not defined. Hence our new equation holds for all real numbers s for which cot s and csc s are defined and is thus an identity. The third Pythagorean identity can be obtained by dividing by cos2 s on both sides of the first Pythagorean identity: sin2 s cos2 s 1 2 2 cos s cos s cos2 s tan2 s 1 sec2 s . Dividing by cos2 s Simplifying The identities we have developed hold no matter what symbols are used for the variables. For example, we could write sin2 s cos2 s 1, sin2 cos2 1, or sin2 x cos2 x 1. Pythagorean Identities sin2 x cos2 x 1, 1 cot 2 x csc2 x , 1 tan2 x sec2 x It is often helpful to express the Pythagorean identities in equivalent forms. PYTHAGOREAN IDENTITIES sin2 x cos2 x 1 1 cot 2 x csc2 x 1 tan2 x sec2 x EQUIVALENT FORMS sin2 x 1 cos2 x cos2 x 1 sin2 x 1 csc2 x cot 2 x cot 2 x csc2 x 1 1 sec2 x tan2 x tan2 x sec2 x 1 Simplifying Trigonometric Expressions We can factor, simplify, and manipulate trigonometric expressions in the same way that we manipulate strictly algebraic expressions. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 533 Section 6.1 Study Tip EXAMPLE 1 The examples in each section were chosen to prepare you for success with the exercise set. Study the step-by-step annotated solutions of the examples, noting that substitutions are highlighted in red. The time you spend understanding the examples will save you valuable time when you do your assignment. Solution • Identities: Pythagorean and Sum and Difference Multiply and simplify: cos x tan x sec x. cos x tan x sec x cos x tan x cos x sec x 1 sin x cos x cos x cos x cos x Multiplying Recalling the identities tan x and sec x sin x 1 533 sin x cos x 1 and substituting cos x Simplifying There is no general procedure for manipulating trigonometric expressions, but it is often helpful to write everything in terms of sines and cosines, as we did in Example 1. We also look for the Pythagorean identity, sin2 x cos2 x 1, within a trigonometric expression. EXAMPLE 2 Factor and simplify: sin2 x cos2 x cos4 x . Solution sin2 x cos2 x cos4 x cos2 x sin2 x cos2 x cos2 x 1 cos2 x GCM y1 cos x [tan x (1/cos x)], y2 sin x 1 4 Removing a common factor Using sin2 x cos2 x 1 A graphing calculator can be used to perform a partial check of an identity. First, we graph the expression on the left side of the equals sign. Then we graph the expression on the right side using the same screen. If the two graphs are indistinguishable, then we have a partial verification that the equation is an identity. Of course, we can never see the entire graph, so there can always be some doubt. Also, the graphs may not overlap precisely, but you may not be able to tell because the difference between the graphs may be less than the width of a pixel. However, if the graphs are obviously different, we know that a mistake has been made. Consider the identity in Example 1: cos x tan x sec x sin x 1. Recalling that sec x 1cos x , we enter 2 π 2π 4 X Y1 6.283 1 5.498 .2929 4.712 ERROR 3.927 .2929 3.142 1 2.356 1.707 1.571 ERROR X 6.28318530718 TblStart 2 π Tbl π/4 Y2 1 .2929 0 .2929 1 1.707 2 y1 cos x tan x 1cos x and y2 sin x 1. To graph, we first select SEQUENTIAL mode. Then we select the “line”-graph style for y1 and the “path”-graph style, denoted by , for y2. The calculator will graph y1 first. Then as it graphs y2, the circular cursor will trace the leading edge of the graph, allowing us to determine whether the graphs coincide. As you can see in the first screen on the left, the graphs appear to be identical. Thus, cos x tan x sec x sin x 1 is most likely an identity. The TABLE feature can also be used to check identities. Note in the table at left that the function values are the same except for those values of x for which cos x 0. The domain of y1 excludes these values. The domain of y2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 534 Chapter 6 1/10/05 1:30 PM Page 534 • Trigonometric Identities, Inverse Functions, and Equations is the set of all real numbers. Thus all real numbers except 2, 32, 52, . . . are possible replacements for x in the identity. Recall that an identity is an equation that is true for all possible replacements. Suppose that we had simplified incorrectly in Example 1 and had gotten cos x 1 on the right instead of sin x 1. Then two different graphs would have appeared in the window. Thus we would have known that we did not have an identity and that cos x tan x sec x cos x 1. y1 cos x (tan x sec x), y2 cos x 1 4 2π 2π 4 EXAMPLE 3 Simplify each of the following trigonometric expressions. cot csc 2 sin2 t sin t 3 b) 1 cos2 t sin t a) Solution cos cot sin a) csc 1 sin cos sin sin cos cos 2 sin t sin t 3 1 cos2 t sin t 2 sin2 t sin t 3 sin2 t sin t Rewriting in terms of sines and cosines Multiplying by the reciprocal The cosine function is even. 2 b) factoring review section R.4. Substituting sin2 t for 1 cos2 t 2 sin t 3 sin t 1 Factoring in both numerator and denominator sin t sin t 1 2 sin t 3 Simplifying sin t 2 sin t 3 sin t sin t 3 , or 2 3 csc t 2 sin t Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 535 Section 6.1 • Identities: Pythagorean and Sum and Difference 535 We can add and subtract trigonometric rational expressions in the same way that we do algebraic expressions. EXAMPLE 4 Add and simplify: cos x tan x . 1 sin x Solution rational expressions review section R.5. cos x sin x cos x sin x Using tan x tan x cos x 1 sin x 1 sin x cos x cos x sin x 1 sin x cos x 1 sin x cos x cos x 1 sin x cos2 x sin x sin2 x cos x 1 sin x 1 sin x cos x 1 sin x 1 , or sec x cos x Multiplying by forms of 1 Adding Using sin2 x cos2 x 1 Simplifying When radicals occur, the use of absolute value is sometimes necessary, but it can be difficult to determine when to use it. In Examples 5 and 6, we will assume that all radicands are nonnegative. This means that the identities are meant to be confined to certain quadrants. EXAMPLE 5 Multiply and simplify: sin3 x cos x cos x . Solution sin3 x cos x cos x sin3 x cos2 x sin2 x cos2 x sin x sin x cos x sin x EXAMPLE 6 Rationalize the denominator: Solution 2 . tan x tan x 2 tan x tan x 2 tan x tan2 x 2 tan x tan x 2 tan x Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 536 Chapter 6 1/10/05 1:30 PM Page 536 • Trigonometric Identities, Inverse Functions, and Equations Often in calculus, a substitution is a useful manipulation, as we show in the following example. EXAMPLE 7 Express 9 x 2 as a trigonometric function of without using radicals by letting x 3 tan . Assume that 0 2. Then find sin and cos . Solution We have 9 x 2 9 3 tan 2 9 9 tan2 91 tan2 9 sec2 3sec 3 sec . Substituting 3 tan for x Factoring Using 1 tan2 x sec2 x For 0 2, sec 0, so sec sec . We can express 9 x 2 3 sec as sec 9 x 2 x u 3 9 x 2 . 3 In a right triangle, we know that sec is hypotenuseadjacent, when is one of the acute angles. Using the Pythagorean theorem, we can determine that the side opposite is x. Then from the right triangle, we see that sin x 3 and cos . 2 9 x 9 x 2 Sum and Difference Identities We now develop some important identities involving sums or differences of two numbers (or angles), beginning with an identity for the cosine of the difference of two numbers. We use the letters u and v for these numbers. Let’s consider a real number u in the interval 2, and a real number v in the interval 0, 2. These determine points A and B on the unit circle, as shown below. The arc length s is u v , and we know that 0 s . Recall that the coordinates of A are cos u, sin u, and the coordinates of B are cos v, sin v. y (cos v, sin v) u s v B (cos u, sin u) A (1, 0) x Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 537 Section 6.1 • Identities: Pythagorean and Sum and Difference 537 Using the distance formula, we can write an expression for the distance AB : distance formula review section 1.1. AB cos u cos v2 sin u sin v2. This can be simplified as follows: AB cos2 u 2 cos u cos v cos2 v sin2 u 2 sin u sin v sin2 v sin2 u cos2 u sin2 v cos2 v 2cos u cos v sin u sin v 2 2cos u cos v sin u sin v. Now let’s imagine rotating the circle on page 536 so that point B is at 1, 0 as shown at left. Although the coordinates of point A are now cos s, sin s, the distance AB has not changed. Again we use the distance formula to write an expression for the distance AB: y (cos s, sin s) A s B (1, 0) AB cos s 12 sin s 02. x This can be simplified as follows: AB cos2 s 2 cos s 1 sin2 s sin2 s cos2 s 1 2 cos s 2 2 cos s . Equating our two expressions for AB, we obtain 2 2cos u cos v sin u sin v 2 2 cos s . Solving this equation for cos s gives cos s cos u cos v sin u sin v . (1) But s u v , so we have the equation cos u v cos u cos v sin u sin v . (2) Formula (1) above holds when s is the length of the shortest arc from A to B. Given any real numbers u and v, the length of the shortest arc from A to B is not always u v . In fact, it could be v u. However, since cos x cos x , we know that cos v u cos u v. Thus, cos s is always equal to cos u v. Formula (2) holds for all real numbers u and v. That formula is thus the identity we sought: y1 cos (x 3), y2 cos x cos 3 sin x sin 3 2 cos u v cos u cos v sin u sin v . Using a graphing calculator, we can graph 2π 2π y1 cos x 3 and 2 y2 cos x cos 3 sin x sin 3 to illustrate this result. The cosine sum formula follows easily from the one we have just derived. Let’s consider cos u v. This is equal to cos u v, and by the identity above, we have cos u v cos u v cos u cos v sin u sin v. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 538 Chapter 6 1/10/05 1:30 PM Page 538 • Trigonometric Identities, Inverse Functions, and Equations But cos v cos v and sin v sin v , so the identity we seek is the following: y1 cos (x 2), y2 cos x cos 2 sin x sin 2 2 cos u v cos u cos v sin u sin v . Using a graphing calculator, we can graph 2π 2π y1 cos x 2 and 2 y2 cos x cos 2 sin x sin 2 to illustrate this result. EXAMPLE 8 Find cos 512 exactly. Solution We can express 512 as a difference of two numbers whose sine and cosine values are known: 5 9 4 , 12 12 12 3 . 4 3 or Then, using cos u v cos u cos v sin u sin v , we have cos 5 3 cos 12 4 3 cos(5π/12) .2588190451 3 3 cos sin sin 4 3 4 3 2 1 2 3 2 2 2 2 2 6 4 4 6 2 . 4 cos ( (6) (2))/4 .2588190451 We can check using a graphing calculator set in RADIAN mode. Consider cos 2 . We can use the identity for the cosine of a difference to simplify as follows: cos cos cos sin sin 2 2 2 0 cos 1 sin sin . Thus we have developed the identity sin cos . 2 This cofunction identity first appeared in Section 5.1. (3) This identity holds for any real number . From it we can obtain an identity for the cosine function. We first let be any real number. Then we replace in sin cos 2 with 2 . This gives us sin cos 2 2 2 cos , Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 539 Section 6.1 • Identities: Pythagorean and Sum and Difference which yields the identity cos sin . 2 539 (4) Using identities (3) and (4) and the identity for the cosine of a difference, we can obtain an identity for the sine of a sum. We start with identity (3) and substitute u v for : sin cos sin u v cos cos cos 2 Identity (3) u v 2 Substituting u v for u v 2 u cos v sin 2 sin u cos v cos u sin v . u sin v 2 Using the identity for the cosine of a difference Using identities (3) and (4) Thus the identity we seek is sin u v sin u cos v cos u sin v . To find a formula for the sine of a difference, we can use the identity just derived, substituting v for v : sin u v sin u cos v cos u sin v. Simplifying gives us sin u v sin u cos v cos u sin v . EXAMPLE 9 Solution Find sin 105 exactly. We express 105 as the sum of two measures: 105 45 60. Then sin 105 sin 45 60 sin 45 cos 60 cos 45 sin 60 Using sin u v sin u cos v cos u sin v 2 1 2 3 2 2 2 2 2 6 . 4 sin(105) .9659258263 ((2)(6))/4 .9659258263 We can check this result using a graphing calculator set in DEGREE mode. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 540 Chapter 6 1/10/05 1:30 PM Page 540 • Trigonometric Identities, Inverse Functions, and Equations Formulas for the tangent of a sum or a difference can be derived using identities already established. A summary of the sum and difference identities follows. Sum and Difference Identities sin u v sin u cos v cos u sin v , cos u v cos u cos v sin u sin v , tan u tan v tan u v 1 tan u tan v There are six identities here, half of them obtained by using the signs shown in color. Find tan 15 exactly. EXAMPLE 10 Solution We rewrite 15 as 45 30 and use the identity for the tangent of a difference: tan 45 tan 30 1 tan 45 tan 30 1 33 3 3 . 1 1 33 3 3 tan 15 tan 45 30 2 1 EXAMPLE 11 Assume that sin 3 and sin 3 and that and are between 0 and 2. Then evaluate sin . Solution Using the identity for the sine of a sum, we have sin sin cos cos sin 2 1 3 cos 3 cos . To finish, we need to know the values of cos and cos . Using reference triangles and the Pythagorean theorem, we can determine these values from the diagrams: y 3 2 a x 5 cos 5 and 3 cos 22 . 3 Cosine values are positive in the first quadrant. Substituting these values gives us y 2 22 1 5 3 3 3 3 1 42 5 4 2 5, or . 9 9 9 sin 3 b 22 1 x Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 541 Section 6.1 6.1 • Identities: Pythagorean and Sum and Difference Exercise Set Multiply and simplify. Check your result using a graphing calculator. 1. sin x cos x sin x cos x sin2 x cos2 x 2. tan x cos x csc x sin x sec x 3. cos y sin y sec y csc y sin y cos y 4. sin x cos x sec x csc x 2 tan x cot x 5. sin cos 2 1 2 sin cos 23. sin4 x cos4 x sin2 x cos2 x 24. 4 cos3 x sin x sin2 x 4 cos x 25. 5 cos sin2 sin cos 5 cot sin2 sin2 cos2 sin cos 26. tan2 y 3 tan3 y sec y sec y 27. 1 2 sin2 s cos2 s cos s sin s 6. 1 tan x2 sec2 x 2 tan x 7. sin x csc x sin2 x csc2 x 1 sin3 x csc3 x Factor and simplify. Check your result using a graphing calculator. 9. sin x cos x cos2 x cos x sin x cos x 10. tan2 cot 2 tan cot tan cot 11. sin x cos x sin x cos x sin x cos x 4 4 12. 4 sin2 y 8 sin y 4 13. 2 cos x cos x 3 2 4sin y 12 2 cos x 3 cos x 1 14. 3 cot 2 6 cot 3 15. sin3 x 27 3cot 12 sin x 3 sin2 x 3 sin x 9 16. 1 125 tan3 s 1 5 tan s 1 5 tan s 25 tan2 s Simplify and check using a graphing calculator. sin2 x cos x 17. tan x cos2 x sin x 3 30 sin x cos x 6 cos2 x sin x 2 5 sin x , or 5 tan x sin x cos x sin2 x 2 sin x 1 19. sin x 1 sin x 1 cos 1 20. cos 1 2 21. 4 tan t sec t 2 sec t 6 tan t sec t 2 sec t 22. csc x sec x cot x 2 tan t 1 3 tan t 1 2 cos x 4 1 cot y 3 1 2 sin s 2 cos s sin2 s cos2 s 29. sin2 9 10 cos 5 5sin 3 2 cos 1 3 sin 9 3 30. 9 cos2 25 cos2 1 2 cos 2 6 cos 10 sin x cos x 2 1 cos2 x 1 3 cos 5 cos 1 4 Simplify and check using a graphing calculator. Assume that all radicands are nonnegative. 31. sin2 x cos x cos x sin x cos x 32. cos2 x sin x sin x cos x sin x 33. cos sin2 cos3 cos sin cos 34. tan2 x 2 tan x sin x sin2 x tan x sin x 35. 1 sin y sin y 1 1 sin y 36. cos 2 cos sin cos cos 2 sin Rationalize the denominator. sin x sin x cos x 37. 38. cos x cos x 39. cos 1 1 28. 8. 1 sin t 1 sin t cos2 t 18. 541 cos2 y 2 sin2 y 2 cot y 2 Rationalize the numerator. cos x cos x 41. sin x sin x cos x 43. 1 sin y 1 sin y 40. 42. 1 sin y 44. cos y Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley cos x tan x sin x tan x 1 cos sin 1 cos 1 cos sin x cot x cos2 x 2 sin2 x sin x cos x cot x 2 BBEPMC06_0321279115.QXP 542 Chapter 6 1/10/05 1:30 PM Page 542 • Trigonometric Identities, Inverse Functions, and Equations Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that a 0 and 0 2. Then find expressions for the indicated trigonometric functions. 45. Let x a sin in a2 x 2. Then find cos and tan . 46. Let x 2 tan in 4 x 2. Then find sin and cos . 47. Let x 3 sec in x 2 9. Then find sin and cos . 48. Let x a sec in x 2 a2. Then find sin and cos . Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that 0 2. x2 49. Let x sin in . sin tan 1 x 2 50. Let x 4 sec in x 2 16 sin cos . x2 4 Use the sum and difference identities to evaluate exactly. Then check using a graphing calculator. 6 2 6 2 51. sin 52. cos 75 4 4 12 53. tan 105 55. cos 15 6 2 4 54. tan 5 12 56. sin 7 12 6 2 4 First write each of the following as a trigonometric function of a single angle; then evaluate. 57. sin 37 cos 22 cos 37 sin 22 sin 59 0.8572 58. cos 83 cos 53 sin 83 sin 53 cos 30 0.8660 59. cos 19 cos 5 sin 19 sin 5 cos 24 0.9135 60. sin 40 cos 15 cos 40 sin 15 sin 25 0.4226 61. tan 20 tan 32 tan 52 1.2799 1 tan 20 tan 32 62. tan 35 tan 12 tan 23 0.4245 1 tan 35 tan 12 63. Derive the formula for the tangent of a sum. 64. Derive the formula for the tangent of a difference. Assuming that sin u 35 and sin v 45 and that u and v are between 0 and 2, evaluate each of the following exactly. 65. cos u v 0 66. tan u v 247 7 67. sin u v 25 68. cos u v 24 25 Assuming that sin 0.6249 and cos 0.1102 and that both and are first-quadrant angles, evaluate each of the following. 69. tan 1.5789 70. sin 0.7071 71. cos 0.7071 72. cos 0.5351 Simplify. 73. sin sin 2 sin cos 74. cos cos 2 sin sin 75. cos u v cos v sin u v sin v cos u 76. sin u v cos v cos u v sin v sin u Collaborative Discussion and Writing 77. What is the difference between a trigonometric equation that is an identity and a trigonometric equation that is not an identity? Give an example of each. 78. Why is it possible to use a graph to disprove that an equation is an identity but not to prove that one is? Skill Maintenance Solve. 3 79. 2x 3 2 x 2 [2.1] All real numbers 80. x 7 x 3.4 [2.1] No solution Given that sin 31 0.5150 and cos 31 0.8572, find the specified function value. 81. sec 59 [5.1] 1.9417 82. tan 59 [5.1] 1.6645 Synthesis One of the identities gives an easy way to find an angle formed by two lines. Consider two lines with equations l1: y m1x b1 and l2: y m2x b2. Angles Between Lines. Answers to Exercises 45–48, 53, 54, 63, and 64 can be found on p. IA-37. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 543 Section 6.1 y 88. Given that fx sin x , show that l2 u2 u 1 or f l1 u2 The slopes m1 and m2 are the tangents of the angles 1 and 2 that the lines form with the positive direction of the x-axis. Thus we have m1 tan 1 and m2 tan 2. To find the measure of 2 1, or , we proceed as follows: tan tan 2 1 tan 2 tan 1 1 tan 2 tan 1 m2 m1 . 1 m2m1 6, or 30 86. l1: 2x y 4 0, 3 , or 135° l2: y 2x 5 0 0; the lines are parallel 126.87 4 87. Circus Guy Wire. In a circus, a guy wire A is attached to the top of a 30-ft pole. Wire B is used for performers to walk up to the tight wire, 10 ft above the ground. Find the angle between the wires if they are attached to the ground 40 ft from the pole. 22.83 92. sin x sin x φ B 10 ft 94. tan cot 1 2 2 93. cos 2 2 cos 95. cos 6x 6 cos x Find the slope of line l1, where m2 is the slope of line l2 and is the smallest positive angle from l1 to l2. 96. m2 43 , 45 17 6 33 0.0645 9 23 98. Line l1 contains the points 2, 4 and 5, 1. Find the slope of line l2 such that the angle from l1 to l2 is 45. 16 99. Line l1 contains the points 3, 7 and 3, 2. Line l2 contains 0, 4 and 2, 6. Find the smallest positive angle from l1 to l2. 168.7 100. Find an identity for sin 2. (Hint: 2 .) sin 2 2 sin cos 101. Find an identity for cos 2. (Hint: 2 .) Derive the identity. Check using a graphing calculator. 3 102. sin x cos x 2 103. tan x 104. A Show that each of the following is not an identity by finding a replacement or replacements for which the sides of the equation do not name the same number. Then use a graphing calculator to show that the equation is not an identity. sin 5x 90. sin2 sin 91. sin 5 x 97. m2 23 , 30 This formula also holds when the lines are taken in the reverse order. When is acute, tan will be positive. When is obtuse, tan will be negative. Find the measure of the angle from l1 to l2. 84. l1: 3y 3x 3, 83. l1: 2x 3 2y , l2: y 3x 2 l2: x y 5 85. l1: y 3, l2: x y 5 fx h fx cos h 1 sin h . cos x sin x h h h x u1 u1 fx h fx cos h 1 sin h . sin x cos x h h h 89. Given that fx cos x , show that u2 543 • Identities: Pythagorean and Sum and Difference 4 1 tan x 1 tan x sin tan tan cos 1 tan tan 105. sin sin 2 sin cos sin sin sin cos cos sin sin cos cos sin 2 sin cos 40 ft Answers to Exercises 88–95 and 101–104 can be found on pp. IA-37 and IA-38. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 544 Chapter 6 1/10/05 1:30 PM Page 544 • Trigonometric Identities, Inverse Functions, and Equations 6.2 Identities: Cofunction, Double-Angle, and Half-Angle Use cofunction identities to derive other identities. Use the double-angle identities to find function values of twice an angle when one function value is known for that angle. Use the half-angle identities to find function values of half an angle when one function value is known for that angle. Simplify trigonometric expressions using the double-angle and half-angle identities. Cofunction Identities Each of the identities listed below yields a conversion to a cofunction. For this reason, we call them cofunction identities. Cofunction Identities x cos x , sin 2 tan sec cos x cot x , 2 cot x csc x , 2 csc x sin x , 2 x tan x , 2 x sec x 2 We verified the first two of these identities in Section 6.1. The other four can be proved using the first two and the definitions of the trigonometric functions. These identities hold for all real numbers, and thus, for all angle measures, but if we restrict to values such that 0 90, or 0 2, then we have a special application to the acute angles of a right triangle. Comparing graphs can lead to possible identities. On the left below, we see that the graph of y sin x 2 is a translation of the graph of y sin x to the left 2 units. On the right, we see the graph of y cos x . y sin (x 2) 2 y y y sin x y cos x 2 2 1 1 2 x 2 1 1 2 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2 x BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 545 Section 6.2 • Identities: Cofunction, Double-Angle, and Half-Angle 545 Comparing the graphs, we observe a possible identity: sin x 2 cos x . The identity can be proved using the identity for the sine of a sum developed in Section 6.1. Prove the identity sin x 2 cos x . EXAMPLE 1 Solution sin x 2 sin x cos cos x sin 2 2 Using sin u v sin u cos v cos u sin v sin x 0 cos x 1 cos x We now state four more cofunction identities. These new identities that involve the sine and cosine functions can be verified using previously established identities as seen in Example 1. Cofunction Identities for the Sine and Cosine sin x cos x , cos x sin x 2 2 Find an identity for each of the following. EXAMPLE 2 a) tan x Solution a) We have 2 b) sec x 90 tan x 2 sin x cos x cos x sin x cot x . 2 2 Using tan x Using cofunction identities Thus the identity we seek is tan x 2 sin x cos x cot x . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 546 Chapter 6 1/10/05 1:30 PM Page 546 • Trigonometric Identities, Inverse Functions, and Equations b) We have sec x 90 1 1 csc x . cos x 90 sin x Thus, sec x 90 csc x . Double-Angle Identities If we double an angle of measure x, the new angle will have measure 2x. Double-angle identities give trigonometric function values of 2x in terms of function values of x. To develop these identities, we will use the sum formulas from the preceding section. We first develop a formula for sin 2x . Recall that sin u v sin u cos v cos u sin v . We will consider a number x and substitute it for both u and v in this identity. Doing so gives us sin x x sin 2x sin x cos x cos x sin x 2 sin x cos x . Our first double-angle identity is thus sin 2x 2 sin x cos x . We can graph y1 sin 2x , and y2 2 sin x cos x using the “line”-graph style for y1 and the “path”-graph style for y2 and see that they appear to have the same graph. We can also use the TABLE feature. y1 sin 2x, y2 2 sin x cos x 2 2π 2π 2 X 6.283 5.498 4.712 3.927 3.142 2.356 1.571 Y1 2E13 1 0 1 0 1 0 Y2 0 1 0 1 0 1 0 X 1.57079632679 Double-angle identities for the cosine and tangent functions can be derived in much the same way as the identity above: cos 2x cos2 x sin2 x , tan 2x 2 tan x . 1 tan2 x EXAMPLE 3 Given that tan 34 and is in quadrant II, find each of the following. a) sin 2 c) tan 2 b) cos 2 d) The quadrant in which 2 lies Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 547 Section 6.2 Solution 547 • Identities: Cofunction, Double-Angle, and Half-Angle By drawing a reference triangle as shown, we find that sin y 3 5 (4, 3) and 5 u 3 cos 4 . 5 4 Thus we have the following. a) sin 2 2 sin cos 2 3 4 5 5 2 tan 1 tan 2 2 4 5 b) cos 2 cos2 sin2 c) tan 2 2 4 3 x 1 2 3 5 24 25 9 7 16 25 25 25 3 3 4 2 2 1 9 16 3 2 16 7 24 7 Note that tan 2 could have been found more easily in this case by simply dividing: tan 2 sin 2 cos 2 24 25 7 25 24 7 . d) Since sin 2 is negative and cos 2 is positive, we know that 2 is in quadrant IV. Two other useful identities for cos 2x can be derived easily, as follows. cos 2x cos2 x sin2 x 1 sin2 x sin2 x 1 2 sin2 x Double-Angle Identities sin 2x 2 sin x cos x , 2 tan x tan 2x 1 tan2 x cos 2x cos2 x sin2 x cos2 x 1 cos2 x 2 cos2 x 1 cos 2x cos2 x sin2 x 1 2 sin2 x 2 cos2 x 1 Solving the last two cosine double-angle identities for sin2 x and cos2 x , respectively, we obtain two more identities: sin2 x 1 cos 2x 1 cos 2x and cos2 x . 2 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 548 Chapter 6 1/10/05 1:30 PM Page 548 • Trigonometric Identities, Inverse Functions, and Equations Using division and these two identities gives us the following useful identity: tan2 x EXAMPLE 4 1 cos 2x . 1 cos 2x Find an equivalent expression for each of the following. a) sin 3 in terms of function values of b) cos3 x in terms of function values of x or 2x, raised only to the first power Solution a) sin 3 sin 2 sin 2 cos cos 2 sin 2 sin cos cos 2 cos2 1 sin Using sin 2 2 sin cos and cos 2 2 cos2 1 2 sin cos2 2 sin cos2 sin 4 sin cos2 sin We could also substitute cos2 sin2 or 1 2 sin2 for cos 2. Each substitution leads to a different result, but all results are equivalent. b) cos3 x cos2 x cos x 1 cos 2x cos x 2 cos x cos x cos 2x 2 Half-Angle Identities If we take half of an angle of measure x, the new angle will have measure x2. Half-angle identities give trigonometric function values of x2 in terms of function values of x. To develop these identities, we replace x with x2 and take square roots. For example, sin2 x 1 cos 2x 2 1 cos 2 x 2 2 x 1 cos x sin2 2 2 sin2 sin x 2 Solving the identity cos 2x 1 2 sin2 x for sin2 x x 2 1 cos x . 2 Substituting x for x 2 Taking square roots The formula is called a half-angle formula. The use of and depends on the quadrant in which the angle x2 lies. Half-angle identities for the Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 549 Section 6.2 • Identities: Cofunction, Double-Angle, and Half-Angle 549 cosine and tangent functions can be derived in a similar manner. Two additional formulas for the half-angle tangent identity are listed below. Half-Angle Identities sin x 2 1 cos x , 2 cos x 2 1 cos x , 2 tan x 2 1 cos x 1 cos x 1 cos x sin x 1 cos x sin x EXAMPLE 5 Find tan 8 exactly. Then check the answer using a graphing calculator in RADIAN mode. Solution 4 tan tan 8 2 4 2 2 2 2 2 1 cos 1 4 2 2 2 2 2 2 2 2 2 2 2 2 2 1 tan(π/8) .4142135624 ((2)1 .4142135624 sin 2 2 The identities that we have developed are also useful for simplifying trigonometric expressions. EXAMPLE 6 a) sin x cos x 1 2 cos 2x Simplify each of the following. b) 2 sin2 x cos x 2 Solution a) We can obtain 2 sin x cos x in the numerator by multiplying the ex2 pression by 2 : sin x cos x 2 sin x cos x 2 sin x cos x 1 1 2 cos 2x 2 cos 2x 2 cos 2x sin 2x Using sin 2x 2 sin x cos x cos 2x tan 2x . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 550 Chapter 6 1/10/05 1:30 PM Page 550 • Trigonometric Identities, Inverse Functions, and Equations b) We have 2 sin2 x 1 cos x cos x 2 2 2 Using sin cos x x 2 1 cos x 1 cos x x , or sin2 2 2 2 1 cos x cos x 1. We can check this result using a graph or a table. y1 2 sin2 x cos x, y2 1 2 2 2π 2π 2 Y1 X 6.283 1 5.498 1 4.712 1 3.927 1 3.142 1 2.356 1 1.571 1 X 6.28318530718 Y2 1 1 1 1 1 1 1 Tbl π/4 Answers to Exercises 1–8 can be found on p. IA-38. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 550 Chapter 6 6.2 1/10/05 1:30 PM Page 550 • Trigonometric Identities, Inverse Functions, and Equations Exercise Set 1. Given that sin 310 0.8090 and cos 310 0.5878, find each of the following. a) The other four function values for 310 b) The six function values for 5 2. Given that sin 2 3 12 2 and cos 2 3 , 12 2 find exact answers for each of the following. a) The other four function values for 12 b) The six function values for 512 1 3 3. Given that sin and that the terminal side is in quadrant II, find exact answers for each of the following. a) The other function values for b) The six function values for 2 c) The six function values for 2 4 4. Given that cos 5 and that the terminal side is in quadrant IV, find exact answers for each of the following. a) The other function values for b) The six function values for 2 c) The six function values for 2 Find an equivalent expression for each of the following. 5. sec x 6. cot x 2 2 7. tan x 2 8. csc x 2 Find the exact value of sin 2, cos 2, tan 2, and the quadrant in which 2 lies. 4 9. sin , in quadrant I 5 24 7 24 sin 2 25, cos 2 25, tan 2 7 ; II Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley Answers to Exercises 1–8 can be found on p. IA-38. BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 551 Section 6.2 • Identities: Cofunction, Double-Angle, and Half-Angle 5 , in quadrant I 13 sin 2 120 , cos 2 119 , tan 2 120 ; II 169 169 119 3 11. cos , in quadrant III 5 7 , cos 2 25 , tan 2 24 ; II sin 2 24 25 7 15 12. tan , in quadrant II 8 sin 2 240 , cos 2 161 , tan 2 240 ; III 289 289 161 5 13. tan , in quadrant II 12 , cos 2 119 , tan 2 120 ; IV sin 2 120 169 169 119 10 14. sin , in quadrant IV 10 sin 2 35 , cos 2 45 , tan 2 34 ; IV 15. Find an equivalent expression for cos 4x in terms of function values of x. c) cot x 10. cos 2 19. sin 112.5 2 2 8 2 5 2 3 22. sin 12 2 21. tan 75 2 3 Given that sin 0.3416 and is in quadrant I, find each of the following using identities. 23. sin 2 0.6421 24. cos 0.9848 2 25. sin 2 26. sin 4 0.9845 0.1735 y 1 cos x y cos x sin x 4 2π 4 2π 2π 4 2π 2π 2π 4 4 x 2 a) sin x csc x tan x c) 2cos2 x sin2 x b) sin x 2 cos x d) 1 cos x sin 2x 2 cos x a) cos x c) cos x sin x b) tan x d) sin x 28. 2 cos2 29. 30. 2 sin cos 2 2 2 d) sin cos a) cos2 b) sin c) sin Simplify. Check your results using a graphing calculator. x 1 cos x 31. 2 cos2 2 32. cos4 x sin4 x cos 2x 33. sin x cos x2 sin 2x 1 1 sin 2x 35. 2 sec2 x sec2 x 36. 1 sin 2x cos 2x 1 sin 2x cos 2x cos 2x cot x 37. 4 cos x sin x 2 cos 2x2 2 cos 2x 4 sin x cos x2 8 38. 2 sin x cos3 x 2 sin3 x cos x 4 2π 4 2π 34. sin x cos x2 In Exercises 27– 30, use a graphing calculator to determine which of the following expressions asserts an identity. Then derive the identity algebraically. cos 2x 27. cos x sin x a) 1 cos x b) cos x sin x y sin x (cot x 1) 4 2 2 20. cos 2 d) sin x cot x 1 y cot x 16. Find an equivalent expression for sin4 in terms of function values of , 2, or 4, raised only to the first power. Use the half-angle identities to evaluate exactly. 17. cos 15 2 3 18. tan 67.5 2 1 551 1 2 sin 4x Collaborative Discussion and Writing 39. Discuss and compare the graphs of y sin x , y sin 2x , and y sin x2. Answers to Exercises 15, 16, and 27–30 can be found on pp. IA-38 and IA-39. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 552 Chapter 6 1/10/05 1:30 PM Page 552 • Trigonometric Identities, Inverse Functions, and Equations 40. Find all errors in the following: 2 sin 2x cos 4x 22 sin x cos x2 2 cos 2x 8 sin2 x cos2 x 2cos2 x sin2 x 8 sin2 x cos2 x 2. 2 In Exercises 41–48, answer “True” or “False.” 41. 1 cos2 x sin2 x 42. sec2 x tan2 x 1 [6.1] True 44. 1 cot 2 x csc2 x 45. csc2 x cot 2 x 1 46. 1 tan2 x sec2 x 47. 1 sin2 x cos2 x 48. sec2 x 1 tan2 x [6.1] False [6.1] False [6.1] False [6.1] True [6.1] True [6.1] True Consider the following functions (a)– (f ). Without graphing them, answer questions 49– 52 below. 1 x 2 2 1 cos 2x 2 4 2 1 d) fx sin x 2 e) fx 2 cos 4x f) fx cos 2 x 8 c) fx sin 2 x 2 49. Which functions have a graph with an amplitude of 2? [5.5] (a), (e) 2 cos x1 cot x x sin x 2 1 sin x 56. cos x cos x tan x 1 tan x cos2 y sin y 57. sin2 y sin 2 y 2 cot 2 y Find sin , cos , and tan under the given conditions. 7 3 58. cos 2 , 2 2 12 2 59. tan 2 55. cos x cot x sin x [6.1] False 43. sin2 x 1 cos2 x b) fx cos x sin Skill Maintenance a) fx 2 sin Simplify. Check your results using a graphing calculator. 54. sin x sec x cos x sin2 x 2 5 3 , 2 3 2 60. Nautical Mile. Latitude is used to measure north–south location on the earth between the equator and the poles. For example, Chicago has latitude 42N. (See the figure.) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth’s radius. Since the earth is flattened slightly at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, by the function N 6066 31 cos 2, where is the latitude in degrees. 90 N 50. Which functions have a graph with a period of ? 42 N [5.5] (b), (c), (f) 51. Which functions have a graph with a period of 2 ? [5.5] (d) 52. Which functions have a graph with a phase shift of ? [5.5] (e) 4 Synthesis 53. Given that cos 51 0.6293, find the six function values for 141. Equator 0 90 S a) What is the length of a British nautical mile at Chicago? 6062.76 ft b) What is the length of a British nautical mile at the North Pole? 6097 ft Answers to Exercises 53, 58, and 59 can be found on p. IA-39. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 1:30 PM Page 553 Section 6.2 553 • Identities: Cofunction, Double-Angle, and Half-Angle c) Express N in terms of cos only. That is, do not use the double angle. N 6097 62 cos2 M 72° I WISCONSIN 42° AN 44° IG where g is measured in meters per second per second at sea level. a) Chicago has latitude 42N. Find g. 9.80359 msec2 b) Philadelphia has latitude 40N. Find g. 46° H g 9.780491 0.005288 sin2 0.000006 sin2 2, g 9.780491 0.005264 sin 0.000024 sin C 61. Acceleration Due to Gravity. The acceleration due to gravity is often denoted by g in a formula such as S 12 gt 2, where S is the distance that an object falls in time t. The number g relates to motion near the earth’s surface and is usually considered constant. In fact, however, g is not constant, but varies slightly with latitude. If stands for latitude, in degrees, g is given with good approximation by the formula c) Express g in terms of sin only. That is, eliminate the double angle. 2 4 N 1/10/05 A AD VT. CA BBEPMC06_0321279115.QXP NEW YORK MASS. CONN. Chicago PENNSYLVANIA Philadelphia 40° INDIANA ILLINOIS OHIO N.J. MD. DEL. W.VA. 38° VIRGINIA KENTUCKY 90° 88° 86° 84° 82° 80° 9.80180 msec2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 78° 76° 74° BBEPMC06_0321279115.QXP 572 Chapter 6 1/10/05 1:30 PM Page 572 • Trigonometric Identities, Inverse Functions, and Equations Solve trigonometric equations. 6.5 Solving Trigonometric Equations When an equation contains a trigonometric expression with a variable, such as cos x , it is called a trigonometric equation. Some trigonometric equations are identities, such as sin2 x cos2 x 1. Now we consider equations, such as 2 cos x 1, that are usually not identities. As we have done for other types of equations, we will solve such equations by finding all values for x that make the equation true. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley Answers to Exercises 81–86 can be found on p. IA-44. BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 573 Section 6.5 EXAMPLE 1 Solution • Solving Trigonometric Equations 573 Solve: 2 cos x 1. We first solve for cos x : 2 cos x 1 1 cos x . 2 1 The solutions are numbers that have a cosine of 2 . To find them, we use the unit circle (see Section 5.5). 1 There are just two points on the unit circle for which the cosine is 2 , as shown in the figure at left. They are the points corresponding to 23 and 43. These numbers, plus any multiple of 2, are the solutions: q , 3 2 o i (1, 0) 2 2k 3 and 4 2k, 3 where k is any integer. In degrees, the solutions are q , 3 2 120 k 360 and 240 k 360, where k is any integer. To check the solution to 2 cos x 1, we can graph y1 2 cos x and y2 1 on the same set of axes and find the first coordinates of the points of intersection. Using 3 as the Xscl facilitates our reading of the solutions. First, let’s graph these equations on the interval from 0 to 2, as shown in the figure on the left below. The only solutions in 0, 2 are 23 and 43. y1 2 cos x, y2 1 y1 2 cos x, y2 1 4 4 4π 3 2π 0 (23,π 1) 4 (43,π 1) π Xscl 3 2π 3 3π 3π 8π 3 2π 3 4 Yscl 1 4π 8π 3 3 π Xscl 3 Next, let’s change the viewing window to 3, 3, 4, 4 and graph again. Since the cosine function is periodic, there is an infinite number of solutions. A few of these appear in the graph on the right above. From the graph, we see that the solutions are 23 2k and 43 2k, where k is any integer. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 574 Chapter 6 1/10/05 1:30 PM Page 574 • Trigonometric Identities, Inverse Functions, and Equations EXAMPLE 2 Solution Solve: 4 sin2 x 1. We begin by solving for sin x : 4 sin2 x 1 1 sin2 x 4 sin x 1 . 2 Again, we use the unit circle to find those numbers having a sine of 21 or 12 . The solutions are S 23 , q 23 , q F A 23 , q (1, 0) G 23 , q 11 5 7 2k, 2k, 2k, and 2k, 6 6 6 6 where k is any integer. In degrees, the solutions are 30 k 360, 150 k 360, 210 k 360, and 330 k 360, where k is any integer. The general solutions listed above could be condensed using odd as well as even multiples of : 5 k and k, 6 6 or, in degrees, 30 k 180 and 150 k 180, where k is any integer. Let’s do a partial check using a graphing calculator, checking only the solutions in 0, 2 . We graph y1 4 sin2 x and y2 1 and note that the solutions in 0, 2 are 6, 56, 76, and 116. y1 4 sin2 x, y2 1 4 0 (56,π 1) π , 1 6 ( ) 4 (76,π 1) ( 11π , 1 6 ) 2π π Xscl 6 In most applications, it is sufficient to find just the solutions from 0 to 2 or from 0 to 360. We then remember that any multiple of 2, or 360, can be added to obtain the rest of the solutions. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 575 Section 6.5 • Solving Trigonometric Equations 575 We must be careful to find all solutions in 0, 2 when solving trigonometric equations involving double angles. EXAMPLE 3 Solution Solve 3 tan 2x 3 in the interval 0, 2 . We first solve for tan 2x: 3 tan 2x 3 tan 2x 1. We are looking for solutions x to the equation for which 0 x 2. Multiplying by 2, we get 0 2x 4, f, which is the interval we use when solving tan 2x 1. Using the unit circle, we find points 2x in 0, 4 for which tan 2x 1. These values of 2x are as follows: 11p 4 2x , 2 2 2 2 (1, 0) 3 7 11 , , , 4 4 4 and 15 . 4 Thus the desired values of x in 0, 2 are each of these values divided 2 , 2 by 2. Therefore, 2 2 j, 15p 4 x 3 7 11 , , , 8 8 8 and 15 . 8 Calculators are needed to solve some trigonometric equations. Answers can be found in radians or degrees, depending on the mode setting. EXAMPLE 4 Solution cos1(0.4216) 65.06435654 65.06 294.94 65.06 (1, 0) Solve 1 cos 1 1.2108 in 0, 360. 2 We have 1 cos 1 1.2108 2 1 cos 0.2108 2 cos 0.4216. Using a calculator set in DEGREE mode (see window at left), we find that the reference angle, cos1 0.4216, is 65.06. Since cos is positive, the solutions are in quadrants I and IV. The solutions in 0, 360 are 65.06 and 360 65.06 294.94. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 576 Chapter 6 1/10/05 1:30 PM Page 576 • Trigonometric Identities, Inverse Functions, and Equations EXAMPLE 5 Solve 2 cos2 u 1 cos u in 0, 360. Algebraic Solution We use the principle of zero products: 2 cos2 u 1 cos u 2 cos u cos u 1 0 2 cos u 1 cos u 1 0 2 cos u 1 0 or cos u 1 0 2 cos u 1 or cos u 1 1 cos u or cos u 1. 2 2 Thus, u 60, 300 or u 180. The solutions in 0, 360 are 60, 180, and 300. Graphical Solution We can use either the Intersect method or the Zero method to solve trigonometric equations. Here we illustrate by solving the equation using both methods. We set the calculator in DEGREE mode. Intersect Method. We graph the equations y1 2 cos2 x and y2 1 cos x and use the INTERSECT feature to find the first coordinates of the points of intersection. y1 2 cos2 x, y2 1 cos x 3 0 Intersection X 60 3 360 Y .5 Xscl 60 The leftmost solution is 60. Using the INTERSECT feature two more times, we find the other solutions, 180 and 300. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 577 Section 6.5 • Solving Trigonometric Equations 577 Zero Method. We write the equation in the form 2 cos2 u cos u 1 0. Then we graph y 2 cos2 x cos x 1 and use the ZERO feature to determine the zeros of the function. y 2 cos2 x cos x 1 3 0 Zero X 60 3 360 Y0 Xscl 60 The leftmost zero is 60. Using the ZERO feature two more times, we find the other zeros, 180 and 300. The solutions in 0, 360 are 60, 180, and 300. EXAMPLE 6 Solution y sin2 x sin x 2 2π 0 1 π Xscl 4 Solve sin2 sin 0 in 0, 2 . We factor and use the principle of zero products: sin2 sin 0 sin sin 1 0 Factoring sin 0 or sin 1 0 sin 0 or sin 1 0, or . 2 The solutions in 0, 2 are 0, 2, and . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 578 Chapter 6 1/10/05 1:30 PM Page 578 • Trigonometric Identities, Inverse Functions, and Equations If a trigonometric equation is quadratic but difficult or impossible to factor, we use the quadratic formula. Solve 10 sin2 x 12 sin x 7 0 in 0, 360. EXAMPLE 7 Solution This equation is quadratic in sin x with a 10, b 12, and c 7. Substituting into the quadratic formula, we get sin x sin1(0.4296) 25.44217782 b b 2 4ac 2a Using the quadratic formula 12 122 410 7 2 10 Substituting 12 144 280 12 424 20 20 12 20.5913 20 sin x 1.6296 or sin x 0.4296. 25.44 205.44 25.44 Since sine values are never greater than 1, the first of the equations has no solution. Using the other equation, we find the reference angle to be 25.44. Since sin x is negative, the solutions are in quadrants III and IV. Thus the solutions in 0, 360 are 334.56 180 25.44 205.44 and 360 25.44 334.56. Trigonometric equations can involve more than one function. EXAMPLE 8 y1 2 cos2 x tan x, y2 tan x 2 Solution Using a graphing calculator, we can determine that there are six solutions. If we let Xscl 4, the solutions are read more easily. In the figures at left, we show the Intersect and Zero methods of solving graphically. Each illustrates that the solutions in 0, 2 are 2π 0 0, 2 y1 2 cos2 x tan x tan x 2 2 7 3 5 , , , , and . 4 4 4 4 We can verify these solutions algebraically, as follows: π Xscl 4 2π 0 Solve 2 cos2 x tan x tan x in 0, 2 . 2 cos2 x tan x tan x 2 cos2 x tan x tan x 0 tan x 2 cos2 x 1 0 tan x 0 or 2 cos2 x 1 0 1 cos2 x 2 2 2 3 5 7 x , , , . 4 4 4 4 cos x π Xscl 4 x 0, or Thus, x 0, 4, 34, , 54, and 74. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 579 Section 6.5 • Solving Trigonometric Equations 579 When a trigonometric equation involves more than one function, it is sometimes helpful to use identities to rewrite the equation in terms of a single function. EXAMPLE 9 Solve sin x cos x 1 in 0, 2 . Algebraic Solution We have sin x cos x 1 sin x cos x2 12 Squaring both sides 2 sin x 2 sin x cos x cos2 x 1 Using sin2 x cos2 x 1 2 sin x cos x 1 1 2 sin x cos x 0 Using 2 sin x cos x sin 2x sin 2x 0. We are looking for solutions x to the equation for which 0 x 2. Multiplying by 2, we get 0 2x 4, which is the interval we consider to solve sin 2x 0. These values of 2x are 0, , 2, and 3. Thus the desired values of x in 0, 2 satisfying this equation are 0, 2, , and 32. Now we check these in the original equation sin x cos x 1: sin 0 cos 0 0 1 1, sin cos 1 0 1, 2 2 sin cos 0 1 1, 3 3 cos 1 0 1. sin 2 2 We find that and 32 do not check, but the other values do. Thus the solutions in 0, 2 are 0 and . 2 When the solution process involves squaring both sides, values are sometimes obtained that are not solutions of the original equation. As we saw in this example, it is important to check the possible solutions. Graphical Solution We can graph the left side and then the right side of the equation as seen in the first window below. Then we look for points of intersection. We could also rewrite the equation as sin x cos x 1 0, graph the left side, and look for the zeros of the function, as illustrated in the second window below. In each window, we see the solutions in 0, 2 as 0 and 2. y1 sin x cos x, y2 1 3 2π 0 3 Xscl π 2 y sin x cos x 1 3 2π 0 3 Xscl π 2 This example illustrates a valuable advantage of the calculator— that is, with a graphing calculator, extraneous solutions do not appear. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 580 Chapter 6 1/10/05 1:30 PM Page 580 • Trigonometric Identities, Inverse Functions, and Equations EXAMPLE 10 Solve cos 2x sin x 1 in 0, 2 . Algebraic Solution Graphical Solution We have We graph y1 cos 2x sin x 1 and look for the zeros of the function. cos 2x sin x 1 1 2 sin2 x sin x 1 Using the identity cos 2x 1 2 sin2 x 2 sin x sin x 0 Factoring sin x 2 sin x 1 0 sin x 0 or 2 sin x 1 0 2 sin x 0 x 0, or or y cos 2x sin x 1 2 Principle of zero products 1 2 5 , . x 6 6 sin x 2 All four values check. The solutions in 0, 2 are 0, 6, 56, and . EXAMPLE 11 Study Tip 2π 0 π Xscl 6 The solutions in 0, 2 are 0, 6, 56, and . Solve tan2 x sec x 1 0 in 0, 2 . Algebraic Solution Check your solutions to the oddnumbered exercises in the exercise sets with the step-by-step annotated solutions in the Student’s Solutions Manual. If you are still having difficulty with the concepts of this section, make time to view the content video that corresponds to the section. We have tan2 x sec x 1 0 sec2 x 1 sec x 1 0 Using the identity 1 tan2 x sec2 x , or tan2 x sec2 x 1 sec2 x sec x 2 0 Factoring sec x 2 sec x 1 0 Principle of zero sec x 2 or sec x 1 products 1 cos x 2 2 4 x , 3 3 or cos x 1 or Using the identity cos x 1sec x x 0. All these values check. The solutions in 0, 2 are 0, 23, and 43. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 581 Section 6.5 • Solving Trigonometric Equations 581 Graphical Solution We graph y tan2 x sec x 1, but we enter this equation in the form y1 tan2 x 1 1. cos x We use the ZERO feature to find zeros of the function. y tan2 x sec x 1 3 2π 0 3 π Xscl 3 The solutions in 0, 2 are 0, 23, and 43. Sometimes we cannot find solutions algebraically, but we can approximate them with a graphing calculator. EXAMPLE 12 Solve each of the following in 0, 2 . a) x 1.5 cos x b) sin x cos x cot x 2 Solution a) In the screen on the left below, we graph y1 x 2 1.5 and y2 cos x and look for points of intersection. In the screen on the right, we graph y1 x 2 1.5 cos x and look for the zeros of the function. y1 x 2 1.5, y2 cos x y1 x 2 1.5 cos x 2 2 2π 0 Intersection X 1.3215996 Y .24662556 2 π Xscl 4 2π 0 Zero X 1.3215996 Y 0 2 π Xscl 4 We determine the solution in 0, 2 to be approximately 1.32. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 582 Chapter 6 1/10/05 1:30 PM Page 582 • Trigonometric Identities, Inverse Functions, and Equations b) In the screen on the left, we graph y1 sin x cos x and y2 cot x and determine the points of intersection. In the screen on the right, we graph the function y1 sin x cos x cot x and determine the zeros. y1 sin x cos x, y2 1/tan x y1 sin x cos x 1/tan x 3 3 2π 0 3 Intersection X 1.1276527 X 5.661357 Y .47462662 Y 1.395337 2π 0 3 Zero X 1.1276527 X 5.661357 Y0 Y0 Each method leads to the approximate solutions 1.13 and 5.66 in 0, 2 . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 584 Chapter 6 6.5 1/10/05 1:30 PM Page 584 • Trigonometric Identities, Inverse Functions, and Equations Exercise Set Solve, finding all solutions. Express the solutions in both radians and degrees. 3 2 1. cos x 2. sin x 2 2 3. tan x 3 5. sin x 1 2 4. cos x 1 2 6. tan x 1 2 7. cos x 2 3 8. sin x 2 Solve, finding all solutions in 0, 2 or 0, 360. Verify your answer using a graphing calculator. 9. 2 cos x 1 1.2814 98.09, 261.91 10. sin x 3 2.0816 246.69, 293.31 4 5 11. 2 sin x 3 0 , 3 3 19. 6 cos2 5 cos 1 0 20. 2 sin t cos t 2 sin t cos t 1 0 21. sin 2x cos x sin x 0 22. 5 sin2 x 8 sin x 3 198.28, 341.72 23. cos2 x 6 cos x 4 0 139.81, 220.19 24. 2 tan2 x 3 tan x 7 70.12, 250.12 25. 7 cot 2 x 4 cot x 37.22, 169.35, 217.22, 349.35 26. 3 sin2 x 3 sin x 2 207.22, 332.78 Solve, finding all solutions in 0, 2 . 27. cos 2x sin x 1 0, , 7 , 11 6 6 y1 cos 2x sin x, y2 1 4 0 2p y1 2 sin x 3 4 4 0 2p 28. 2 sin x cos x sin x 0 0, 2 4 , , 3 3 y 2 sin x cos x sin x 4 4 12. 2 tan x 4 1 68.20, 248.20 0 y1 2 tan x 4, y2 1 4 4 0 2p 29. sin 4x 2 sin 2x 0 0, 30. tan x sin x tan x 0 4 13. 2 cos2 x 1 2p 14. csc2 x 4 0 3 , , 2 2 0, 31. sin 2x cos x sin x 0 0, 32. cos 2x sin x sin x 0 0, 2, , 3 2 15. 2 sin2 x sin x 1 33. 2 sec x tan x 2 sec x tan x 1 0 3 4, 7 4 34. sin 2x sin x cos 2x cos x cos x 0, 2, , 3 2 18. 2 sin2 7 sin 4 5 35. sin 2x sin x 2 cos x 1 0 2 3, 4 3, 3 2 16. cos2 x 2 cos x 3 0 3 11 , , , 17. 2 cos2 x 3 cos x 0 6 2 2 6 6 , 6 Answers to Exercises 1–8, 13–15, and 19–21 can be found on p. IA-44. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:30 PM Page 585 Section 6.5 36. tan2 x 4 2 sec2 x tan x 2.034, 5.176, 4, 5 4 37. sec2 x 2 tan2 x 0 4, 3 4, 5 4, 7 4 • Solving Trigonometric Equations 585 52. Daylight Hours. The data in the following table give the number of daylight hours for certain days in Fairbanks, Alaska. 38. cot x tan 2x 3 6, 5 6, 7 6, 11 6 39. 2 cos x 2 sin x 6 12, 5 12 40. 3 cos x sin x 1 6, 3 2 41. sec2 x 2 tan x 6 0.967, 1.853, 4.109, 4.994 42. 5 cos 2x sin x 4 0.379, 2.763, 3.416, 6.009 43. cos x sin x 44. 2 1 sin2 x 1 2 1 2 cos x 1 2 2 4 , 3 3 3 , 4 4 Solve using a calculator, finding all solutions in 0, 2 . 45. x sin x 1 1.114, 2.773 46. x 2 2 sin x No solutions in 0, 2 47. 2 cos2 x x 1 0.515 48. x cos x 2 0 5.114 49. cos x 2 x 2 3x 0.422, 1.756 GCM 50. sin x tan 0, 2, 3 2 x 2 Some graphing calculators can use regression to fit a trigonometric function to a set of data. 51. Sales. Sales of certain products fluctuate in cycles. The data in the following table show the total sales of skis per month for a business in a northern climate. MONTH, x August, 8 November, 11 February, 2 May, 5 August, 8 TOTAL SALES, y (IN THOUSANDS) $ 0 7 14 7 0 y 7 sin 2.6180x 0.5236 7 a) Using the SINE REGRESSION feature on a graphing calculator, fit a sine function of the form y A sin Bx C D to this set of data. b) Approximate the total sales for December and for July. $10,500; $13,062 NUMBER OF DAYLIGHT HOURS, y DAY, x January 10, February 19, March 3, April 28, May 14, June 11, July 17, August 22, September 19, October 1, November 14, December 28, 10 50 62 118 134 162 198 234 262 274 318 362 4.7 9.0 10.3 16.7 18.5 21.4 19.9 15.8 12.7 11.3 6.4 3.8 Source: Astronomical Applications Department; U.S. Naval Observatory, Washington, DC a) Using the SINE REGRESSION feature on a graphing calculator, model these data with an equation of the form y A sin Bx C D. b) Approximate the number of daylight hours in Fairbanks for April 22 (x 112), July 4 (x 185), and December 15 (x 349). c) Determine on which day of the year there will be about 10.5 hr of daylight. 64th day (March 5th) Answers to Exercises 52(a) and 52(b) can be found on p. IA-44. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 586 Chapter 6 1/10/05 1:30 PM Page 586 • Trigonometric Identities, Inverse Functions, and Equations Collaborative Discussion and Writing 53. Jan lists her answer to a problem as 6 k, for any integer k, while Jacob lists his answer as 6 2k and 76 2k , for any integer k. Are their answers equivalent? Why or why not? 54. Under what circumstances will a graphing calculator give exact solutions of a trigonometric equation? Skill Maintenance Solve the right triangle. 55. C 201 b [5.2] B 35, b 140.7, c 245.4 B 67. Temperature During an Illness. The temperature T, in degrees Fahrenheit, of a patient t days into a 12-day illness is given by Tt 101.6 3 sin t . 8 Find the times t during the illness at which the patient’s temperature was 103. 1.24 days, 6.76 days 68. Satellite Location. A satellite circles the earth in such a manner that it is y miles from the equator (north or south, height from the surface not considered) t minutes after its launch, where y 5000 cos t 10 . 45 At what times t in the interval 0, 240, the first 4 hr, is the satellite 3000 mi north of the equator? c 55 23.28 min, 113.28 min, 203.28 min 69. Nautical Mile. (See Exercise 60 in Exercise Set 6.2.) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth’s radius. Since the earth is flattened at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, by the function A 56. T 3.8 14.2 [5.2] R 15.5, T 74.5, t 13.7 S N 6066 31 cos 2, t where is the latitude in degrees. At what latitude north is the length of a British nautical mile found to be 6040 ft? 16.5N R Solve. x 4 57. 27 3 [2.1] 36 58. 0.01 0.2 0.7 h [2.1] 14 Synthesis Solve in 0, 2 . 3 59. sin x 2 60. cos x 2 4 5 , , , 3 3 3 3 g 9.780491 0.005288 sin2 0.000006 sin2 2 , 2 4 5 , , , 3 3 3 3 1 2 where g is measured in meters per second per second at sea level. At what latitude north does g 9.8? 4 61. tan x 3 3, 4 3 37.95615N 62. 12 sin x 7sin x 1 0 0.063, 0.111, 3.031, 63. ln cos x 0 0 3.079 64. e sin x 1 0, 65. sin ln x 1 e 3 /22k, where k (an integer) 1 66. e ln sin x 1 2 70. Acceleration Due to Gravity. (See Exercise 61 in Exercise Set 6.2.) The acceleration due to gravity is often denoted by g in a formula such as S 12 gt 2, where S is the distance that an object falls in t seconds. The number g is generally considered constant, but in fact it varies slightly with latitude. If stands for latitude, in degrees, an excellent approximation of g is given by the formula Solve. 71. cos1 x cos1 35 sin1 1 72. sin 1 1 x tan 3 4 5 tan1 12 1 22 73. Suppose that sin x 5 cos x . Find sin x cos x . 0.1923 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:31 PM Page 587 Chapter 6 • Summary and Review Chapter 6 Summary and Review Important Properties and Formulas Double-Angle Identities sin 2x 2 sin x cos x , cos 2x cos2 x sin2 x 1 2 sin2 x 2 cos2 x 1, 2 tan x tan 2x 1 tan2 x Basic Identities sin x 1 tan x , , sin x csc x cos x cos x 1 cot x cos x , , sec x sin x 1 tan x , cot x sin x sin x , cos x cos x , tan x tan x Half-Angle Identities Pythagorean Identities sin2 x cos2 x 1, 1 cot 2 x csc2 x , 1 tan2 x sec2 x Sum and Difference Identities sin u v sin u cos v cos u sin v , cos u v cos u cos v sin u sin v , tan u tan v tan u v 1 tan u tan v Cofunction Identities sin x cos x , 2 tan sec x cot x , 2 sin x 2 cos x , cos x 2 sin x sin x 2 1 cos x , 2 cos x 2 1 cos x , 2 tan x 2 1 cos x 1 cos x sin x 1 cos x 1 cos x sin x x csc x 2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 587 BBEPMC06_0321279115.QXP 588 Chapter 6 1/10/05 1:31 PM Page 588 • Trigonometric Identities, Inverse Functions, and Equations Inverse Trigonometric Functions FUNCTION DOMAIN y sin1 x 1, 1 y cos1 x 1, 1 y tan1 x , RANGE , 2 2 0, , 2 2 Composition of Trigonometric Functions The following are true for any x in the domain of the inverse function: sin sin1 x x , cos cos1 x x , tan tan1 x x . The following are true for any x in the range of the inverse function: sin1 sin x x , cos1 cos x x , tan1 tan x x . Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 588 Chapter 6 1/10/05 1:31 PM Page 588 • Trigonometric Identities, Inverse Functions, and Equations Review Exercises Complete the Pythagorean identity. 1. 1 cot 2 x [6.1] csc2 x 11. 2. sin2 x cos2 x [6.1] 1 Multiply and simplify. Check using a graphing calculator. 3. tan y cot y tan y cot y [6.1] tan2 y cot 2 y 4. cos x sec x2 [6.1] cos2 x 12 cos2 x 3 2 cos y sin y sin2 y cos2 y 12. 13. 4 sin x cos2 x 16 sin2 x cos x cot x csc x 2 1 csc2 x [6.1] 1 [6.1] 1 cot x 4 14. Simplify. Assume the radicand is nonnegative. Factor and simplify. Check using a graphing calculator. 5. sec x csc x csc2 x [6.1] csc x sec x csc x 6. 3 sin2 y 7 sin y 20 [6.1] 3 sin y 5 sin y 4 sin2 x 2 cos x sin x cos2 x [6.1] sin x cos x 7. 1000 cos3 u [6.1] 10 cos u 100 10 cos u cos2 u Simplify and check using a graphing calculator. sec4 x tan4 x 2 sin2 x cos x 8. 2 [6.1] 1 9. 2 sec x tan x cos31 x 2 sin x 2 [6.1] 2 sec x 3 sin x cos2 x cos x sin x 3 tan x 10. [6.1] sin x cos x cos2 x sin2 x cos2 x 15. Rationalize the denominator: 16. Rationalize the numerator: 1 sin x . 1 sin x cos x . tan x 17. Given that x 3 tan , express 9 x 2 as a trigonometric function without radicals. Assume that 0 2. [6.1] 3 sec Answers to Exercises 11, 15, and 16 can be found on p. IA-44. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:31 PM Page 589 Chapter 6 Use the sum and difference formulas to write equivalent expressions. You need not simplify. 3 18. cos x 19. tan 45 30 2 Prove the identity. 1 sin x cos x 36. cos x 1 sin x 20. Simplify: cos 27 cos 16 sin 27 sin 16. 37. 1 cos 2 cot sin 2 38. tan y sin y y cos2 2 tan 2 39. sin x cos x tan2 x 1 cos2 x sin x cos x 21. Find cos 165 exactly. 22. Given that tan 3 and sin 22 and that and are between 0 and 2, evaluate tan exactly. [6.1] 2 3 23. Assume that sin 0.5812 and cos 0.2341 and that both and are first-quadrant angles. Evaluate cos . [6.1] 0.3745 Complete the cofunction identity. 24. cos x 25. cos 2 [6.2] sin x 26. sin x 2 x 2 [6.2] sin x [6.2] cos x 27. Given that cos 35 and that the terminal side is in quadrant III: a) Find the other function values for . b) Find the six function values for 2 . c) Find the six function values for 2. 28. Find an equivalent expression for csc x [6.2] sec x . 2 29. Find tan 2, cos 2, and sin 2 and the quadrant in which 2 lies, where cos 45 and is in quadrant III. 30. Find sin 2 2 exactly. [6.2] 2 8 31. Given that sin 0.2183 and is in quadrant I, find sin 2, cos , and cos 4. 2 Simplify and check using a graphing calculator. x 32. 1 2 sin2 [6.2] cos x 2 33. sin x cos x2 sin 2x [6.2] 1 34. 2 sin x cos x 2 sin x cos x 3 35. 3 2 cot x [6.2] tan 2x cot 2 x 1 [6.2] sin 2x 589 • Review Exercises In Exercises 40 – 43, use a graphing calculator to determine which expression (A)–(D) on the right can be used to complete the identity. Then prove the identity algebraically. csc x 40. csc x cos x cot x A. sec x 1 cos x 41. B. sin x sin x cos x sin x 2 cot x 1 C. 42. sin x 1 tan x sin x cos x cos x 1 sin x D. 43. 1 sin2 x sin x cos x 1 Find each of the following exactly in both radians and degrees. 1 3 44. sin1 45. cos1 2 2 [6.4] 6, 30 46. tan1 1 [6.4] 4, 45 [6.4] 6, 30 47. sin1 0 [6.4] 0, 0 Use a calculator to find each of the following in radians, rounded to four decimal places, and in degrees, rounded to the nearest tenth of a degree. 48. cos1 0.2194 49. cot 1 2.381 [6.4] 1.7920, 102.7 Evaluate. [6.4] 0.3976, 22.8 2 53. cos sin1 7 2 [6.4] 3 3 [6.4] 2 2 [6.4] 257 50. cos cos1 1 2 [6.4] 12 52. sin1 sin 7 [6.4] Find. 54. cos tan1 Answers to Exercises 18–21, 27, 29, 31, and 36 –43 can be found on pp. IA-44 and IA-45. b 3 [6.4] 51. tan1 tan 3 b 2 3 3 55. cos 2 sin1 9 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4 5 BBEPMC06_0321279115.QXP 590 Chapter 6 1/10/05 1:31 PM Page 590 • Trigonometric Identities, Inverse Functions, and Equations Solve, finding all solutions. Express the solutions in both radians and degrees. 2 56. cos x 57. tan x 3 2 Solve, finding all solutions in 0, 2 . 58. 4 sin2 x 1 y 4 y 1 3 2 1 2 x 1 y 4 sin2 x 68. Prove the identity 2 cos2 x 1 cos4 x sin4 x in three ways: a) Start with the left side and deduce the right (method 1). b) Start with the right side and deduce the left (method 1). c) Work with each side separately until you deduce the same expression (method 2). Then determine the most efficient method and explain why you chose that method. 69. Why are the ranges of the inverse trigonometric functions restricted? Synthesis 70. Find the measure of the angle from l1 to l2: 59. sin 2x sin x cos x 0 l1: x y 3 y 2 Collaborative Discussion and Writing y sin 2x sin x − cos x 71. Find an identity for cos u v involving only cosines. 1 2 x 1 y 0 2 60. 2 cos2 x 3 cos x 1 [6.5] 23, , 43 61. sin2 x 7 sin x 0 [6.5] 0, 62. csc2 x 2 cot 2 x 0 [6.5] 4, 34, 54, 74 72. Simplify: cos 2 [6.2] cos x x csc x sin x. 2 73. Find sin , cos , and tan under the given conditions: sin 2 1 , 2 . 5 2 74. Prove the following equation to be an identity: ln e sin t sin t . 63. sin 4x 2 sin 2x 0 [6.5] 0, 2, , 32 75. Graph: y sec1 x . 64. 2 cos x 2 sin x 2 76. Show that [6.5] 712, 2312 65. 6 tan2 x 5 tan x sec2 x [6.5] 0.864, 2.972, 4.006, 6.114 Solve using a graphing calculator, finding all solutions in 0, 2 . 66. x cos x 1 [6.5] 4.917 l2: 2x y 5. [6.1] 108.4 tan1 x is not an identity. sin1 x cos1 x 77. Solve e cos x 1 in 0, 2. [6.5] 2, 32 67. 2 sin2 x x 1 [6.5] No solution in 0, 2 Answers to Exercises 56–59, 68, 69, 71, and 73–76 can be found on pp. IA-45 and IA-46. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC06_0321279115.QXP 1/10/05 1:31 PM Page 591 Chapter 6 • Test 591 Chapter 6 Test Simplify. 2 cos2 x cos x 1 1. cos x 1 2. sec x tan x 2 1 tan2 x Prove each of the following identities. 13. csc x cos x cot x sin x [6.1] 2 cos x 1 14. sin x cos x2 1 sin 2x 15. csc cot 2 [6.1] 1 3. Rationalize the denominator: 1 sin cos . [6.1] 1 sin 1 sin 4. Given that x 2 sin , express 4 x 2 as a trigonometric function without radicals. Assume 0 2. [6.1] 2 cos Use the sum or difference identities to evaluate exactly. 5. sin 75 6. tan 12 7. Assuming that cos u 135 and cos v 12 13 and that u and v are between 0 and 2, evaluate cos u v exactly. [6.1] 120 169 23 8. Given that cos and that the terminal side is in quadrant II, find cos 2 . [6.2] 53 9. Given that sin 45 and is in quadrant III, find , II sin 2 and the quadrant in which 2 lies. [6.2] 24 25 10. Use a half-angle identity to evaluate cos exactly. 12 11. Given that sin 0.6820 and that is in quadrant I, find cos 2. [6.2] 0.9304 12. Simplify: sin x cos x2 1 2 sin 2x . [6.2] 3 sin 2x 16. 1 cos 1 cos 1 sin tan 1 csc sec [6.4] 45 17. Find sin1 2 2 exactly in degrees. 18. Find tan1 3 exactly in radians. [6.4] 3 19. Use a calculator to find cos1 0.6716 in radians, rounded to four decimal places. [6.4] 2.3072 20. Evaluate cos sin1 21. Find tan sin1 1 3 . [6.4] 2 2 5 5 . [6.4] 2 x x 25 22. Evaluate cos sin1 12 cos1 1 2 . [6.4] 0 Solve, finding all solutions in 0, 2 . 23. 4 cos2 x 3 [6.5] 6, 56, 76, 116 24. 2 sin2 x 2 sin x [6.5] 0, 4, 34, 11 25. 3 cos x sin x 1 [6.5] , 2 6 Synthesis 26. Find cos , given that cos 2 [6.2] 11 12 Answers to Exercises 5, 6, 10, and 13–16 can be found on p. IA-46. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 5 3 , 2. 6 2 BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 593 Applications of Trigonometry 7.1 7.2 7.3 7.4 7.5 7.6 7 The Law of Sines The Law of Cosines Complex Numbers: Trigonometric Form Polar Coordinates and Graphs Vectors and Applications Vector Operations SUMMARY AND REVIEW TEST A P P L I C A T I O N A n eagle flies from its nest 7 mi in the direction northeast, where it stops to rest on a cliff. It then flies 8 mi in the direction S30°W to land on top of a tree. Place an xy-coordinate system so that the origin is the bird’s nest, the x-axis points east, and the y-axis points north. (a) At what point is the cliff located? (b) At what point is the tree located? This problem appears as Exercise 55 in Exercise Set 7.5. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 594 Chapter 7 1/10/05 1:35 PM Page 594 • Applications of Trigonometry 2.1 7.1 The Law Polynomial of Sines Functions and Modeling Use the law of sines to solve triangles. Find the area of any triangle given the lengths of two sides and the measure of the included angle. To solve a triangle means to find the lengths of all its sides and the measures of all its angles. We solved right triangles in Section 5.2. For review, let’s solve the right triangle shown below. We begin by listing the known measures. Q 37.1 W 90 Z? Z q W w 37.1 6.3 q? w? z 6.3 Q Since the sum of the three angle measures of any triangle is 180, we can immediately find the measure of the third angle: Z 180 90 37.1 52.9. Then using the tangent and cosine ratios, respectively, we can find q and w : q , or 6.3 q 6.3 tan 37.1 4.8, tan 37.1 and 6.3 , or w 6.3 w 7.9. cos 37.1 cos 37.1 Now all six measures are known and we have solved triangle QWZ. Q 37.1 W 90 Z 52.9 q 4.8 w 7.9 z 6.3 Solving Oblique Triangles The trigonometric functions can also be used to solve triangles that are not right triangles. Such triangles are called oblique. Any triangle, right or oblique, can be solved if at least one side and any other two measures are known. The five possible situations are illustrated on the next page. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 595 Section 7.1 1. AAS: Two angles of a triangle and a side opposite one of them are known. • The Law of Sines 595 224 100 2. ASA: Two angles of a triangle and the included side are known. 25 AAS 31 51 37.5 ASA 3. SSA: Two sides of a triangle and an angle opposite one of them are known. (In this case, there may be no solution, one solution, or two solutions. The latter is known as the ambiguous case.) 38 q 115.7 20 ~ SSA 4. SAS: Two sides of a triangle and the included angle are known. 82.14 58 19.05 SAS 5. SSS: All three sides of the triangle are known. 210 75 172 SSS The list above does not include the situation in which only the three angle measures are given. The reason for this lies in the fact that the angle measures determine only the shape of the triangle and not the size, as shown with the following triangles. Thus we cannot solve a triangle when only the three angle measures are given. 10 10 30 140 10 30 140 30 140 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 596 Chapter 7 1/10/05 1:35 PM Page 596 • Applications of Trigonometry In order to solve oblique triangles, we need to derive the law of sines and the law of cosines. The law of sines applies to the first three situations listed above. The law of cosines, which we develop in Section 7.2, applies to the last two situations. The Law of Sines We consider any oblique triangle. It may or may not have an obtuse angle. Although we look at only the acute-triangle case, the derivation of the obtuse-triangle case is essentially the same. In acute ABC at left, we have drawn an altitude from vertex C. It has length h. From ADC , we have C b A c h D a sin A B h , b or h b sin A. or h a sin B. From BDC , we have sin B h , a With h b sin A and h a sin B, we now have a sin B b sin A a sin B b sin A sin A sin B sin A sin B a b . sin A sin B Dividing by sin A sin B Simplifying There is no danger of dividing by 0 here because we are dealing with triangles whose angles are never 0 or 180. Thus the sine value will never be 0. If we were to consider altitudes from vertex A and vertex B in the triangle shown above, the same argument would give us b c sin B sin C and a c . sin A sin C We combine these results to obtain the law of sines. The Law of Sines In any triangle ABC, b c a . sin A sin B sin C Thus in any triangle, the sides are proportional to the sines of the opposite angles. B a c A b Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley C BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 597 Section 7.1 • The Law of Sines 597 Study Tip Solving Triangles (AAS and ASA) Maximize the learning that you accomplish during a lecture by preparing for the class. Your time is valuable; let each lecture become a positive learning experience. Review the lesson from the previous class and read the section that will be covered in the next lecture. Write down questions you want answered and take an active part in class discussion. When two angles and a side of any triangle are known, the law of sines can be used to solve the triangle. EXAMPLE 1 triangle. Solution In EFG , e 4.56, E 43, and G 57. Solve the We first make a drawing. We know three of the six measures. E 43 F? G 57 F g E 4.56 43 57 f e 4.56 f? g? G From the figure, we see that we have the AAS situation. We begin by finding F : F 180 43 57 80. We can now find the other two sides, using the law of sines: f e sin F sin E f 4.56 sin 80 sin 43 4.56 sin 80 f sin 43 f 6.58; g e sin G sin E g 4.56 sin 57 sin 43 4.56 sin 57 g sin 43 g 5.61. Substituting Solving for f Substituting Solving for g Thus, we have solved the triangle: E 43, F 80, G 57, e 4.56, f 6.58, g 5.61. The law of sines is frequently used in determining distances. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 598 Chapter 7 1/10/05 1:35 PM Page 598 • Applications of Trigonometry EXAMPLE 2 Rescue Mission. During a rescue mission, a Marine fighter pilot receives data on an unidentified aircraft from an AWACS plane and is instructed to intercept the aircraft. The diagram shown below appears on the screen, but before the distance to the point of interception appears on the screen, communications are jammed. Fortunately, the pilot remembers the law of sines. How far must the pilot fly? Z y 115 X (Unidentified aircraft) x 27 500 km Y (Pilot) Solution We let x represent the distance that the pilot must fly in order to intercept the aircraft and Z represent the point of interception. We first find angle Z : Z 180 115 27 38. Because this application involves the ASA situation, we use the law of sines to determine x : x z sin X sin Z x 500 sin 115 sin 38 500 sin 115 x sin 38 x 736. Substituting Solving for x Thus the pilot must fly approximately 736 km in order to intercept the unidentified aircraft. Solving Triangles (SSA) When two sides of a triangle and an angle opposite one of them are known, the law of sines can be used to solve the triangle. Suppose for ABC that b, c, and B are given. The various possibilities are as shown in the eight cases below: five cases when B is acute and three cases when B is obtuse. Note that b c in cases 1, 2, 3, and 6; b c in cases 4 and 7; and b c in cases 5 and 8. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 599 Section 7.1 Angle B Is Acute Case 1: No solution b c ; side b is too short to reach the base. No triangle is formed. Case 2 : One solution b c ; side b just reaches the base and is perpendicular to it. A 599 • The Law of Sines Case 3 : Two solutions b c ; an arc of radius b meets the base at two points. (This case is called the ambiguous case.) A A b c c B B Case 4 : One solution b c ; an arc of radius b meets the base at just one point, other than B. c Angle B Is Obtuse Case 6 : No solution b c ; side b is too short to reach the base. No triangle is formed. b B C Case 7 : No solution b c ; an arc of radius b meets the base only at point B. No triangle is formed. b c B Case 8: One solution b c ; an arc of radius b meets the base at just one point. A A b c C A C A b Case 5 : One solution b c ; an arc of radius b meets the base at just one point. b B b B C A c c b b c B B C The eight cases above lead us to three possibilities in the SSA situation: no solution, one solution, or two solutions. Let’s investigate these possibilities further, looking for ways to recognize the number of solutions. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 600 Chapter 7 1/10/05 1:35 PM Page 600 • Applications of Trigonometry EXAMPLE 3 No solution. Solve the triangle. Solution In QRS , q 15, r 28, and Q 43.6. We make a drawing and list the known measures. Q 43.6 R? S? S 15 28 q 15 r 28 s? 43.6 Q s ?R? We observe the SSA situation and use the law of sines to find R: q r sin Q sin R 15 28 sin 43.6 sin R 28 sin 43.6 sin R 15 sin R 1.2873. Substituting Solving for sin R Since there is no angle with a sine greater than 1, there is no solution. EXAMPLE 4 One solution. 39.7. Solve the triangle. Solution In XYZ , x 23.5, y 9.8, and X We make a drawing and organize the given information. X z Y 23.5 39.7 9.8 X 39.7 Y? Z? x 23.5 y 9.8 z? Z We see the SSA situation and begin by finding Y with the law of sines: y 15.4° 164.6° x x y sin X sin Y 23.5 9.8 sin 39.7 sin Y 9.8 sin 39.7 sin Y 23.5 sin Y 0.2664. Substituting Solving for sin Y There are two angles less than 180 with a sine of 0.2664. They are 15.4 and 164.6, to the nearest tenth of a degree. An angle of 164.6 cannot be an angle of this triangle because it already has an angle of 39.7 and these Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 601 Section 7.1 • The Law of Sines 601 two angles would total more than 180. Thus, 15.4 is the only possibility for Y. Therefore, Z 180 39.7 15.4 124.9. We now find z: z x sin Z sin X z 23.5 sin 124.9 sin 39.7 23.5 sin 124.9 z sin 39.7 z 30.2. Substituting Solving for z We now have solved the triangle: X 39.7, Y 15.4, Z 124.9, x 23.5, y 9.8, z 30.2. The next example illustrates the ambiguous case in which there are two possible solutions. EXAMPLE 5 Two solutions. Solve the triangle. In ABC , b 15, c 20, and B 29. Solution We make a drawing, list the known measures, and see that we again have the SSA situation. A? B 29 C? A 20 B 15 29 a a? b 15 c 20 C We first find C: y sin 40 sin 140 140 40 x b c sin B sin C 15 20 sin 29 sin C 20 sin 29 0.6464. sin C 15 Substituting Solving for sin C There are two angles less than 180 with a sine of 0.6464. They are 40 and 140, to the nearest degree. This gives us two possible solutions. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 602 Chapter 7 1/10/05 1:35 PM Page 602 • Applications of Trigonometry Possible Solution I. Possible Solution II. If C 40, then If C 140, then A 180 29 40 111. A 180 29 140 11. Then we find a: Then we find a: a b sin A sin B a 15 sin 111 sin 29 15 sin 111 a 29. sin 29 a b sin A sin B a 15 sin 11 sin 29 15 sin 11 a 6. sin 29 These measures make a triangle as shown below; thus we have a solution. These measures make a triangle as shown below; thus we have a second solution. A 20 B 111 A 15 40 29 29 20 29 B 6 C C 15 140 11 Examples 3 – 5 illustrate the SSA situation. Note that we need not memorize the eight cases or the procedures in finding no solution, one solution, or two solutions. When we are using the law of sines, the sine value leads us directly to the correct solution or solutions. The Area of a Triangle The familiar formula for the area of a triangle, A 12 bh, can be used only when h is known. However, we can use the method used to derive the law of sines to derive an area formula that does not involve the height. Consider a general triangle ABC , with area K, as shown below. B c A B a h h C D D A is acute. a c A A is obtuse. C Note that in the triangle on the right, sin A sin 180 A. Then in each ADB , sin A h , c or h c sin A. 1 Substituting into the formula K 2 bh, we get 1 K 2 bc sin A. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 603 Section 7.1 • The Law of Sines 603 Any pair of sides and the included angle could have been used. Thus we also have K 12 ab sin C and 1 K 2 ac sin B. The Area of a Triangle The area K of any ABC is one half the product of the lengths of two sides and the sine of the included angle: K 1 1 1 bc sin A ab sin C ac sin B. 2 2 2 EXAMPLE 6 Area of the Peace Monument. Through the Mentoring in the City Program sponsored by Marian College, in Indianapolis, Indiana, children have turned a vacant downtown lot into a monument for peace.* This community project brought together neighborhood volunteers, businesses, and government in hopes of showing children how to develop positive, nonviolent ways of dealing with conflict. A landscape architect† used the children’s drawings and ideas to design a triangularshaped peace garden. Two sides of the property, formed by Indiana Avenue and Senate Avenue, measure 182 ft and 230 ft, respectively, and together form a 44.7 angle. The third side of the garden, formed by an apartment building, measures 163 ft. What is the area of this property? 163 ft 182 ft 44.7 Solution formula: 230 ft Since we do not know a height of the triangle, we use the area K 12 bc sin A K 12 182 ft 230 ft sin 44.7 K 14,722 ft 2. The area of the property is approximately 14,722 ft 2. *The Indianapolis Star, August 6, 1995, p. J8. †Alan Day, a landscape architect with Browning Day Mullins Dierdorf, Inc., donated his time to this project. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 604 Chapter 7 7.1 1/10/05 1:35 PM Page 604 • Applications of Trigonometry Exercise Set Solve the triangle, if possible. 1. B 38, C 21, b 24 A 121, a 33, c 14 2. A 131, C 23, b 10 or at least approximate, the area of the yard. Find the area of the yard to the nearest square foot. 787 ft 2 B 26, a 17, c 9 3. A 36.5, a 24, b 34 B 57.4, C 86.1, c 40, or B 122.6, C 20.9, c 14 4. B 118.3, C 45.6, b 42.1 C A 16.1, a 13.3, c 34.2 B 42 ft 5. C 6110, c 30.3, b 24.2 135° 53 ft B 4424, A 7426, a 33.3 6. A 126.5, a 17.2, c 13.5 A C 39.1, B 14.4, b 5.3 7. c 3 mi, B 37.48, C 32.16 A 110.36, a 5 mi, b 3 mi 8. a 2345 mi, b 2345 mi, A 124.67 No solution 9. b 56.78 yd, c 56.78 yd, C 83.78 B 83.78, A 12.44, a 12.30 yd 10. A 12932, C 1828, b 1204 in. B 32, a 1752 in., c 720 in. 11. a 20.01 cm, b 10.07 cm, A 30.3 B 14.7, C 135.0, c 28.04 cm 12. b 4.157 km, c 3.446 km, C 5148 13. A 89, a 15.6 in., b 18.4 in. No solution 14. C 4632, a 56.2 m, c 22.1 m No solution 24. Boarding Stable. A rancher operates a boarding stable and temporarily needs to make an extra pen. He has a piece of rope 38 ft long and plans to tie the rope to one end of the barn (S) and run the rope around a tree (T) and back to the barn (Q). The tree is 21 ft from where the rope is first tied, and the rope from the barn to the tree makes an angle of 35 with the barn. Does the rancher have enough rope if he allows 4 12 ft at each end to fasten the rope? No 15. a 200 m, A 32.76, C 21.97 B 125.27, b 302 m, c 138 m 16. B 115, c 45.6 yd, b 23.8 yd No solution Find the area of the triangle. 17. B 42, a 7.2 ft, c 3.4 ft 8.2 ft 2 18. A 1712, b 10 in., c 13 in. 19. C 8254, a 4 yd, b 6 yd Q 2 19 in 12 yd2 S 35° 21 ft T 20. C 75.16, a 1.5 m, b 2.1 m 1.5 m 2 21. B 135.2, a 46.12 ft, c 36.74 ft 596.98 ft 2 22. A 113, b 18.2 cm, c 23.7 cm 198.5 cm2 Solve. 23. Area of Back Yard. A new homeowner has a triangular-shaped back yard. Two of the three sides measure 53 ft and 42 ft and form an included angle of 135. To determine the amount of fertilizer and grass seed to be purchased, the owner has to know, Answer to Exercise 12 can be found on p. IA-46. 25. Rock Concert. In preparation for an outdoor rock concert, a stage crew must determine how far apart to place the two large speaker columns on stage (see the figure at the top of the next page). What generally works best is to place them at 50 angles to the center of the front row. The distance from the center of the front row to each of the speakers is 10 ft. How far apart does the crew need to place the speakers on stage? About 12.86 ft, or 12 ft 10 in. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 605 Section 7.1 50° Speaker 605 29. Fire Tower. A ranger in fire tower A spots a fire at a direction of 295. A ranger in fire tower B, located 45 mi at a direction of 045 from tower A, spots the same fire at a direction of 255. How far from tower A is the fire? from tower B? From A: about 35 mi; 50° from B: about 66 mi Stage 10 ft • The Law of Sines N 10 ft B 255° 26. Lunar Crater. Points A and B are on opposite sides of a lunar crater. Point C is 50 m from A. The measure of BAC is determined to be 112 and the measure of ACB is determined to be 42. What is the width of the crater? About 76.3 m C Fire N 45° 45 mi 295° A 27. Length of Pole. A pole leans away from the sun at an angle of 7 to the vertical. When the angle of elevation of the sun is 51, the pole casts a shadow 47 ft long on level ground. How long is the pole? About 51 ft 7° P 51° 47 ft In Exercises 28 – 31, keep in mind the two types of bearing considered in Sections 5.2 and 5.3. 28. Reconnaissance Airplane. A reconnaissance airplane leaves its airport on the east coast of the United States and flies in a direction of 085. Because of bad weather, it returns to another airport 230 km to the north of its home base. For the return trip, it flies in a direction of 283. What is the total distance that the airplane flew? About 1467 km About 2.4 km or 10.6 km 31. Mackinac Island. Mackinac Island is located 18 mi N3120W of Cheboygan, Michigan, where the Coast Guard cutter Mackinaw is stationed. A freighter in distress radios the Coast Guard cutter for help. It radios its position as S7840E of Mackinac Island and N6410W of Cheboygan. How far is the freighter from Cheboygan? About 22 mi N New airport 30. Lighthouse. A boat leaves lighthouse A and sails 5.1 km. At this time it is sighted from lighthouse B, 7.2 km west of A. The bearing of the boat from B is N6510E. How far is the boat from B? Mackinac Island B N 230 km 85 283 Original airport Cheboygan Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 606 Chapter 7 1/10/05 1:35 PM Page 606 • Applications of Trigonometry 32. Gears. Three gears are arranged as shown in the figure below. Find the angle . 89 Synthesis 45. Prove the following area formulas for a general triangle ABC with area represented by K. a2 sin B sin C 2 sin A 2 c sin A sin B K 2 sin C 2 b sin C sin A K 2 sin B r 28 ft K r 22 ft f 41 46. Area of a Parallelogram. Prove that the area of a parallelogram is the product of two adjacent sides and the sine of the included angle. r 36 ft s1 Collaborative Discussion and Writing S 33. Explain why the law of sines cannot be used to find the first angle when solving a triangle given three sides. 34. We considered eight cases of solving triangles given two sides and an angle opposite one of them. Describe the relationship between side b and the height h in each. s2 47. Area of a Quadrilateral. Prove that the area of a quadrilateral is one half the product of the lengths of its diagonals and the sine of the angle between the diagonals. c a b d Skill Maintenance Find the acute angle A, in both radians and degrees, for the given function value. 35. cos A 0.2213 [5.1] 1.348, 77.2° 48. Find d. d 18.8 in. 36. cos A 1.5612 [5.1] No angle Convert to decimal degree notation. 37. 181420 [5.1] 18.24° d 11 in. 38. 125342 [5.1] 125.06° 39. Find the absolute value: 5. [R.1] 5 Find the values. 3 40. cos [5.3] 2 6 42. sin 300 [5.3] 3 2 41. sin 45 [5.3] 43. cos 2 3 50 12 in. 2 2 [5.3] 12 44. Multiply: 1 i 1 i. [2.2] 2 Answers to Exercises 45–47 can be found on pp. IA-46 and IA-47. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 15 in. BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 607 Section 7.2 • The Law of Cosines 607 Use the law of cosines to solve triangles. Determine whether the law of sines or the law of cosines should be applied to solve a triangle. 7.2 The Law of Cosines The law of sines is used to solve triangles given a side and two angles (AAS and ASA) or given two sides and an angle opposite one of them (SSA). A second law, called the law of cosines, is needed to solve triangles given two sides and the included angle (SAS) or given three sides (SSS). The Law of Cosines To derive this property, we consider any ABC placed on a coordinate system. We position the origin at one of the vertices — say, C — and the positive half of the x-axis along one of the sides — say, CB. Let x, y be the coordinates of vertex A. Point B has coordinates a, 0 and point C has coordinates 0, 0. y (x, y) A c y b (a, 0) x C (0, 0) a Then cos C x , so b x b cos C and sin C y , b y b sin C . so B x Thus point A has coordinates b cos C, b sin C. Next, we use the distance formula to determine c 2: y or (x, y), or (b cos C, b sin C) A y Now we multiply and simplify: c b (a, 0) x c 2 x a2 y 02, c 2 b cos C a2 b sin C 02. C (0, 0) a B x c 2 b 2 cos2 C 2ab cos C a2 b 2 sin2 C a2 b 2sin2 C cos2 C 2ab cos C a2 b 2 2ab cos C . Using the identity sin2 x cos2 x 1 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 608 Chapter 7 1/10/05 1:35 PM Page 608 • Applications of Trigonometry Had we placed the origin at one of the other vertices, we would have obtained a2 b 2 c 2 2bc cos A b 2 a2 c 2 2ac cos B . or The Law of Cosines In any triangle ABC, or C a b c 2bc cos A, b 2 a2 c 2 2ac cos B, c 2 a2 b 2 2ab cos C . 2 2 2 b A a B c Thus, in any triangle, the square of a side is the sum of the squares of the other two sides, minus twice the product of those sides and the cosine of the included angle. When the included angle is 90, the law of cosines reduces to the Pythagorean theorem. Solving Triangles (SAS) When two sides of a triangle and the included angle are known, we can use the law of cosines to find the third side. The law of cosines or the law of sines can then be used to finish solving the triangle. EXAMPLE 1 Solve ABC if a 32, c 48, and B 125.2. Solution We first label a triangle with the known and unknown measures. C b A? B 125.2 C? a 32 b? c 48 32 A 48 125.2 B We can find the third side using the law of cosines, as follows: b 2 a2 c 2 2ac cos B b 2 322 48 2 2 32 48 cos 125.2 b 2 5098.8 b 71. Substituting We now have a 32, b 71, and c 48, and we need to find the other two angle measures. At this point, we can find them in two ways. One way uses the law of sines. The ambiguous case may arise, however, and we would have to be alert to this possibility. The advantage of using Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 609 Section 7.2 • The Law of Cosines 609 the law of cosines again is that if we solve for the cosine and find that its value is negative, then we know that the angle is obtuse. If the value of the cosine is positive, then the angle is acute. Thus we use the law of cosines to find a second angle. Let’s find angle A. We select the formula from the law of cosines that contains cos A and substitute: a2 b 2 c 2 2bc cos A 322 712 48 2 2 71 48 cos A 1024 5041 2304 6816 cos A 6321 6816 cos A cos A 0.9273768 A 22.0. Substituting The third angle is now easy to find: C 180 125.2 22.0 32.8. Thus, A 22.0, B 125.2, C 32.8, a 32, b 71, c 48. Due to errors created by rounding, answers may vary depending on the order in which they are found. Had we found the measure of angle C first in Example 1, the angle measures would have been C 34.1 and A 20.7. Variances in rounding also change the answers. Had we used 71.4 for b in Example 1, the angle measures would have been A 21.5 and C 33.3. Suppose we used the law of sines at the outset in Example 1 to find b. We were given only three measures: a 32, c 48, and B 125.2. When substituting these measures into the proportions, we see that there is not enough information to use the law of sines: a b 32 b l , sin A sin B sin A sin 125.2 b c b 48 l , sin B sin C sin 125.2 sin C c 32 48 a l . sin A sin C sin A sin C In all three situations, the resulting equation, after the substitutions, still has two unknowns. Thus we cannot use the law of sines to find b. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 610 Chapter 7 1/10/05 1:35 PM Page 610 • Applications of Trigonometry Solving Triangles (SSS) When all three sides of a triangle are known, the law of cosines can be used to solve the triangle. EXAMPLE 2 Solution Solve RST if r 3.5, s 4.7, and t 2.8. We sketch a triangle and label it with the given measures. R? S? T? R 4.7 2.8 S 3.5 r 3.5 s 4.7 t 2.8 T Since we do not know any of the angle measures, we cannot use the law of sines. We begin instead by finding an angle with the law of cosines. We choose to find S first and select the formula that contains cos S : s 2 r 2 t 2 2rt cos S 4.72 3.52 2.82 23.5 2.8 cos S 3.52 2.82 4.72 cos S 23.5 2.8 cos S 0.1020408 S 95.86. Substituting Similarly, we find angle R: r 2 s 2 t 2 2st cos R 3.52 4.72 2.82 24.7 2.8 cos R 4.72 2.82 3.52 cos R 24.7 2.8 cos R 0.6717325 R 47.80. Then T 180 95.86 47.80 36.34. Thus, R 47.80, S 95.86, T 36.34, r 3.5, s 4.7, t 2.8. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 611 Section 7.2 0.5 cm B C 2 cm 2 cm A • The Law of Cosines 611 EXAMPLE 3 Knife Bevel. Knifemakers know that the bevel of the blade (the angle formed at the cutting edge of the blade) determines the cutting characteristics of the knife. A small bevel like that of a straight razor makes for a keen edge, but is impractical for heavy-duty cutting because the edge dulls quickly and is prone to chipping. A large bevel is suitable for heavy-duty work like chopping wood. Survival knives, being universal in application, are a compromise between small and large bevels. The diagram at left illustrates the blade of a hand-made Randall Model 18 survival knife. What is its bevel? (Source : Randall Made Knives, P.O. Box 1988, Orlando, FL 32802) Solution We know three sides of a triangle. We can use the law of cosines to find the bevel, angle A. a2 b 2 c 2 2bc cos A 0.52 22 22 2 2 2 cos A 0.25 4 4 8 cos A 4 4 0.25 cos A 8 cos A 0.96875 A 14.36. Thus the bevel is approximately 14.36. CONNECTING THE CONCEPTS CHOOSING THE APPROPRIATE LAW The following summarizes the situations in which to use the law of sines and the law of cosines. To solve an oblique triangle: Use the law of sines for: Use the law of cosines for: AAS ASA SSA SAS SSS The law of cosines can also be used for the SSA situation, but since the process involves solving a quadratic equation, we do not include that option in the list above. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 612 Chapter 7 1/10/05 1:35 PM Page 612 • Applications of Trigonometry EXAMPLE 4 In ABC , three measures are given. Determine which law to use when solving the triangle. You need not solve the triangle. a) b) c) d) e) f) a 14, b 23, c 10 a 207, B 43.8, C 57.6 A 112, C 37, a 84.7 B 101, a 960, c 1042 b 17.26, a 27.29, A 39 A 61, B 39, C 80 Solution It is helpful to make a drawing of a triangle with the given information. The triangle need not be drawn to scale. The given parts are shown in color. FIGURE a) SITUATION C A b) A Law of Cosines ASA Law of Sines AAS Law of Sines SAS Law of Cosines SSA Law of Sines AAA Cannot be solved B C A d) SSS B C c) LAW TO USE B C Study Tip The InterAct Math Tutorial software that accompanies this text provides practice exercises that correlate at the objective level to the odd-numbered exercises in the text. Each practice exercise is accompanied by an example and guided solution designed to involve students in the solution process. This software is available in your campus lab or on CD-ROM. A e) B C A f) B C A B Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 613 Section 7.2 7.2 • The Law of Cosines 613 Exercise Set Solve the triangle, if possible. 1. A 30, b 12, c 24 a 15, B 24, C 126 2. B 133, a 12, c 15 b 25, A 20, C 27 3. a 12, b 14, c 20 A 36.18, B 43.53, C 100.29 4. a 22.3, b 22.3, c 36.1 A 35.96, B 35.96, C 108.08 5. B 7240, c 16 m, a 78 m a rangefinder. She observes some poachers, and the rangefinder indicates that they are 500 ft from her position. They are headed toward big game that she knows to be 375 ft from her position. Using her compass, she finds that the poachers’ azimuth (the direction measured as an angle from north) is 355 and that of the big game is 42. What is the distance between the poachers and the game? About 367 ft b 75 m, A 9451, C 1229 6. C 22.28, a 25.4 cm, b 73.8 cm c 51.2 cm, A 10.82, B 146.90 Poachers N 7. a 16 m, b 20 m, c 32 m A 24.15, B 30.75, C 125.10 8. B 72.66, a 23.78 km, c 25.74 km Game b 29.38 km, A 50.59, C 56.75 9. a 2 ft, b 3 ft, c 8 ft No solution 500 ft 10. A 9613, b 15.8 yd, c 18.4 yd 42° a 25.5 yd, B 382, C 4545 11. a 26.12 km, b 21.34 km, c 19.25 km 355° A 79.93, B 53.55, C 46.52 12. C 2843, a 6 mm, b 9 mm 375 ft Rangers c 5 mm, A 3857, B 11220 13. a 60.12 mi, b 40.23 mi, C 48.7 c 45.17 mi, A 89.3, B 42.0 14. a 11.2 cm, b 5.4 cm, c 7 cm A 128.71, B 22.10, C 29.19 15. b 10.2 in., c 17.3 in., A 53.456 a 13.9 in., B 36.127, C 90.417 No solution 16. a 17 yd, b 15.4 yd, c 1.5 yd Determine which law applies. Then solve the triangle. 17. A 70, B 12, b 21.4 Law of sines; C 98, a 96.7, c 101.9 18. a 15, c 7, B 62 Law of cosines; b 13, A 92, C 26 19. a 3.3, b 2.7, c 2.8 Law of cosines; A 73.71, B 51.75, C 54.54 20. a 1.5, b 2.5, A 58 No solution 26. Circus Highwire Act. A circus highwire act walks up an approach wire to reach a highwire. The approach wire is 122 ft long and is currently anchored so that it forms the maximum allowable angle of 35 with the ground. A greater approach angle causes the aerialists to slip. However, the aerialists find that there is enough room to anchor the approach wire 30 ft back in order to make the approach angle less severe. When this is done, how much farther will they have to walk up the approach wire, and what will the new approach angle be? 26 ft farther, about 28° 21. A 40.2, B 39.8, C 100 Cannot be solved C 22. a 60, b 40, C 47 Law of cosines; c 44, A 91, B 42 High wire 23. a 3.6, b 6.2, c 4.1 Law of cosines; A 33.71, B 107.08, C 39.21 24. B 11030, C 810, c 0.912 Law of sines; A 6120, a 5.633, b 6.014 Solve. 25. Poachers. A park ranger establishes an observation post from which to watch for poachers. Despite losing her map, the ranger does have a compass and 35° A 30 ft B Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley Ground D BBEPMC07_0321279115.QXP 614 Chapter 7 1/10/05 1:35 PM Page 614 • Applications of Trigonometry 27. In-line Skater. An in-line skater skates on a fitness trail along the Pacific Ocean from point A to point B. As shown below, two streets intersecting at point C also intersect the trail at A and B. In his car, the skater found the lengths of AC and BC to be approximately 0.5 mi and 1.3 mi, respectively. From a map, he estimates the included angle at C to be 110. How far did he skate from A to B? 30. Survival Trip. A group of college students is learning to navigate for an upcoming survival trip. On a map, they have been given three points at which they are to check in. The map also shows the distances between the points. However, to navigate they need to know the angle measurements. Calculate the angles for them. S 112.5, T 27.2, U 40.3 About 1.5 mi S C 31.6 km 110° 0.5 mi 22.4 km 1.3 mi Start T A B 28. Baseball Bunt. A batter in a baseball game drops a bunt down the first-base line. It rolls 34 ft at an angle of 25 with the base path. The pitcher’s mound is 60.5 ft from home plate. How far must the pitcher travel to pick up the ball? (Hint: A baseball diamond is a square.) About 30.8 ft Pitcher 34 ft 25 Batter 29. Ships. Two ships leave harbor at the same time. The first sails N15W at 25 knots (a knot is one nautical mile per hour). The second sails N32E at 20 knots. After 2 hr, how far apart are the ships? U 50 45.2 km 50 31. Airplanes. Two airplanes leave an airport at the same time. The first flies 150 kmh in a direction of 320. The second flies 200 kmh in a direction of 200. After 3 hr, how far apart are the planes? About 912 km 32. Slow-pitch Softball. A slow-pitch softball diamond is a square 65 ft on a side. The pitcher’s mound is 46 ft from home plate. How far is it from the pitcher’s mound to first base? About 46 ft 33. Isosceles Trapezoid. The longer base of an isosceles trapezoid measures 14 ft. The nonparallel sides measure 10 ft, and the base angles measure 80. a) Find the length of a diagonal. About 16 ft b) Find the area. About 122 ft 2 34. Area of Sail. A sail that is in the shape of an isosceles triangle has a vertex angle of 38. The angle is included by two sides, each measuring 20 ft. Find the area of the sail. About 124 ft 2 35. Three circles are arranged as shown in the figure below. Find the length PQ. About 4.7 cm About 37 nautical mi r 1.1 cm P d N15W 109 N32E r 1.8 cm Q Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley r 1.4 cm BBEPMC07_0321279115.QXP 1/10/05 1:35 PM Page 615 Section 7.2 36. Swimming Pool. A triangular swimming pool measures 44 ft on one side and 32.8 ft on another side. These sides form an angle that measures 40.8. How long is the other side? About 28.8 ft 615 • The Law of Cosines elevation of the ends of the bridge are 78 and 72. How deep is the canyon? About 9386 ft 5045 ft Collaborative Discussion and Writing 37. Try to solve this triangle using the law of cosines. Then explain why it is easier to solve it using the law of sines. h 78° C a 11.1 19 A 28.5 B 38. Explain why we cannot solve a triangle given SAS with the law of sines. Skill Maintenance Classify the function as linear, quadratic, cubic, quartic, rational, exponential, logarithmic, or trigonometric. 39. fx 34 x 4 [3.1] Quartic 40. y 3 17x [1.3] Linear 41. y sin x 3 sin x 2 42. fx 2x1/2 43. fx [5.5] Trigonometric [4.2] Exponential x 2x 3 [3.5] Rational x1 2 44. fx 27 x 3 [3.1] Cubic 45. y e x e x 4 [4.2] Exponential 50. Heron’s Formula. If a, b, and c are the lengths of the sides of a triangle, then the area K of the triangle is given by K ss a s b s c, 1 where s 2 a b c. The number s is called the semiperimeter. Prove Heron’s formula. (Hint: Use 1 the area formula K 2 bc sin A developed in Section 7.1.) Then use Heron’s formula to find the area of the triangular swimming pool described in Exercise 36. 51. A 12 a 2 sin ; when 90° 51. Area of Isosceles Triangle. Find a formula for the area of an isosceles triangle in terms of the congruent sides and their included angle. Under what conditions will the area of a triangle with fixed congruent sides be maximum? 52. Reconnaissance Plane. A reconnaissance plane patrolling at 5000 ft sights a submarine at bearing 35 and at an angle of depression of 25. A carrier is at bearing 105 and at an angle of depression of 60. How far is the submarine from the carrier? About 10,106 ft 46. y log2 x 2 log2 x 3 [4.3] Logarithmic 105 47. fx cos x 3 [5.5] Trigonometric 48. y 1 2 2x 2x 2 [2.3] Quadratic 72° 60 5000 ft E 25 35 N x Synthesis y 49. Canyon Depth. A bridge is being built across a canyon. The length of the bridge is 5045 ft. From the deepest point in the canyon, the angles of z E Answer to Exercise 50 can be found on p. IA-47. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley N BBEPMN00_0321279115.QXP 1/7/05 3:53 PM Page A-37 A-37 Chapters 4 – 5 12. [4.2] 13. [4.3] y 4 4 4 2 2 2 2 2 4 f (x) 4 x ex 2 2 2 3 4 4 x f(x) ln (x 2) 4 17. [4.3] 15 20. [4.3] 2.7726 x 2z 21. [4.3] 0.5331 22. [4.3] 1.2851 23. [4.4] loga y 1 2 24. [4.4] 5 ln x 5 ln y 25. [4.4] 0.656 26. [4.4] 4t 14. [4.3] 5 15. [4.3] 1 16. [4.3] 0 18. [4.3] x e 4 19. [4.3] x log3 5.4 27. [4.5] 12 28. [4.5] 1 29. [4.5] 1 30. [4.5] 4.174 31. [4.6] 1.54% 32. [4.6] (a) 4.5%; (b) Pt 1000e 0.045t ; (c) $1433.33; (d) 15.4 yr 27 33. [4.5] 8 Chapter 5 Exercise Set 5.1 1. sin cot 158 15 17 , cos 2 3 19. 2 21. 23. 12 25. 1 27. 2 2 3 29. 62.4 m 31. 9.72 33. 35.01 35. 3.03 37. 49.65 39. 0.25 41. 5.01 43. 1736 45. 83130 47. 1145 49. 474936 51. 54 53. 3927 55. 0.6293 57. 0.0737 59. 1.2765 61. 0.7621 63. 0.9336 65. 12.4288 67. 1.0000 69. 1.7032 71. 30.8 73. 12.5 75. 64.4 77. 46.5 79. 25.2 81. 38.6 83. 45 85. 60 87. 45 89. 60 91. 30 1 93. cos 20 sin 70 sec 20 1 95. tan 52 cot 38 cot 52 97. sin 25 0.4226, cos 25 0.9063, tan 25 0.4663, csc 25 2.3662, sec 25 1.1034, cot 25 2.1445 99. sin 184955 0.3228, cos 184955 0.9465, tan 184955 0.3411, csc 184955 3.0979, sec 184955 1.0565, cot 184955 2.9317 1 1 101. sin 8 q, cos 8 p, tan 8 , csc 8 , r q 1 sec 8 , cot 8 r 103. Discussion and Writing p 105. [4.2] 104. [4.2] 17. y y y 8 17 , tan 15 8 , csc 17 15 , sec 17 8 , 1 23 3 , cos , tan 3, csc , 2 2 3 3 sec 2, cot 3 765 465 7 5. sin , cos , tan , 65 65 4 4 65 65 , sec , cot csc 7 4 7 25 35 3 3 2 7. csc , or ; sec ; cot , or 5 2 5 5 5 25 7 9. cos 257 , tan 247 , csc 25 , , sec cot 24 7 24 25 5 5 11. sin , cos , csc , sec 5, 5 5 2 1 cot 2 2 25 35 5 13. sin , cos , tan , sec , 3 3 5 5 5 cot 2 25 5 15. sin , tan 2, csc , sec 5, 5 2 1 cot 2 5 4 5 4 3 2 3. sin 3 2 1 1 f(x) 2x f(x) e x/2 3 2 1 2 3 3 2 1 1 x 106. [4.3] 107. [4.3] y y 2 3 x 3 2 1 3 2 1 1 1 2 3 1 1 2 3 4 5 x g(x) log 2 x 1 1 2 3 1 2 3 4 5 h(x) ln x 108. [4.5] 9.21 109. [4.5] 4 110. [4.5] 101 97 111. [4.5] 343 113. 0.6534 115. Area 12 ab. But a c sin A, so Area 12 bc sin A. Exercise Set 5.2 1. F 60, d 3, f 5.2 3. A 22.7, a 52.7, c 136.6 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley x BBEPMN00_0321279115.QXP A-38 1/7/05 3:53 PM Page A-38 Answers 5. P 4738, n 34.4, p 25.4 7. B 217, b 0.39, c 9.74 9. A 77.2, B 12.8, a 439 11. B 42.42, a 35.7, b 32.6 13. B 55, a 28.0, c 48.8 15. A 62.4, B 27.6, a 3.56 17. About 62.2 ft 19. About 2.5 ft 21. About 606 ft 23. About 92.9 cm 25. About 599 ft 27. About 8 km 29. About 275 ft 31. About 24 km 33. Discussion and Writing 35. [1.1] 102, or about 14.142 36. [1.1] 310, or about 9.487 37. [4.3] 103 0.001 38. [4.3] ln t 4 39. 3.3 41. Cut so that 79.38 43. 27 77. Positive: all 79. sin 319 0.6561, cos 319 0.7547, tan 319 0.8693, csc 319 1.5242, sec 319 1.3250, cot 319 1.1504 81. sin 115 0.9063, cos 115 0.4226, tan 115 2.1445, csc 115 1.1034, sec 115 2.3663, cot 115 0.4663 83. East: about 130 km; south: 75 km 85. About 223 km 87. 1.1585 89. 1.4910 91. 0.8771 93. 0.4352 95. 0.9563 97. 2.9238 99. 275.4 101. 200.1 103. 288.1 105. 72.6 107. Discussion and Writing y 109. [3.4] 1 Exercise Set 5.3 1. III 3. III 5. I 7. III 9. II 11. II 13. 434, 794, 286, 646 15. 475.3, 835.3, 244.7, 604.7 17. 180, 540, 540, 900 19. 72.89, 162.89 21. 775646, 1675646 23. 44.8, 134.8 5 13 25. sin 135 , cos 12 13 , tan 12 , csc 5 , 13 12 sec 12 , cot 5 27 23 21 27. sin , cos , tan , 7 7 3 7 21 3 , sec , cot csc 2 3 2 213 313 2 29. sin , cos , tan 13 13 3 541 441 5 31. sin , cos , tan 41 41 4 22 2 33. cos , tan , csc 3, 3 4 32 , cot 22 sec 4 25 1 5 35. sin , cos , tan , 5 5 2 5 csc 5, sec 2 37. sin 45 , tan 43 , csc 54 , sec 53 , 3 cot 4 3 2 39. 30; 41. 45; 1 43. 0 45. 45; 2 2 3 47. 30; 2 49. 30; 3 51. 30; 3 53. Not defined 55. 1 57. 60; 3 2 59. 45; 61. 45; 2 63. 1 65. 0 67. 0 2 69. 0 71. Positive: cos, sec; negative: sin, csc, tan, cot 73. Positive: tan, cot; negative: sin, csc, cos, sec 75. Positive: sin, csc; negative: cos, sec, tan, cot 6 6 x 5 x x5 f(x) 1 x 2 25 y 110. [3.1] 25 20 15 10 5 43 1 5 1 2 3 4 x 10 g(x) x3 2x 1 15 20 25 111. [1.2], [3.5] Domain: x x 2; range: x x 1 112. [1.2], [3.1] Domain: x x 32 and x 5 ; range: all real numbers 113. [2.1] 12 114. [2.3] 2, 3 115. [2.1] 12, 0 116. [2.3] 2, 0, 3, 0 117. 19.625 in. Exercise Set 5.4 1. 11p (c) f ; (e) 4 y 3. y (a) d (f) 17p 4 (b) i (d) p 14p (e) 6 (a) A x x (c) F (b) w Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley (f) 10p (d) 6 23p 4 BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-39 A-39 Chapter 5 5. M: 2 4 3 5 3 11 , ; N: , ; P: , ; Q: , 3 3 2 2 4 4 6 6 5. y 7. (a) 2.4 2 2 , 2 2 25. 33. 41. 43. (c) 32 x (d) 320 17. 9 7 , 4 4 19 5 4 8 , 13. , 6 6 3 3 2 15. Complement: ; supplement: 6 3 5 17. Complement: ; supplement: 8 8 5 11 19. Complement: ; supplement: 12 12 5 10 214.6 5 21. 23. 25. 27. 29. 12 9 180 72 17 31. 33. 4.19 35. 1.05 37. 2.06 9 39. 0.02 41. 6.02 43. 1.66 45. 135 47. 1440 49. 57.30 51. 134.47 53. 225 55. 5156.62 57. 51.43 59. 0 0 radians, 30 , 45 , 60 , 6 4 3 3 5 3 , 135 , 180 , 225 , 270 , 90 2 4 4 2 7 , 360 2 315 4 61. 2.29 63. 3.2 yd 65. 1.1; 63 67. 3.2 yd 5 cm 69. , or about 5.24 71. 3150 3 min 73. About 18,852 revolutions per hour 75. 1047 mph 77. 10 mph 79. About 202 81. Discussion and Writing 83. [4.1] One-to-one 84. [5.1] Cosine of 85. [4.2] Exponential function 86. [3.5] Horizontal asymptote 87. [1.7] Odd function 88. [4.3] Natural 89. [4.1] Horizontal line; inverse 90. [4.3] Logarithm 91. 111.7 km; 69.8 mi 93. (a) 53730; (b) 194115 95. 1.676 radianssec 97. 1.46 nautical miles 1. (a) 3. (a) 3 7 , ; (b) 4 4 3 7 , ; (c) 4 4 3 7 , 4 4 2 21 2 2 21 21 , ; (b) , ; (c) , 5 5 5 5 5 5 (b) q i S p x y sin x 1 y 1 p f q d y sin (x) d q p x f 1 (c) same as (b); (d) the same 45. (a) See Exercise 43(a); (b) y 1 p f q y sin (x p) d d q p x f 1 (c) same as (b); (d) the same y 47. (a) y cos x 1 f q pSi d f u A Adu q i S p x 1 (b) y y cos (x p) 1 p f q f Adu d d 1 (c) same as (b); (d) the same Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley q f p x 3 2 23. 0 31. 1 39. 0 d pSiq u A 11. Exercise Set 5.5 21. 0 1 f 9. 13. 11. 0 3 2 19. 2 2 0.4816 27. 1.3065 29. 2.1599 1.1747 35. 1 37. 0.7071 0.8391 y (a) 15. Not defined (b) 7.5 9. 3 7. 0 BBEPMN00_0321279115.QXP A-40 1/7/05 3:54 PM Page A-40 Answers 49. Even: cosine, secant; odd: sine, tangent, cosecant, cotangent 51. Positive: I, III; negative: II, IV 53. Positive: I, IV; negative: II, III 55. Discussion and Writing y 57. [1.7] 6 f (x) x 2 77. 79. 2k, 2k , k 2 2 xx k , k 2 y 81. 2 4 2 2 2 4 2p w p x g(x) 2x 2 3 q y 83. g(x) (x 2)2 2p w 2 2 4 1 p 2 4 q x 85. (a) 4 Shift the graph of f right 2 units. 59. [1.7] f (x) |x| y g(x) 8 1 2 |x 4| 1 4 4 4 8 x 4 8 y f (x) x3 4 x 2 2 2 2 4 w p 2p x (b) OPA ODB ; OD OB OP OA 1 OD 1 cos OD sec (d) OAP ECO CE CO AO AP CE 1 cos sin cos CE sin CE cot OE csc 87. 1 4 4 OPA ODB ; BD AP Thus, OA OB sin BD cos 1 tan BD (c) OAP ECO ; OE CO PO AP 1 OE 1 sin Shift the graph of f to the right 4 units, shrink it vertically, then shift it up 1 unit. 60. [1.7] q 1 2 2 8 y sin x cos x 2 y 4 x w p 3 4 Stretch the graph of f vertically, then shift it down 3 units. f(x) x 2 2p q 4 58. [1.7] y 3 sin x 4 3 2 1 4 g(x) x 3 Reflect the graph of f across the x-axis. 1 61. [1.7] y x 23 1 62. [1.7] y 3 4x 63. cos x 65. sin x 67. sin x 69. cos x 71. sin x 73. (a) 2k, k ; (b) 2k, 2 75. Domain: , ; range: 0, 1; k ; (c) k, k period: ; amplitude: 12 Visualizing the Graph 1. J 8. A 2. H 9. C 3. E 10. I 4. F 5. B 6. D 7. G Exercise Set 5.6 1. Amplitude: 1; period: 2 ; phase shift: 0 y 2 1 2p w p q 1 2 q p w 2p y sin x 1 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley x BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-41 A-41 Chapter 5 3. Amplitude: 3; period: 2 ; phase shift: 0 13. Amplitude: 3; period: 2 ; phase shift: y 3 2 1 w 2p p q w q 1 2 2p p x w 2p p q y 3 cos x 1 5. Amplitude: 2 ; period: 2 ; phase shift: 0 y 3 cos (x p) 3 2 1 w q 1 2 p 2p x 2p x 15. Amplitude: 13 ; period: 2 ; phase shift: 0 y y 2 1 1 2p w p 2 1 2 x q 1 2 3 1 y 2 cos x 2 q p w y a sin x 4 5 17. Amplitude: 1; period: 2 ; phase shift: 0 7. Amplitude: 1; period: ; phase shift: 0 y y 3 2 2 1 1 2 2 x 2p w p 1 2 y 2 1 2p w p q 1 q 11. Amplitude: p w 2p x y 2 sin qx 2 1 q 1 p w 2p x y cos (x ) 2 19. Amplitude: 2; period: 4 ; phase shift: 1 21. Amplitude: ; period: ; phase shift: 2 4 3 23. Amplitude: 3; period: 2; phase shift: 25. Amplitude: 12 ; period: 1; phase shift: 0 27. Amplitude: 1; period: 4 ; phase shift: 29. Amplitude: 1; period: 1; phase shift: 0 4 1 31. Amplitude: ; period: 2; phase shift: 4 33. (b) 35. (h) 37. (a) 39. (f ) 1 41. y 2 cos x 1 43. y cos x 2 2 y 45. 1 ; period: 2 ; phase shift: 2 2 y w q y sin (2x) 9. Amplitude: 2; period: 4 ; phase shift: 0 2p y y q sin x q 4 q p q 1 p w 2p x 2p p 1 2 3 p 2p y 2 cos x cos 2x Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley x BBEPMN00_0321279115.QXP A-42 1/7/05 3:54 PM Page A-42 Answers y 47. 2p p 49. 2p p 1 2 y 75. y sin x cos 2x 2 3 2 1 x 2p w p y 2 1 2p 77. p 2p p 2 4 y sin x cos x y 3 cos x sin 2x 79. p 2p p 3 10 2p y cos 2x 2x q p w 2p x y 3 2 1 2p 2p w p 10 2p 83. 2p 10 Discussion and Writing [3.5] Rational 64. [4.3] Logarithmic [3.1] Quartic 66. [1.3] Linear [5.6] Trigonometric 68. [4.2] Exponential [1.3] Linear 70. [5.6] Trigonometric [3.1] Cubic 72. [4.2] Exponential Maximum: 8; minimum: 4 q 1 y 2 sec (x p) q p w 3 y 4 cos 2x 2 sin x 10 61. 63. 65. 67. 69. 71. 73. 81. 59. 10 q 1 2 3 3 10 2p y 2 cot x 2 10 10 2p y tan x q p w 2p x 2p p q y cos x x 57. x 1 y x sin x 2p 2p y 2 tan qx 2 55. 2p w y x 2 3 4 53. p 3 2 1 x 1 2p q y 2p w p y 51. q 1 2 3 y 10 8 6 4 2 r w q w x y 2 csc qx f 85. 9.42, 6.28, 3.14, 3.14, 6.28, 9.42 87. 3.14, 0, 3.14 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2p x BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-43 Chapter 5 89. (a) y 101.6 3 sin p x 8 (b) 104.6, 98.6 40. [5.4] A-43 y 105 k 0 97 373 873 3 1. [5.1] sin , cos , tan , 73 73 8 8 73 73 , sec , cot csc 3 8 3 3 1091 91 2. [5.1] cos , tan , csc , 10 3 91 10 391 sec , cot 3 91 2 3 2 3. [5.1] 4. [5.1] 5. [5.3] 6. [5.3] 12 2 3 2 23 7. [5.3] Not defined 8. [5.3] 3 9. [5.1] 3 10. [5.1] 1 11. [5.1] 221612 12. [5.1] 47.56 13. [5.3] 0.4452 14. [5.3] 1.1315 15. [5.3] 0.9498 16. [5.3] 0.9092 17. [5.3] 1.5282 18. [5.3] 0.2778 19. [5.3] 205.3 20. [5.3] 47.2 21. [5.1] 60 22. [5.1] 60 23. [5.1] 45 24. [5.1] 30 25. [5.1] sin 30.9 0.5135, cos 30.9 0.8581, tan 30.9 0.5985, csc 30.9 1.9474, sec 30.9 1.1654, cot 30.9 1.6709 26. [5.2] b 4.5, A 58.1, B 31.9 27. [5.2] A 38.83, b 37.9, c 48.6 28. [5.2] 1748 m 29. [5.2] 14 ft 30. [5.3] II 31. [5.3] I 32. [5.3] IV 5 33. [5.3] 425, 295 34. [5.4] , 3 3 35. [5.3] Complement: 76.6; supplement: 166.6 5 36. [5.4] Complement: ; supplement: 3 6 313 213 3 37. [5.3] sin , cos , tan , 13 13 2 2 13 13 , sec , cot csc 3 2 3 2 5 5 38. [5.3] sin , cos , cot , 3 3 2 35 3 , csc 39. [5.3] About 1743 mi sec 5 2 u f 91. Amplitude: 3000; period: 90; phase shift: 10 93. 4 in. Review Exercises: Chapter 5 x F 12 , 0.52 6 [5.4] 270 44. [5.4] 171.89 45. [5.4] 257.83 7 [5.4] 1980 47. [5.4] , or 5.5 cm 4 [5.4] 2.25, 129 49. [5.4] About 37.9 ftmin [5.4] 497,829 radianshr [5.5] 35 , 45 , 35 , 45 , 35 , 45 3 [5.5] 1 53. [5.5] 1 54. [5.5] 55. [5.5] 12 2 3 [5.5] 57. [5.5] 1 58. [5.5] 0.9056 3 [5.5] 0.9218 60. [5.5] Not defined 61. [5.5] 4.3813 [5.5] 6.1685 63. [5.5] 0.8090 [5.5] 41. [5.4] 121 150 , 2.53 43. 46. 48. 50. 51. 52. 56. 59. 62. 64. y 2 1 2p p 42. [5.4] y y sin x p 2 2p p 2p x 2 1 2 y cos x p y 3 2 1 2p w p q 1 2 3 y tan x q p y q 1 2p x y cot x 3 2 1 2p w p w q p Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley w 2p x 2p x BBEPMN00_0321279115.QXP A-44 1/7/05 3:54 PM Page A-44 Answers y 1 ; period: ; phase shift: 2 4 69. [5.6] Amplitude: 3 2 y 4 2p w p q 1 2 3 q p 2p w p 3 y sec x 2 w 2p y 3 qcos 2x q 1 y q x 2p w p y csc x 3 2 1 70. [5.6] (d) 74. [5.6] y w q 1 2p p q q 71. [5.6] (a) p w 2p 72. [5.6] (c) x 73. [5.6] (b) 3 x 2 1 FUNCTION DOMAIN RANGE Sine , 1, 1 Cosine , 1, 1 xx Tangent k, k 2 , 67. [5.3] FUNCTION I II III IV Sine Cosine Tangent 68. [5.6] Amplitude: 1; period: 2 ; phase shift: y 2 2p p 1 2 y sin x q p 2p x 2 q 1 65. [5.5] Period of sin, cos, sec, csc: 2 ; period of tan, cot: 66. [5.5] 2p x p 2 3 y 3 cos x sin x 75. Discussion and Writing [5.1], [5.4] Both degrees and radians are units of angle measure. A degree is defined to be 1 360 of one complete positive revolution. Degree notation has been in use since Babylonian times. Radians are defined in terms of intercepted arc length on a circle, with one radian being the measure of the angle for which the arc length equals the radius. There are 2 radians in one complete revolution. 76. Discussion and Writing [5.5] The graph of the cosine function is shaped like a continuous wave, with “high” points at y 1 and “low” points at y 1. The maximum value of the cosine function is 1, and it occurs at all points where x 2k, k . 77. Discussion and Writing [5.5] No; sin x is never greater than 1. 78. Discussion and Writing [5.6] When x is very large or very small, the amplitude of the function becomes small. The dimensions of the window must be adjusted to be able to see the shape of the graph. Also, when x is 0, the function is undefined, but this may not be obvious from the graph. 79. [5.5] All values 80. [5.6] Domain: , ; range: 3, 3; period 4 y y 3 sin 3 x 2 2 1 1 2p p 4p 3p x 2 3 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-45 Chapters 5 – 6 2 2 82. [5.6] The domain consists of the intervals 2k, 2k , k . 2 2 83. [5.3] cos x 0.7890, tan x 0.7787, cot x 1.2842, sec x 1.2674, csc x 1.6276 81. [5.6] y2 2 sin x Test: Chapter 5 465 4 7 765 , or ; cos , or ; 65 65 65 65 4 7 65 65 ; sec ; cot tan ; csc 7 4 7 4 3 2. [5.3] 3. [5.3] 1 4. [5.4] 1 5. [5.4] 2 2 6. [5.1] 38.47 7. [5.3] 0.2419 8. [5.3] 0.2079 9. [5.4] 5.7588 10. [5.4] 0.7827 11. [5.1] 30 12. [5.1] sin 61.6 0.8796; cos 61.6 0.4756; tan 61.6 1.8495; csc 61.6 1.1369; sec 61.6 2.1026; cot 61.6 0.5407 13. [5.2] B 54.1, a 32.6, c 55.7 14. [5.3] Answers may vary; 472, 248 15. [5.4] 6 5 4 41 16. [5.3] cos ; tan ; csc ; 5 4 41 7 5 41 sec ; cot 17. [5.4] 18. [5.4] 135 5 4 6 16 16.755 cm 19. [5.4] 20. [5.5] 1 21. [5.5] 2 3 22. [5.5] 23. [5.6] (c) 24. [5.2] About 444 ft 2 25. [5.2] About 272 mi 26. [5.4] 18 56.55 mmin 2k, k an integer 27. [5.5] x 2k x 2 2 1. [5.1] sin Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley A-45 BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-45 Chapters 5 – 6 Chapter 6 Exercise Set 6.1 1. sin2 x cos2 x 3. sin y cos y 5. 1 2 sin cos 7. sin3 x csc 3 x 9. cos x sin x cos x 11. sin x cos x sin x cos x 13. 2 cos x 3 cos x 1 15. sin x 3 sin2 x 3 sin x 9 17. tan x 2 tan t 1 19. sin x 1 21. 23. 1 3 tan t 1 1 2 sin s 2 cos s 5 cot 25. 27. sin cos sin2 s cos2 s A-45 5sin 3 31. sin x cos x 3 33. cos sin cos 35. 1 sin y cos x sin x cos x 2 cot y 37. 39. 41. cos x 2 sin x cos x 1 sin y x a 2 x 2 43. 45. cos , tan 2 cos y a a x 2 3 x 2 9 47. sin , cos 49. sin tan x x 6 2 3 1 51. 53. , or 2 3 4 1 3 6 2 55. 57. sin 59 0.8572 4 59. cos 24 0.9135 61. tan 52 1.2799 sin 63. tan cos sin cos cos sin cos cos sin sin 1 sin cos cos sin cos cos cos cos sin sin 1 cos cos sin sin cos cos sin sin 1 cos cos tan tan 1 tan tan 65. 0 67. 257 69. 1.5789 71. 0.7071 73. 2 sin cos 75. cos u 77. Discussion and Writing 79. [2.1] All real numbers 80. [2.1] No solution 81. [5.1] 1.9417 82. [5.1] 1.6645 83. 0; the lines are 3 parallel 85. , or 135 87. 22.83 4 cos x h cos x 89. h cos x cos h sin x sin h cos x h cos x cos h cos x sin x sin h h h cos h 1 sin h sin x cos x h h sin 5x sin 91. Let x . Then 0 sin 5. 5 x 5 Answers may vary. 93. Let . Then cos 2 cos 0, but 4 2 2 cos 2 cos 2. Answers may vary. 4 29. BBEPMN00_0321279115.QXP A-46 1/7/05 3:54 PM Page A-46 Answers 1 cos cos 6x . Then 6. 6 cos x 32 cos 6 6 33 Answers may vary. 97. 0.0645 9 23 99. 168.7 101. cos 2 cos2 sin2 , or 1 2 sin2 , or 2 cos2 1 tan x tan 1 tan x 4 103. tan x 4 1 tan x 1 tan x tan 4 105. sin sin sin cos cos sin sin cos cos sin 2 sin cos 95. Let x Exercise Set 6.2 3 3 3 1.3763, csc 1.2361, sec 1.7013, 10 10 10 3 cot 0.7266; (b) sin 0.5878, cos 0.8090, 10 5 5 tan 0.7266, csc 1.7013, sec 1.2361, 5 5 5 1.3763 cot 5 22 2 3. (a) cos , tan , csc 3, 3 4 32 , cot 22; sec 4 22 1 (b) sin , cos , 2 3 2 3 1. (a) tan tan sec 22, csc 2 3, cot 2 (c) sin 2 17. tan 2 22, csc sec 2 3, cot 2 2 2 4 csc x cot x 7. tan x 2 2 24 7 24 9. sin 2 25 , cos 2 25 , tan 2 7 ; II 5. sec x 24 7 24 11. sin 2 25 , cos 2 25 , tan 2 7 ; II 120 119 120 13. sin 2 169 , cos 2 169 , tan 2 119 ; IV 2 2 15. cos 4x 1 8 sin x cos x , or cos4 x 6 sin2 x cos2 x sin4 x , or 8 cos4 x 8 cos2 x 1 2 21. 2 3 Exercise Set 6.3 1. sec x sin x tan x sin x 1 sin x cos x cos x 1 sin2 x cos x cos2 x cos x cos x 1 , 3 32 , 4 2 2 25. 0.1735 cos2 x sin2 x cos 2x 27. (d); cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x sin x cos x sin x sin x cos x sin x sin x sin x sin x sin x cot x 1 2 sin x cos x sin 2x 29. (d); 31. cos x 33. 1 sin x 2 cos x 2 cos x 35. cos 2x 37. 8 39. Discussion and Writing 41. [6.1] True 42. [6.1] False 43. [6.1] False 44. [6.1] True 45. [6.1] False 46. [6.1] True 47. [6.1] False 48. [6.1] True 49. [5.5] (a), (e) 50. [5.5] (b), (c), (f ) 51. [5.5] (d) 52. [5.5] (e) 53. sin 141 0.6293, cos 141 0.7772, tan 141 0.8097, csc 141 1.5891, sec 141 1.2867, 55. cos x 1 cot x 57. cot 2 y cot 141 1.2350 8 15 59. sin 15 , , cos tan 17 17 8 61. (a) 9.80359 msec 2; (b) 9.80180 msec 2; (c) g 9.780491 0.005264 sin2 0.000024 sin4 2 ; 2 4 19. 2 23. 0.6421 32 , 2 4 22 , cos 3 2 2 3 cos x 3. 1 cos x sin x sin x 1 cos x 1 cos x sin x 1 cos x 1 cos x sin x 1 cos x 1 cos2 x sin x 1 cos x sin2 x 1 cos x sin x BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-47 Chapter 6 5. A-47 11. 1 tan 1 cot 1 tan 1 cot cos sin 1 1 cos sin sin cos 1 1 cos sin cos sin sin cos cos sin cos sin sin cos cos sin cos cos sin cos cos sin sin sin cos sin sin cos cos sin sin cos cos sin sin cos cos sin cos sin cos sin cos sin 0 1 cos 5 cos 3 sin 5 sin 3 1 cos 5 cos 3 sin 5 sin 3 1 cos 5 3 1 cos 2 0 13. 2 sin cos3 2 sin3 cos 2 sin cos cos2 sin2 2 sin cos 1 sin 1 cos sin 1 cos sin sin cos 1 sin sin cos 1 sin sin cos 1 sin cos 1 cos cos 9. 2 tan 1 tan2 2 tan sec 2 2 sin cos2 cos 1 2 sin cos sin 2 2 sin cos tan x sin x 2 tan x sin x sin x 1 cos x 2 sin x cos x sin x sin x cos x cos x cos x sin x 1 cos x 2 1 2 sin2 x 2 1 cos x 2 17. sin sin sin cos sin cos cos sin cos sin 2 2 sin cos cos2 sin2 cos2 1 cos2 cos2 1 cos2 cos2 cos2 cos2 cos2 cos2 cos2 cos2 cos2 cos2 tan 1 cos2 tan 1 sin cos2 1 cos sin cos2 1 cos sin cos 1 sin cos 1 sin 2 2 sin cos 15. 7. cos2 cot cos2 cot cos cos2 sin cos cos2 sin 1 cos cos sin 2 sin2 1 cos 2 19. tan tan cot tan2 tan cot tan2 1 sec 2 sec 2 21. 1 cos2 x sin2 x cos2 x 1 2 sin x sin2 x csc 2 x cot 2 x csc 2 x csc 2 x 1 2 csc 2 x 1 2 csc 2 x 1 sin2 sin2 1 cos2 1 cos2 cos2 cos2 BBEPMN00_0321279115.QXP A-48 1/7/05 3:54 PM Page A-48 Answers 23. 35. C; 1 sin x sin x 1 1 sin x 1 sin x 1 sin x2 1 sin x2 1 sin2 x 2 1 2 sin x sin x 1 2 sin x sin2 x cos2 x 4 sin x cos2 x 25. cos2 cot 2 cot 2 cos2 1 sin2 cot 2 cos2 cot 2 sin2 sin2 cot 2 cos2 27. 4 sec x tan x 1 sin x cos x cos x 4 sin x cos2 x 4 1 sin 2 sin cos cos cos2 2 sin2 cos2 1 sin2 1 sin2 cos2 sin2 sin2 cos2 1 sin2 cos2 cos2 sin2 1 29. 1 sin x sec x tan x2 1 sin x sin x 2 1 1 sin x 1 sin x 1 sin x 1 sin x cos x cos x 2 1 sin x2 1 sin x 2 1 sin x cos2 x 2 1 sin x cos2 x 31. B; cos x cot x cos x 1 csc x cos x cos x 1 sin x 1 1 sin x sin x sin x cos x cos x sin x sin x 1 cos x sin x 1 sin x 1 cos x 33. A; 2 2 4 sin x cos x 1 1 cot x sin2 x 1 cos x sin2 x sin x 1 cos x sin x tan x cot x sin x cos x cos x sin x sin2 x cos2 x cos x sin x 1 cos x sin x 37. Discussion and Writing y 39. [4.1] (a), (d) 2 2 4 sin3 x cos3 x sin x cos x sin x cos x sin2 x sin x cos x cos2 x sin x cos x sin2 x sin x cos x cos2 x sin x cos x 1 f(x) 3x 2 4 f 1(x) (b) yes; (c) f 1x 2 4 2 x2 3 x x2 3 40. [4.1] (a), (d) f(x) x 3 1 y 4 3 f 1(x) x 1 2 4 2 2 4 x 2 4 (b) yes; (c) f 1x x 1 y 41. [4.1] (a), (d) 3 f(x) x 2 4, x 0 4 f 1(x) x 4 4 2 4 x 2 4 (b) yes; (c) f 1x x 4 y 42. [4.1] (a), (d) 6 f 1(x) x 2 2, x 0 4 f(x) x 2 2 4 2 (b) yes; (c) f 1x x 2 2, x 0 6 x BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-49 Chapter 6 5 7 43. [2.3] 0, 2 44. [2.3] 4, 3 45. [2.3] 2, 3i 46. [2.3] 5 26 47. [2.5] 27 48. [2.5] 9 49. ln tan x ln cot x 1 ln cot x ln 1 ln cot x 0 ln cot x ln cot x 51. log cos x sin x log cos x sin x log cos x sin x cos x sin x log cos2 x sin2 x log cos 2x 1 1 53. Ctan tan sin sin C cos cos 1 sin cos sin cos C cos cos cos cos C sin Exercise Set 6.4 , 45 7. 0, 0 4 9. , 30 11. , 30 13. , 30 6 6 6 15. , 30 17. , 90 19. , 60 6 2 3 21. 0.3520, 20.2 23. 1.2917, 74.0 25. 2.9463, 168.8 27. 0.1600, 9.2 29. 0.8289, 47.5 31. 0.9600, 55.0 33. sin1: 1, 1; cos1: 1, 1; tan1: , 2000 35. sin1 37. 0.3 39. 41. d 4 5 11 43. 45. 12 47. 1 49. 51. 3 3 33 a p q 2 p 2 53. 55. 57. 59. 2 6 p 3 a 9 2 3 61. 63. 65. xy 1 x 2 1 y 2 2 10 67. 0.9861 69. Discussion and Writing 71. Discussion and Writing 72. [5.5] Periodic 73. [5.4] Radian measure 74. [5.1] Similar 75. [5.2] Angle of depression 76. [5.4] Angular speed 77. [5.3] Supplementary 78. [5.5] Amplitude 79. [5.1] Acute 80. [5.5] Circular 1. , 60 3 3. , 45 4 5. A-49 81. sin1 x cos1 x sin sin1 x cos1 x sin sin1 x cos cos1 x cos sin1 x sin cos1 x x x 1 x 2 1 x 2 x2 1 x2 1 83. x sin1 x tan1 1 x 2 x sin sin1 x sin tan1 1 x 2 x x 85. sin1 x cos1 1 x 2 1 sin sin x sin cos1 1 x 2 x x y h y 87. tan1 tan1 ; 38.7 x x 2 sin 2 1 Visualizing the Graph 1. D 8. J 2. G 3. C 9. F 10. B 4. H 5. I 6. A 7. E Exercise Set 6.5 11 2k, 2k, or 30 k 360, 330 k 360 6 6 2 3. k, or 120 k 180 3 5 5. 2k, 2k, or 30 k 360, 150 k 360 6 6 3 5 7. 2k, 2k, or 135 k 360, 225 k 360 4 4 4 5 3 5 7 9. 98.09, 261.91 11. , 13. , , , 3 3 4 4 4 4 5 3 3 11 15. , , 17. , , , 6 6 2 6 2 2 6 19. 109.47, 120, 240, 250.53 3 5 7 21. 0, , , , , 23. 139.81, 220.19 4 4 4 4 7 11 25. 37.22, 169.35, 217.22, 349.35 27. 0, , , 6 6 3 3 7 29. 0, , , 31. 0, 33. , 2 2 4 4 2 4 3 3 5 7 5 35. , , 37. , , , 39. , 3 3 2 4 4 4 4 12 12 1. BBEPMN00_0321279115.QXP A-50 1/7/05 3:54 PM Page A-50 Answers 2 4 , 3 3 45. 1.114, 2.773 47. 0.515 49. 0.422, 1.756 51. (a) y 7 sin 2.6180x 0.5236 7; (b) $10,500, $13,062 53. Discussion and Writing 55. [5.2] B 35, b 140.7, c 245.4 56. [5.2] R 15.5, T 74.5, t 13.7 57. [2.1] 36 2 4 5 4 58. [2.1] 14 59. , , , 61. , 3 3 3 3 3 3 3/22k 63. 0 65. e , where k (an integer) 1 67. 1.24 days, 6.76 days 69. 16.5N 71. 1 73. 0.1923 41. 0.967, 1.853, 4.109, 4.994 43. 29. [6.2] tan 2 247 , cos 2 257 , sin 2 24 25 ; I 30. [6.2] cos cot csc 4 , tan 2 5 4 , sec 2 3 3 , 2 4 5 , 2 4 5 ; (c) sin 2 3 2 cos 2 cot 2 csc 2 4 , tan 5 2 4 , sec 3 2 5 3 3 , 4 5 , 4 28. [6.2] sec x 3 , 5 2 31. [6.2] sin 2 0.4261, cos 32. [6.2] cos x 35. [6.2] tan 2x 36. [6.3] 33. [6.2] 1 1 sin x cos x 1 sin x cos x cos x cos x cos x sin x cos x cos2 x Review Exercises: Chapter 6 1. [6.1] csc 2 x 2. [6.1] 1 3. [6.1] tan2 y cot 2 y 2 2 cos x 1 4. [6.1] 5. [6.1] csc x sec x csc x cos2 x 6. [6.1] 3 sin y 5 sin y 4 7. [6.1] 10 cos u 100 10 cos u cos2 u 8. [6.1] 1 3 tan x 1 9. [6.1] 2 sec x 10. [6.1] sin x cos x 3 cos y 3 sin y 2 11. [6.1] 12. [6.1] 1 cos2 y sin2 y 1 13. [6.1] 4 cot x 14. [6.1] sin x cos x cos x cos x 15. [6.1] 16. [6.1] 17. [6.1] 3 sec 1 sin x sin x 3 3 18. [6.1] cos x cos sin x sin 2 2 tan 45 tan 30 19. [6.1] 1 tan 45 tan 30 6 2 20. [6.1] cos 27 16, or cos 11 21. [6.1] 4 22. [6.1] 2 3 23. [6.1] 0.3745 24. [6.2] sin x 25. [6.2] sin x 26. [6.2] cos x 4 4 3 27. [6.2] (a) sin , tan , cot , 5 3 4 5 5 3 sec , csc ; (b) sin , 3 4 2 5 2 2 0.9940, cos 4 0.6369 2 34. [6.2] sin 2x cos x 1 sin x cos x 1 sin x 1 sin x 1 sin x cos x sin x cos x 1 sin2 x cos x sin x cos x cos2 x 37. [6.3] 1 cos 2 sin 2 1 2 cos2 1 2 sin cos cos sin cot cos sin 38. [6.3] 1 2 1 2 tan y sin y 2 tan y sin y sin y cos y cos y sin y cos y cos2 y 2 1 cos y 2 sin y 1 cos y cos y cos y sin y 1 cos y 2 39. [6.3] sin x cos x cos2 x tan2 x 1 sin x cos x sin2 x 1 cos2 x sin x cos x 1 sin2 x cos2 x cos2 x sin x cos x sin x cos x cos2 x BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-51 Chapter 6 40. [6.3] B; csc x cos x cot x cos x 1 cos x sin x sin x 1 cos2 x sin x sin2 x sin x sin x 49. [6.4] 0.3976, 22.8 sin x cos x 1 sin x cos x sin x cos2 x 1 sin x cos x sin x cos x 1 cos2 x sin x cos x sin2 x sin x cos x sin x cos x sin x cos x 1 sin2 x sin x cos x cos2 x sin x cos x 42. [6.3] A; cot x 1 1 tan x cos x sin x sin x sin x cos x sin x cos x cos x cos x cos x sin x sin x cos x sin x cos x sin x csc x sec x 1 sin x 1 cos x 1 cos x sin x 1 cos x sin x 43. [6.3] C; sin x 2 cos x 1 sin x cos x 1 sin x cos x 12 sin2 x sin x cos x 1 cos2 x 2 cos x 1 sin2 x sin x cos x 1 2 cos x 2 sin x cos x 1 2cos x 1 sin x cos x 1 2 sin x 44. [6.4] , 30 45. [6.4] , 30 6 6 46. [6.4] , 45 47. [6.4] 0, 0 48. [6.4] 1.7920, 102.7 4 51. [6.4] 3 3 3 2 54. [6.4] 2 b 2 9 3 5 55. [6.4] 257 56. [6.5] 2k, 2k, or 4 4 135 k 360, 225 k 360 57. [6.5] k, or 60 k 180 3 5 7 11 58. [6.5] , , , 6 6 6 6 3 5 3 7 2 4 59. [6.5] , , , , , 60. [6.5] , , 4 2 4 4 2 4 3 3 3 5 7 61. [6.5] 0, 62. [6.5] , , , 4 4 4 4 3 7 23 63. [6.5] 0, , , 64. [6.5] , 2 2 12 12 65. [6.5] 0.864, 2.972, 4.006, 6.114 66. [6.5] 4.917 67. [6.5] No solution in 0, 2 68. [6.3] Discussion and Writing (a) 2 cos2 x 1 cos 2x cos2 x sin2 x 1 cos2 x sin2 x cos2 x sin2 x cos2 x sin2 x cos4 x sin4 x ; (b) cos4 x sin4 x cos2 x sin2 x cos2 x sin2 x 1 cos2 x sin2 x cos2 x sin2 x cos 2x 2 cos2 x 1; (c) cos4 x sin4 x 2 cos2 x 1 cos 2x cos2 x sin2 x cos2 x sin2 x 1 cos2 x sin2 x cos2 x sin2 x cos 2x Answers may vary. Method 2 may be the more efficient because it involves straightforward factorization and simplification. Method 1(a) requires a “trick” such as multiplying by a particular expression equivalent to 1. 69. Discussion and Writing [6.4] The ranges of the inverse trigonometric functions are restricted in order that they might be functions. 70. [6.1] 108.4 71. [6.1] cos u v cos u cos v sin u sin v u cos v cos u cos v cos 2 2 2 72. [6.2] cos x 1 1 6 6 73. [6.2] sin ; cos ; 2 5 2 5 5 26 tan 5 26 74. [6.3] ln e sin t loge e sin t sin t 52. [6.4] 41. [6.3] D; 7 50. [6.4] 12 A-51 53. [6.4] BBEPMN00_0321279115.QXP A-52 1/7/05 3:54 PM Page A-52 Answers 75. [6.4] 16. [6.3] y p y sec1 1 sin 1 csc x 1 sin 1 1 sin 1 sin sin 1 sin sin q 3 2 1 76. [6.4] Let x sin1 1 cos 1 2 3 x 2 2 . Then tan1 0.6155 and 2 2 4 1. 2 2 4 2 2 77. [6.5] 3 , 2 2 17. [6.4] 45 tan sec sin cos 1 cos sin 18. [6.4] 3 19. [6.4] 2.3072 5 3 21. [6.4] 22. [6.4] 0 2 2 x 25 5 7 11 3 23. [6.5] , , , 24. [6.5] 0, , , 6 6 6 6 4 4 11 25. [6.5] , 26. [6.2] 11 12 2 6 20. [6.4] Test: Chapter 6 1. [6.1] 2 cos x 1 4. [6.1] 2 cos 7. [6.1] 120 169 10. [6.2] 12. 13. 14. 15. 2. [6.1] 1 5. [6.1] 8. [6.2] 2 6 4 5 3 cos 1 sin 3 3 6. [6.1] 3 3 3. [6.1] Chapter 7 9. [6.2] 24 25 , II 2 3 11. [6.2] 0.9304 2 [6.2] 3 sin 2x [6.3] csc x cos x cot x sin x cos x 1 cos x sin x sin x 1 cos2 x sin x sin2 x sin x sin x [6.3] sin x cos x2 2 sin x 2 sin x cos x cos2 x 1 2 sin x cos x 1 sin 2x [6.3] 1 cos csc cot 2 1 cos 2 1 cos 1 cos sin sin 1 cos Exercise Set 7.1 1 sin 2x 2 1 cos 2 1 cos2 1 cos 2 sin2 1 cos 2 sin2 1 cos sin 1 cos 1 cos 1. A 121°, a 33, c 14 3. B 57.4°, C 86.1°, c 40, or B 122.6°, C 20.9°, c 14 5. B 44°24, A 74°26, a 33.3 7. A 110.36°, a 5 mi, b 3 mi 9. B 83.78°, A 12.44°, a 12.30 yd 11. B 14.7°, C 135.0°, c 28.04 cm 13. No solution 15. B 125.27°, b 302 m, c 138 m 17. 8.2 ft 2 19. 12 yd2 21. 596.98 ft 2 23. 787 ft 2 25. About 12.86 ft, or 12 ft, 10 in. 27. About 51 ft 29. From A: about 35 mi; from B: about 66 mi 31. About 22 mi 33. Discussion and Writing 35. [5.1] 1.348, 77.2° 36. [5.1] No angle 37. [5.1] 18.24° 38. [5.1] 125.06° 39. [R.1] 5 3 3 2 40. [5.3] 41. [5.3] 42. [5.3] 2 2 2 BBEPMN00_0321279115.QXP 1/7/05 3:54 PM Page A-53 A-53 Chapters 6 – 7 1 44. [2.2] 2 45. Use the formula for the 2 area of a triangle and the law of sines. 43. [5.3] 1 c sin B bc sin A and b , 2 sin C 2 c sin A sin B so K . 2 sin C 1 a sin B K ab sin C and b , 2 sin A 2 a sin B sin C so K . 2 sin A 1 b sin C K bc sin A and c , 2 sin B 2 b sin A sin C so K . 2 sin B K 47. A a Discussion and Writing 39. [3.1] Quartic [1.3] Linear 41. [5.5] Trigonometric [4.2] Exponential 43. [3.4] Rational [3.1] Cubic 45. [4.2] Exponential [4.3] Logarithmic 47. [5.5] Trigonometric [2.3] Quadratic 49. About 9386 ft 1 51. A a 2 sin ; when 90° 2 Exercise Set 7.3 1. 5; 3. 1; Imaginary Imaginary 4 3i 4 2 2 4 2 b 4 2 4 2 B c u 37. 40. 42. 44. 46. 48. C Real 4 2 2 4 Real 4 Real 2 4 d i 4 5. 17; 7. 3; Imaginary Imaginary D For the quadrilateral ABCD, we have 1 1 Area bd sin ac sin 2 2 1 1 adsin 180° bc sin 180° 2 2 Exercise Set 7.2 1. a 15, B 24°, C 126° 3. A 36.18°, B 43.53°, C 100.29° 5. b 75 m, A 94°51, C 12°29 7. A 24.15°, B 30.75°, C 125.10° 9. No solution 11. A 79.93°, B 53.55°, C 46.52° 13. c 45.17 mi, A 89.3°, B 42.0° 15. a 13.9 in., B 36.127°, C 90.417° 17. Law of sines; C 98°, a 96.7, c 101.9 19. Law of cosines; A 73.71°, B 51.75°, C 54.54° 21. Cannot be solved 23. Law of cosines; A 33.71°, B 107.08°, C 39.21° 25. About 367 ft 27. About 1.5 mi 29. About 37 nautical mi 31. About 912 km 33. (a) About 16 ft; (b) about 122 ft2 35. About 4.7 cm 4 2 2 3 4 2 2 4i 2 Note: sin sin 180° . 1 bd ac ad bc sin 2 1 a b c d sin 2 1 d 1d 2 sin , 2 where d 1 a b and d 2 c d. 4 Real 4 2 2 2 4 4 7 7 i sin , or 4 4 32 cos 315° i sin 315° , or 4cos 90° i sin 90° i sin 11. 4i; 4 cos 2 2 9. 3 3i; 32 cos 13. 2 cos 15. 3 cos 7 7 i sin , or 2cos 315° i sin 315° 4 4 3 3 , or 3cos 270° i sin 270° i sin 2 2 i sin , or 2cos 30° i sin 30° 6 6 2 2 cos 0 i sin 0, or cos 0° i sin 0° 19. 5 5 17. 2 cos 21. 6 cos 23. 5 5 i sin , or 6cos 225° i sin 225° 4 4 3 33 i 2 2 25. 10i 27. 2 2i 29. 2i Summer 2012 L.E.A.D. Ambassador Team 3 (Statistics Course Content) “Algebra 2 and Trigonometry” Ann Xavier Gantert Amsco School Publications, Inc. (2009) 14411C15.pgs 8/14/08 10:36 AM Page 587 CHAPTER STATISTICS Every year the admission officers of colleges choose, from thousands of applicants, those students who will be offered a place in the incoming class for the next year. An attempt is made by the college to choose students who will be best able to succeed academically and who best fit the profile of the student body of that college. Although this choice is based not only on academic standing, the scores on standardized tests are an important part of the selection. Statistics establishes the validity of the information obtained from standardized tests and influence the interpretation of the data obtained from them. 15 CHAPTER TABLE OF CONTENTS 15-1 Gathering Data 15-2 Measures of Central Tendency 15-3 Measures of Central Tendency for Grouped Data 15-4 Measures of Dispersion 15-5 Variance and Standard Deviation 15-6 Normal Distribution 15-7 Bivariate Statistics 15-8 Correlation Coefficient 15-9 Non-Linear Regression 15-10 Interpolation and Extrapolation Chapter Summary Vocabulary Review Exercises Cumulative Review 587 14411C15.pgs 8/14/08 588 10:36 AM Page 588 Statistics 15-1 GATHERING DATA Important choices in our lives are often made by evaluating information, but in order to use information wisely, it is necessary to organize and condense the multitude of facts and figures that can be collected. Statistics is the science that deals with the collection, organization, summarization, and interpretation of related information called data. Univariate statistics consists of one number for each data value. Collection of Data Where do data come from? Individuals, government organizations, businesses, and other political, scientific, and social groups usually keep records of their activities. These records provide factual data. In addition to factual data, the outcome of an event such as an election, the sale of a product, or the success of a movie often depends on the opinion or choices of the public. Common methods of collecting data include the following: 1. Censuses: every ten years, the government conducts a census to determine the U.S. population. Each year, almanacs are published that summarize and update data of general interest. 2. Surveys: written questionnaires, personal interviews, or telephone requests for information can be used when experience, preference, or opinions are sought. 3. Controlled experiments: a structured study that usually consists of two groups: one that makes use of the subject of the study (for example, a new medicine) and a control group that does not. Comparison of results for the two groups is used to indicate effectiveness. 4. Observational studies: similar to controlled experiments except that the researcher does not apply the treatment to the subjects. For example, to determine if a new drug causes cancer, it would be unethical to give the drug to patients. A researcher observes the occurrence of cancer among groups of people who previously took the drug. When information is gathered, it may include data for all cases to which the result of the study is to be applied. This source of information is called the population. When the entire population can be examined, the study is a census. For example, a study on the age and number of accidents of every driver insured by an auto insurance company would constitute a census if all of the company’s records are included. However, when it is not possible to obtain information from every case, a sample is used to determine data that may then be applied to every case. For example, in order to determine the quality of their product, the quality-control department of a business may study a sample of the product being produced. 14411C15.pgs 8/14/08 10:36 AM Page 589 Gathering Data 589 In order that the sample reflect the properties of the whole group, the following conditions should exist: 1. The sample must be representative of the group being studied. 2. The sample must be large enough to be effective. 3. The selection should be random or determined in such a way as to eliminate any bias. If a new medicine being tested is proposed for use by people of all ages, of different ethnic backgrounds, and for use by both men and women, then the sample must be made up of people who represent these differences in sufficient number to be effective. A political survey to be effective must include people of different cultural, ethnic, financial, geographic, and political backgrounds. Potential Pitfalls of Surveys Surveys are a very common way of collecting data. However, if not done correctly, the results of the survey can be invalid. One potential problem is with the wording of the survey questions. For example, the question, “Do you agree that teachers should make more money?” will likely lead to a person answering “Yes.” A more neutral form of this question would be, “Do you believe that teachers’ salaries are too high, too low, or just about right?” Questions can also be too vague, “loaded” (that is, use words with unintended connotations), or confusing. For example, for many people, words that invoke race will likely lead to an emotional response. Another potential problem with surveys is the way that participants are selected. For example, a magazine would like to examine the typical teenager’s opinion on a pop singer in a given city. The magazine editors conduct a survey by going to a local mall. The problem with this survey is that teenagers who go to the mall are not necessarily representative of all teenagers in the city. A better survey would be done by visiting the high schools of the city. Many surveys often rely on volunteers. However, volunteers are likely to have stronger opinions than the general population. This is why the selection of participants, if possible, should be random. EXAMPLE 1 A new medicine intended for use by adults is being tested on five men whose ages are 22, 24, 25, 27, and 30. Does the sample provide a valid test? Solution No: • The sample is too small. • The sample includes only men. • The sample does not include adults over 30. 14411C15.pgs 8/14/08 590 10:36 AM Page 590 Statistics EXAMPLE 2 The management of a health club has received complaints about the temperature of the water in the swimming pool. They want to sample 50 of the 200 members of the club to determine if the temperature of the pool should be changed. How should this sample be chosen? Solution One suggestion might be to poll the first 50 people who use the pool on a given day. However, this will only include people who use the pool and who can therefore tolerate the water temperature. Another suggestion might be to place 50 questionnaires at the entrance desk and request members to respond. However, this includes only people who choose to respond and who therefore may be more interested in a change. A third suggestion might be to contact by phone every fourth person on the membership list and ask for a response. This method will produce a random sample but will include people who have no interest in using the pool. This sample may be improved by eliminating the responses of those people. Organization of Data In order to be more efficiently presented and more easily understood and interpreted, the data collected must be organized and summarized. Charts and graphs such as the histogram are useful tools. The stem-and-leaf diagram is an effective way of organizing small sets of data. For example, the heights of Heights of Children the 20 children in a seventhgrade class are shown to the 61 71 58 72 60 53 74 61 68 65 right. To draw a stem-and-leaf 72 67 64 48 70 56 65 67 59 61 diagram, choose the tens digit as the stem and the units digit as the leaf. (1) Draw a vertical line and list the tens digits, 4, 5, 6, and 7 (or the stem), from bottom to top to the left of the line: Stem (2) Enter each height by writing the leaf, the units digit, to the right of the line, following the appropriate stem: Stem 7 6 5 4 7 6 5 4 Leaf 12420 1018574571 8369 8 14411C15.pgs 8/14/08 10:36 AM Page 591 Gathering Data (3) Arrange the leaves in numerical order after each stem: (4) Add a key to indicate the meaning of the numbers in the diagram: Stem 591 Leaf 7 01224 6 0111455778 5 3689 4 8 Key: 4 8 5 48 For larger sets of data, a frequency distribution table can be drawn. Information is grouped and the frequency, the number of times that a particular value or group of values occurs, is stated for each group. For example, the table to the right lists the scores of 250 students on a test administered to all tenth graders in a school district. The table tells us that 24 students scored between 91 and 100, that 82 scored between 81 and 90, that 77 scored between 71 and 80. The table also tells us that the largest number of students scored in the 80’s and that ten students scored 50 or below. Note that unlike the stem-and-leaf diagram, the table does not give us the individual scores in each interval. Score Frequency 91–100 24 81–90 82 71–80 77 61–70 36 51–60 21 41–50 8 31–40 2 EXAMPLE 2 The prices of a gallon of milk in 15 stores are listed below. $3.15 $3.39 $3.28 $2.98 $3.25 $3.45 $3.58 $3.24 $3.35 $3.29 $3.29 $3.30 $3.25 $3.40 $3.29 a. Organize the data in a stem-and-leaf diagram. b. Display the data in a frequency distribution table. c. If the 15 stores were chosen at random from the more than 100 stores that sell milk in Monroe County, does the data set represent a population or a sample? Solution a. Use the first two digits of the price as the stem. Use the last digit as the leaf. Stem Stem 3.5 3.4 3.3 3.2 3.1 3.0 2.9 3.5 3.4 3.3 3.2 3.1 3.0 2.9 Leaf 8 50 950 8549959 5 8 Write the leaves in numerical order. Stem Leaf 3.5 8 3.4 05 3.3 059 3.2 4558999 3.1 5 3.0 2.9 8 Key: 2.9 8 5 2.98 8/14/08 592 10:36 AM Page 592 Statistics b. Divide the data into groups of length $0.10 starting with $2.90. These groups correspond with the stems of the stem-and-leaf diagram. The frequencies can be determined by the use of a tally to represent each price. Stem Leaf 3.5 8 3.4 05 3.3 059 3.2 4558999 3.1 5 3.0 2.9 8 Key: 2.9 8 5 2.98 Price Tally $3.50–$3.59 | 1 $3.40–$3.49 || 2 $3.30–$3.39 ||| 3 $3.20–$3.29 |||| || 7 $3.10–$3.19 | 1 Frequency $3.00–$3.09 $2.90–$2.99 0 | 1 c. The data set is obtained from a random selection of stores from all of the stores in the study and is therefore a sample. Answer Once the data have been organized, a graph can be used to visualize the intervals and their frequencies. A histogram is a vertical bar graph where each interval is represented by the width of the bar and the frequency of the interval is represented by the height of the bar. The bars are placed next to each other to show that, as one interval ends, the next interval begins. The histogram below shows the data of Example 2: PRICE OF A GALLON OF MILK Frequency 14411C15.pgs 8 7 6 5 4 3 2 1 0 .99 $2 2 –$ 0 9 . .19 .09 $3 .00 3 –$ .10 $3 3 –$ .20 $3 Price .49 .39 .29 3 –$ .30 $3 3 –$ .40 $3 .59 3 –$ .50 $3 3 –$ 14411C15.pgs 8/14/08 10:36 AM Page 593 Gathering Data 593 A graphing calculator can be used to display a histogram from the data on a frequency distribution table. (1) Clear L1 and L2, the lists to be used, of existing data. ENTER: STAT 4 2nd L1 , 2nd 1 to edit the lists. Enter the (2) Press STAT minimum value of each interval in L1 and the frequencies into L2. L2 ENTER L1 L2 3.5 1 3.4 2 3.3 3 3.2 7 3.1 1 3 0 2.9 1 L3(1)= L3 (3) Clear any functions in the Y menu. (4) Turn on Plot1 from the STAT PLOT menu and select for histogram. Make sure to also set Xlist to L1 and Freq to L2. ENTER: 2nd STAT PLOT 䉲 䉴 䉴 L1 䉲 2nd ENTER Plot1 Plot2 Plot3 On Off . Ty p e : ...... ENTER 1 䉲 2nd Xlist:L1 Freq:L2 L2 (5) In the WINDOW menu, enter Xmin as 2.8, the length of one interval less than the smallest interval value, and Xmax as 3.7, the length of one interval more than the largest interval value. Enter Xscl as 0.10, the length of the interval. The Ymin is 0 and Ymax is 9 to be greater than the largest frequency. (6) Press GRAPH to graph the histogram. We can view the frequency associated with each interval by pressing TRACE . Use the left and right arrows to move from one interval to the next. WINDOW Xmin =2.8 Xmax=3.7 Xscl =.1 Ymin =0 Ymax=9 Yscl =1 Xres=1 3 14411C15.pgs 8/14/08 594 10:36 AM Page 594 Statistics Exercises Writing About Mathematics 1. In a controlled experiment, two groups are formed to determine the effectiveness of a new cold remedy. One group takes the medicine and one does not. Explain why the two groups are necessary. 2. In the experiment described in Exercise 1, explain why it is necessary that a participant does not know to which group he or she belongs. Developing Skills In 3–5, organize the data in a stem-and-leaf diagram. 3. The grades on a chemistry test: 95 90 84 85 74 67 78 86 54 82 75 67 92 66 90 68 88 85 76 87 4. The weights of people starting a weight-loss program: 173 210 182 190 175 169 236 192 203 196 201 187 205 195 224 177 195 207 188 184 196 155 5. The heights, in centimeters, of 25 ten-year-old children: 137 134 130 144 131 141 136 140 137 129 139 137 144 127 147 143 132 132 142 142 131 129 138 151 137 In 6–8, organize the data in a frequency distribution table. 6. The numbers of books read during the summer months by each of 25 students: 2 2 5 1 3 0 7 2 4 3 3 1 8 5 7 3 4 1 0 6 3 4 1 1 2 7. The sizes of 26 pairs of jeans sold during a recent sale: 8 12 14 10 12 16 14 6 10 9 8 13 12 8 12 10 12 14 10 12 16 10 11 15 8 14 8. The number of siblings of each of 30 students in a class: 2 1 1 5 1 0 2 2 1 3 4 0 6 2 0 3 1 2 2 1 1 1 0 2 1 0 1 1 2 3 14411C15.pgs 8/14/08 10:36 AM Page 595 595 Gathering Data In 9–11, graph the histogram of each set of data. 9. 10. xi fi xi 35–39 13 101–110 30–34 19 25–29 fi 11. xi fi 3 $55–$59 20 91–100 6 $50–$54 15 10 81–90 10 $45–$49 12 20–24 13 71–80 13 $40–$44 5 15–19 8 61–70 14 $35–$39 10 10–14 19 51–60 2 $30–$34 12 5–9 15 41–50 2 $25–$29 16 Applying Skills In 12–18, suggest a method that might be used to collect data for each study. Tell whether your method uses a population or a sample. 12. Average temperature for each month for a given city 13. Customer satisfaction at a restaurant 14. Temperature of a patient in a hospital over a period of time 15. Grades for students on a test 16. Population of each of the states of the United States 17. Heights of children entering kindergarten 18. Popularity of a new movie 19. The grades on a math test of 25 students are listed below. 86 92 77 84 75 95 66 88 84 53 98 87 83 74 61 82 93 98 87 77 86 58 72 76 89 a. Organize the data in a stem-and-leaf diagram. b. Organize the data in a frequency distribution table. c. How many students scored 70 or above on the test? d. How many students scored 60 or below on the test? 20. The stem-and-leaf diagram at the right shows the ages of 30 people in an exercise class. Use the diagram to answer the following questions. a. How many people are 45 years old? b. How many people are older than 60? c. How many people are younger than 30? d. What is the age of the oldest person in the class? e. What is the age of the youngest person in the class? Stem Leaf 7 2 6 015 5 136679 4 2245567 3 9 2 1335788 1 02269 Key: 1 9 5 19 14411C15.pgs 8/14/08 596 10:36 AM Page 596 Statistics Hands-On Activity In this activity, you will take a survey of 25 people. You will need a stopwatch or a clock with a second hand. Perform the following experiment with each participant to determine how each perceives the length of a minute: 1. Indicate the starting point of the minute. 2. Have the person tell you when he or she believes that a minute has passed. 3. Record the actual number of seconds that have passed. After surveying all 25 participants, use a stem-and-leaf diagram to record your data. Keep this data. You will use it throughout your study of this chapter. 15-2 MEASURES OF CENTRAL TENDENCY After data have been collected, it is often useful to represent the data by a single value that in some way seems to represent all of the data. This number is called the measure of central tendency. The most frequently used measures of central tendency are the mean, the median, and the mode. The Mean The mean or arithmetic mean is the most common measure of central tendency. The mean is the sum of all of the data values divided by the number of data values. For example, nine members of the basketball team played during all or part of the last game. The number of points scored by each of the players was: 21, 15, 12, 9, 8, 7, 5, 2, 2 Mean 5 21 1 15 1 12 1 9 19 8 1 7 1 5 1 2 1 2 5 81 9 59 Note that if each of the 9 players had scored 9 points, the total number of points scored in the game would have been the same. The summation symbol, S, is often used to designate the sum of the data values. We designate a data value as xi and the sum of n data values as n c1x n a xi 5 x1 1 x2 1 x3 1 i51 For the set of data given above: n 5 9, x1 5 21, x2 5 15, x3 5 12, x4 5 9, x5 5 8, x6 5 7, x7 5 5, x8 5 2, x9 5 2 9 a xi 5 81 i51 The subscript for each data value indicates its position in a list of data values, not its value, although the value of i and the data value may be the same. 14411C15.pgs 8/14/08 10:36 AM Page 597 597 Measures of Central Tendency Procedure To find the mean of a set of data: 1. Add the data values. 2. Divide the sum by n, the total number of data values. EXAMPLE 1 An English teacher recorded the number of spelling errors in the 40 essays written by students. The table below shows the number of spelling errors and the frequency of that number of errors, that is, the number of essays that contained that number of misspellings. Find the mean number of spelling errors for these essays. Errors 0 1 2 3 4 5 6 7 8 9 10 Frequency 1 3 2 2 6 9 7 5 2 1 2 Solution To find the total number of spelling errors, first multiply each number of errors by the frequency with which that number of errors occurred. For example, since 2 essays each contained 10 errors, there were 20 errors in these essays. Add the products in the fi xi row to find the total number of errors in the essays. Divide this total by the total frequency. Total Errors (xi) 0 1 2 3 4 5 6 7 8 9 10 Frequency (fi) 1 3 2 2 6 9 7 5 2 1 2 40 Errors Frequency (fixi) 0 3 4 6 24 45 42 35 16 9 20 204 10 a fixi Mean 5 i50 40 5 204 40 5 5.1 Note that for this set of data, the data value is equal to i for each xi. Answer 5.1 errors The Median The median is the middle number of a data set arranged in numerical order. When the data are arranged in numerical order, the number of values less than or equal to the median is equal to the number of values greater than or equal to 8/14/08 Page 598 Statistics the median. Consider again the nine members of the basketball team who played during all or part of the last game and scored the following number of points: 2, 2, 5, 7, 8, 9, 12, 15, 21 We can write 9 5 4 1 1 1 4. Therefore, we think of four scores below the median, the median, and four scores above the median. 8, scores less than the median ↑ 9, 12, 15, 21 d 2, 2, 5, 7, d scores greater than the median median Note that the number of data values can be written as 2(4) 1 1. The median is the (4 1 1) or 5th value from either end of the distribution. The median number of points scored is 8. When the number of values in a set of data is even, then the median is the mean of the two middle values. For example, eight members of a basketball team played in a game and scored the following numbers of points. 4, 6, 6, 7, 11, 12, 18, 20 We can separate the eight data values into two groups of 4 values. Therefore, we average the largest score of the four lowest scores and the smallest score of the four highest scores. 4, 6, 6, 7, four lowest scores 7 1 11 2 ↑ , 11, 12, 18, 20 d 598 10:36 AM d 14411C15.pgs four highest scores median median 5 7 12 11 5 9 Note that the number of data values can be written as 2(4). The median is the mean of the 4th and 5th value from either end of the distribution. The median is 9. There are four scores greater than the median and four scores lower than the median. The median is a middle mark. Procedure To find the median of a set of data: 1. Arrange the data in order from largest to smallest or from smallest to largest. 2. a. If the number of data values is odd, write that number as 2n 1 1.The median is the score that is the (n 1 1)th score from either end of the distribution. b. If the number of data values is even, write that number as 2n.The median is the score that is the mean of the nth score and the (n 1 1)th score from either end of the distribution. 10:36 AM Page 599 Measures of Central Tendency 599 The Mode The mode is the value or values that occur most frequently in a set of data. For example, in the set of numbers 3, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 10, the number that occurs most frequently is 6. Therefore, 6 is the mode for this set of data. When a set of numbers is arranged in a Number of frequency distribution table, the mode is the Books Read Frequency entry with the highest frequency. The table to 8 1 the right shows, for a given month, the number of books read by each student in a class. 7 1 The largest number of students, 12, each 6 3 read four books. The mode for this distribu5 6 tion is 4. A data set may have more than one 4 12 mode. For example, in the set of numbers 3, 4, 3 4 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 10, the numbers 5 and 2 2 6 each occur four times, more frequently than any other number. Therefore, 5 and 6 are 1 0 modes for this set of data. The set of data is 0 1 said to be bimodal. Quartiles When a set of data is listed in numerical order, the median separates the data into two equal parts. The quartiles separate the data into four equal parts. To find the quartiles, we first separate the data into two equal parts and then separate each of these parts into two equal parts. For example, the grades of 20 students on a math test are listed below. 58, 60, 65, 70, 72, 75, 76, 80, 80, 81, 83, 84, 85, 87, 88, 88, 90, 93, 95, 98 Since there are 20 or 2(10) grades, the median grade that separates the data into two equal parts is the average of the 10th and 11th grade. 58, 60, 65, 70, 72, 75, 76, 80, 80, 81, Lower half 83, 84, 85, 87, 88, 88, 90, 93, 95, 98 ↑ 82 i 8/14/08 i 14411C15.pgs Upper half Median The 10th grade is 81 and the 11th grade is 83. Therefore, the mean of these 83 two grades, 81 1 or 82, separates the data into two equal parts. This number 2 is the median grade. 8/14/08 Page 600 Statistics Now separate each half into two equal parts. Find the median of the two lower quarters and the median of the two upper quarters. D D ↑ 73.5 D 58, 60, 65, 70, 72 75, 76, 80, 80, 81 83, 84, 85, 87, 88 88, 90, 93, 95, 98 D ↑ 82 ↑ 88 The numbers 73.5, 82, and 88 are the quartiles for this data. • One quarter of the grades are less than or equal to 73.5. Therefore, 73.5 is the first quartile or the lower quartile. • Two quarters of the grades are less than or equal to 82. Therefore, 82 is the second quartile. The second quartile is always the median. • Three quarters of the grades are less than or equal to 88. Therefore, 88 is the third quartile or the upper quartile. Note: The minimum, first quartile, median, third quartile, and maximum make up the five statistical summary of a data set. When the data set has an odd number of values, the median or second quartile will be one of the values. This number is not included in either half of the data when finding the first and third quartiles. For example, the heights, in inches, of 19 children are given below: 37, 39, 40, 42, 42, 43, 44, 44, 44, 45, 46, 47, 47, 48, 49, 49, 50, 52, 53 There are 19 5 2(9) 1 1 data values. Therefore, the second quartile is the 10th height or 45. 45 , 46, 47, 47, 48, 49, 49, 50, 52, 53 I 37, 39, 40, 42, 42, 43, 44, 44, 44, I There are 9 5 2(4) 1 1 heights in the lower half of the data and also in the upper half of the data. The middle height of each half is the 5th height. 45 , 46, 47, 47, 48, 49 , 49, 50, 52, 53 D 42 , 43, 44, 44, 44, D 37, 39, 40, 42, D 600 10:36 AM D 14411C15.pgs ↑ ↑ ↑ Lower quartile Median Upper quartile In this example, the first or lower quartile is 42, the median or second quartile is 45, and the third or upper quartile is 49. Each of these values is one of the data values, and the remaining values are separated into four groups with the same number of heights in each group. Box-and-Whisker Plot A box-and-whisker plot is a diagram that is used to display the quartile values and the maximum and minimum values of a distribution. We will use the data from the set of heights given above. 10:36 AM Page 601 Measures of Central Tendency 601 49 , 49, 50, 52, 53 D 45 , 46, 47, 47, 48, D 42 , 43, 44, 44, 44, 37, 39, 40, 42, D 8/14/08 D 14411C15.pgs 1. Choose a scale that includes the maximum and minimum values of the data. We will use a scale from 35 to 55. 2. Above the scale, place dots to represent the minimum value, the lower quartile, the mean, the upper quartile, and the maximum value. 35 42 37 45 49 53 55 3. Draw a box with opposite sides through the lower and upper quartiles and a vertical line through the median. 35 42 37 45 49 53 55 4. Draw whiskers by drawing a line to join the dot that represents the minimum value to the dot that represents the lower quartile and a line to join the dot that represents the upper quartile to the dot that represents the maximum value. 35 42 37 45 49 53 55 A graphing calculator can be used to find the quartiles and to display the box-and-whisker plot. (1) Enter the data for this set of heights into L1. (2) Turn off all plots and enter the required choices in Plot 1. ENTER: 䉲 2nd 2nd STAT PLOT 䉴 L1 䉴 䉴 䉲 ALPHA On Off . Ty p e : ...... ENTER 1 䉴 Plot1 Plot2 Plot3 ENTER 1 䉲 Xlist:L1 Freq:1 8/14/08 602 10:36 AM Page 602 Statistics (3) Now display the plot by entering 9 . You can press TRACE ZOOM to display the five statistical summary. The five statistical summary of a set of data can also be displayed on the calculator by using 1-Var Stats. ENTER: STAT 䉴 ENTER ENTER The first entry under 1-Var Stats is –x, the value of the mean. The next is a x, the sum of the data values. The next three entries are values that will be used in the sections that follow. The last entry is the number of data values. The arrow tells us that there is more information. Scroll down to display the minimum value, minX 5 37, the lower quartile, Q1 5 42, the median or second quartile, Med 5 45, the upper quartile, Q3 5 49, and the maximum value, maxX 5 53. > 1–Var Stats – x=45.31578947 x=861 x2=39353 Sx=4.321170515 X=4.205918531 n=19 1–Var Stats n=19 minX=37 Q1=42 Med=45 Q3=49 maxX=53 > 14411C15.pgs EXAMPLE 1 Find the mean, the median, and the mode of the following set of grades: 92, 90, 90, 90, 88, 87, 85, 70 n Solution Mean 5 n1 a xi 5 18(92 1 90 1 90 1 90 1 88 1 87 1 85 1 70) 5 692 8 5 86.5 i51 88 Median 5 the average of the 4th and 5th grades 5 90 1 5 89 2 Mode 5 the grade that appears most frequently 5 90 EXAMPLE 2 The following list shows the length of time, in minutes, for each of 35 employees to commute to work. 25 12 20 18 35 25 40 35 27 30 60 22 36 20 18 27 35 42 35 55 27 30 15 22 10 35 27 15 57 18 25 45 24 27 25 14411C15.pgs 8/14/08 10:36 AM Page 603 Measures of Central Tendency 603 a. Organize the data in a stem-and-leaf diagram. b. Find the median, lower quartile, and upper quartile. c. Draw a box-and-whisker plot. Solution a. (1) Choose the tens digit as the stem and enter the units digit as the leaf for each value. Stem (2) Write the leaves in numerical order from smallest to largest. Stem b. (1) The median is the middle value of the 35 data values when the values are arranged in order. Stem 35 5 2(17) 1 1 The median is the 18th value. Separate the data into groups of 17 from each end of the distribution. The median is 27. (2) There are 17 5 2(8) 1 1 data values below the median. The lower quartile is the 9th data value from the lower end. The upper quartile is the 9th data value from the upper end. 6 5 4 3 2 1 Leaf 0 57 025 55065505 50572077275475 2885058 Leaf 6 0 5 57 4 025 3 00555556 2 00224555577777 1 0255888 Key: 1 0 5 10 6 5 4 3 2 1 Stem 6 5 4 3 2 1 Leaf 0 57 025 005 5 5556 002 2 455557 7 777 0255888 Leaf 0 57 025 005 5 5 5 56 0 0 2 2 455557 7 777 0255888 The lower quartile is 20, the median is 27, and the upper quartile is 35. c. 10 20 27 35 60 14411C15.pgs 8/14/08 604 10:36 AM Page 604 Statistics Note: In the stem-and-leaf diagram of Example 2, the list is read from left to right to find the lower quartile. The list is read from right to left (starting from the top) to find the upper quartile. Exercises Writing About Mathematics 1. Cameron said that the number of data values of any set of data that are less than the lower quartile or greater than the upper quartile is exactly 50% of the number of data values. Do you agree with Cameron? Explain why or why not. 2. Carlos said that for a set of 2n data values or of 2n 1 1 data values, the lower quartile is the median of the smallest n values and the upper quartile is the median of the largest n values. Do you agree with Carlos? Explain why or why not. Developing Skills In 3–8, find the mean, the median, and the mode of each set of data. 3. Grades: 74, 78, 78, 80, 80, 80, 82, 88, 90 4. Heights: 60, 62, 63, 63, 64, 65, 66, 68, 68, 68, 70, 75 5. Weights: 110, 112, 113, 115, 115, 116, 118, 118, 125, 134, 145, 148 6. Number of student absences: 0, 0, 0, 1, 1, 2, 2, 2, 3, 4, 5, 9 7. Hourly wages: $6.90, $7.10, $7.50, $7.50, $8.25, $9.30, $9.50, $10.00 8. Tips: $1.00, $1.50, $2.25, $3.00, $3.30, $3.50, $4.00, $4.75, $5.00, $5.00, $5.00 In 9–14, find the median and the first and third quartiles for each set of data values. 9. 2, 3, 5, 8, 9, 11, 15, 16, 17, 20, 22, 23, 25 10. 34, 35, 35, 36, 38, 40, 42, 43, 43, 43, 44, 46, 48, 50 11. 23, 27, 15, 38, 12, 17, 22, 39, 28, 20, 27, 18, 25, 28, 30, 29 12. 92, 86, 77, 85, 88, 90, 81, 83, 95, 76, 65, 88, 91, 81, 88, 87, 95 13. 75, 72, 69, 68, 66, 65, 64, 63, 63, 61, 60, 59, 59, 58, 56, 54, 52, 50 14. 32, 32, 30, 30, 29, 27, 26, 22, 20, 20, 19, 18, 17 15. A student received the following grades on six tests: 90, 92, 92, 95, 95, x. a. For what value(s) of x will the set of grades have no mode? b. For what value(s) of x will the set of grades have only one mode? c. For what value(s) of x will the set of grades be bimodal? 16. What are the first, second, and third quartiles for the set of integers from 1 to 100? 17. What are the first, second, and third quartiles for the set of integers from 0 to 100? 14411C15.pgs 8/14/08 10:36 AM Page 605 Measures of Central Tendency for Grouped Data Applying Skills 18. The grades on a English test are shown in the stem-and-leaf diagram to the right. a. Find the mean grade. b. Find the median grade. c. Find the first and third quartiles. d. Draw a box-and-whisker plot for this data. Stem 605 Leaf 9 001259 8 022555788 7 35667 6 055 5 5 4 7 Key: 4 7 5 47 19. The weights in pounds of the members of the football team are shown below: 181 199 178 203 211 208 209 202 212 194 185 208 223 206 202 213 202 186 189 203 a. Find the mean. b. Find the median. c. Find the mode or modes. d. Find the first and third quartiles. e. Draw a box-and-whisker plot. 20. Mrs. Gillis gave a test to her two classes of algebra. The mean grade for her class of 20 students was 86 and the mean grade of her class of 15 students was 79. What is the mean grade when she combines the grades of both classes? Hands-On Activity Use the estimates of a minute collected in the Hands-On Activity of the previous section to determine the five statistical summary for your data. Draw a box-and-whisker plot to display the data. 15-3 MEASURES OF CENTRAL TENDENCY FOR GROUPED DATA Most statistical studies involve much larger numbers of data values than can be conveniently displayed in a list showing each data value. Large sets of data are usually organized into a frequency distribution table. Frequency Distribution Tables for Individual Data Values A frequency distribution table records the individual data values and the frequency or number of times that the data value occurs in the data set. The example on page 606 illustrates this method of recording data. Each of an English teacher’s 100 students recently completed a book report. The teacher recorded the number of misspelled words in each report. The table 14411C15.pgs 8/14/08 606 10:36 AM Page 606 Statistics records the number of reports for each number of misspelled words. Let xi represent the number of misspelled words in a report and fi represent the number of reports that contain xi misspelled words. Total xi 0 1 2 3 4 5 6 7 8 9 10 fi 5 7 6 8 19 26 16 7 5 0 1 100 xifi 0 7 12 24 76 130 96 49 40 0 10 444 The mean of this set of data is the total number of misspelled words divided by the number of reports. To find the total number of misspelled words, we must first multiply each number of misspelled words, xi, by the number of reports that contain that number of misspelled words, fi. That is, we must find xi fi for each number of misspelled words. 10 For this set of data, when we sum the fi row, a fi 5 100, and when we sum i50 10 the xi fi row a xifi 5 444. i50 10 a xi fi Mean 5 i50 10 a fi 5 444 100 5 4.44 i50 To find the median and the quartiles for this set of data, we will find the cumulative frequency for each number of misspelled words. The cumulative frequency is the accumulation or the sum of all frequencies less than or equal to a given frequency. For example, the cumulative frequency for 3 misspelled words on a report is the sum of the frequencies for 3 or fewer misspelled words, and the cumulative frequency for 6 misspelled words on a report is the sum of the frequencies for 6 or fewer misspelled words. The third row of the table shows that 0 misspelled words occur 5 times, 1 or fewer occur 7 1 5 or 12 times, 2 or fewer occur 6 1 12 or 18 times. In each case, the cumulative frequency for xi is the frequency for xi plus the cumulative frequency for xi21. The cumulative frequency for the largest data value is always equal to the total number of data values. xi 0 1 2 3 4 5 6 7 8 9 10 fi 5 7 6 8 19 26 16 7 5 0 1 Cumulative Frequency 5 12 18 26 45 71 87 94 99 99 100 ↑ 3 Lower quartile ↑ 5 ↑ 6 Median Upper quartile 14411C15.pgs 8/14/08 10:36 AM Page 607 Measures of Central Tendency for Grouped Data 607 Since there are 100 data values, the median is the average of the 50th and 51st values. To find these values, look at the cumulative frequency column. There are 45 values less than or equal to 4 and 71 less than or equal to 5. Therefore, the 50th and 51st values are both 5 and the median is 5 misspelled words. Similarly, the upper quartile is the average of the 75th and 76th values. Since there are 71 values less than or equal to 5, the 75th and 76th values are both 6 misspelled words. The lower quartile is the average of the 25th and 26th values. Since there are 18 values less than or equal to 2, the 25th and 26th values are both 3 misspelled words. Percentiles A percentile is a number that tells us what percent of the total number of data values lie at or below a given measure. For example, let us use the data from the previous section. The table records the number of reports, fi, that contain each number of misspelled words, xi , on 100 essays. xi 0 1 2 3 4 5 6 7 8 9 10 fi 5 7 6 8 19 26 16 7 5 0 1 Cumulative Frequency 5 12 18 26 45 71 87 94 99 99 100 To find the percentile rank of 7 misspelled words, first find the number of essays with fewer than 7 misspelled words, 87. Add to this half of the essays with 7 misspelled words, 72 or 3.5. Add these two numbers and divide the sum by the number of essays, 100. 87 1 3.5 100 5 90.5 100 5 90.5% Percentiles are usually not written with fractions. We say that 7 misspelled words is at the 90.5th or 91st percentile. That is, 91% of the essays had 7 or fewer misspelled words. Frequency Distribution Tables for Grouped Data Often the number of different data values in a set of data is too large to list each data value separately in a frequency distribution table. In this case, it is useful to list the data in terms of groups of data values rather than in terms of individual data values. The following example illustrates this method of recording data. There are 50 members of a weight-loss program. The weights range from 181 to 285 pounds. It is convenient to arrange these weights in groups of 10 pounds starting with 180–189 and ending with 280–289. The frequency distribution table shows the frequencies of the weights for each interval. 14411C15.pgs 8/14/08 608 10:36 AM Page 608 Statistics Weights Midpoint xi Frequency fi 280–289 270–279 260–269 250–259 240–249 230–239 220–229 210–219 200–209 190–199 180–189 284.5 274.5 264.5 254.5 244.5 234.5 224.5 214.5 204.5 194.5 184.5 1 3 4 8 12 10 5 3 2 1 1 Cumulative Frequency xifi 284.5 823.5 1,058.0 2,036.0 2,934.0 2,345.0 1,122.5 643.5 409.0 194.5 184.5 50 49 46 42 34 22 12 7 4 2 1 In order to find the mean, assume that the weights are evenly distributed throughout the interval. The mean is found by using the midpoint of the weight intervals as representative of each value in the interval groupings. 11 a xi fi Mean 5 i51 11 a fi 12,035 5 50 5 240.7 i51 To find the median for a set of data that is organized in intervals greater than 1, first find 50 12,035 the interval in which the median lies by using the cumulative frequency. There are 50 data values. Therefore, the median is the value between the 25th and the 26th values. The cumulative frequency tells us that there are 22 values less than or equal to 239 and 34 values less than or equal to 249. Therefore, the 25th and 26th values are in the interval 240–249. Can we give a better approximation for the median? The endpoints of an interval are the lowest and highest data values to be entered in that interval. The boundary values are the values that separate intervals. The lower boundary of an interval is midway between the lower endpoint of the interval and the upper endpoint of the interval that precedes it. The lower boundary of the 240–249 is midway between 240 and 239, that is, 239.5. The upper endpoint of this interval is between 249 and 250, that is 249.5. Since there are 34 weights less than or equal to 249 and 22 weights less than 240, the weights in the 240–249 interval are the 23rd through the 34th weights. Think of these 12 weights as being evenly spaced throughout the interval. median 23 24 25 26 27 29 28 30 31 32 239.5 33 34 249.5 3 12 3 The midpoint between the 25th and 26th weights is 12 of the distance between the boundaries of the interval, a difference of 10. 3 Median 5 239.5 1 12 (10) 5 239.5 1 2.5 5 242 The estimated median for the weights is 242 pounds. Thus, if we assume that the data values are evenly distributed within each interval, we can obtain a better approximation for the median. 14411C15.pgs 8/14/08 10:36 AM Page 609 Measures of Central Tendency for Grouped Data 609 EXAMPLE 1 The numbers of pets owned by the children in a sixth-grade class are given in the table. No. of Pets 6 5 4 3 2 1 0 a. Find the mean. b. Find the median. c. Find the mode for this set of data. d. Find the percentile rank of 4 pets. Frequency 2 1 3 5 8 14 7 Solution a. Add to the table the number of pets times the frequency, and the cumulative frequency for each row of the table. No. of Pets xi Frequency fi fi xi Cumulative Frequency 6 5 4 3 2 1 0 2 1 3 5 8 14 7 12 5 12 15 16 14 0 40 38 37 34 29 21 7 40 74 Add the numbers in the fixi column and the numbers in the fi column. 6 6 a fi 5 40 a fixi 5 74 i50 i50 6 a xi fi Mean 5 i50 6 a fi 5 74 40 5 1.85 i50 b. There are 40 data values in this set of data. The median is the average of the 20th and the 21st data values. Since there are 21 students who own 1 or fewer pets, both the 20th and the 21st data value is 1. Therefore, the median number of pets is 1. c. The largest number of students, 14, have 1 pet. The mode is 1. d. There are 34 students with fewer than 4 pets and 3 students with 4 pets. Add 34 and half of 3 and divide the sum by the total number of students, 40. 34 1 32 40 5 35.5 40 5 0.8875 5 88.75% Four pets represents the 89th percentile. 8/14/08 610 10:37 AM Page 610 Statistics Calculator Clear the lists first if necessary by keying in STAT Solution for 2nd ENTER . L2 a, b, and c 4 2nd L2 ENTER L1 , Enter the number of pets in L1. Enter the frequency for each number of pets in L2. Display 1-Var Stats. ENTER: DISPLAY: STAT 䉴 2nd 1 L1 , 2nd 1–Var Stats – x=1.85 x=74 x2=236 Sx=1.594059485 x=1.574007624 n=40 1–Var Stats n=40 minX=0 Q1=1 Med=1 Q3=3 maxX=6 > 14411C15.pgs > – , is the mean, 1.85. Use the down-arrow key, The first entry, x the median, Med. The median is 1. 6 䉲 , to display 6 Note that a x 5 74 is a fixi and n 5 40 is a fi. i50 i50 Answers a. mean 5 1.85 b. median 5 1 c. mode 5 1 d. 4 is the 89th percentile Note: The other information displayed under 1-Var Stats will be used in the sections that follow. EXAMPLE 2 A local business made the summary of the ages of 45 employees shown below. Find the mean and the median age of the employees to the nearest integer. Age Frequency 20–24 25–29 30–34 35–39 40–44 45–49 1 2 5 6 7 10 50–54 55–59 60–64 7 5 Solution Add to the table the midpoint, the midpoint times the frequency, and the cumulative frequency for each interval. 2 14411C15.pgs 8/14/08 10:37 AM Page 611 Measures of Central Tendency for Grouped Data Age Midpoint (xi) Frequency (fi) xifi Cumulative Frequency 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 62 57 52 47 42 37 32 27 22 2 5 7 10 7 6 5 2 1 124 285 364 470 294 222 160 54 22 45 43 38 31 21 14 8 3 1 45 1,995 611 11 a xi fi Mean 5 i51 11 a fi 1,995 5 45 5 44.333 . . . 44 i51 The mean is the middle age of 45 ages or the 23rd age. Since there are 21 ages less than 45, the 23rd age is the second of 10 ages in the 45–49 interval. The boundaries of the 45–49 interval are 44.5 to 49.5 median Median 5 44.5 1 112 10 35 5 44.5 1 0.75 5 45.25 45 22 44.5 1 12 23 24 25 26 49.5 10 Answer mean 5 44, median 5 45 Exercises Writing About Mathematics 1. Adelaide said that since, in Example 2, there are 10 employees whose ages are in the 45–49 interval, there must be two employees of age 45. Do you agree with Adelaide? Explain why or why not. 2. Gail said that since, in Example 2, there are 10 employees whose ages are in the 45–49 interval, there must at least two employees who are the same age. Do you agree with Gail? Explain why or why not. 14411C15.pgs 8/14/08 612 10:37 AM Page 612 Statistics Developing Skills In 3–8, find the mean, the median, and the mode for each set of data. 3. 6. xi fi 5 4 3 2 1 0 6 10 15 11 2 1 xi fi 10 9 8 7 6 5 4 1 1 3 7 6 2 2 4. 7. xi fi 50 40 30 20 10 8 12 17 10 3 5. xi fi $1.10 $1.20 $1.30 $1.40 $1.50 1 5 8 6 6 xi fi 12 11 10 9 8 7 6 5 7 15 13 16 14 15 9 2 xi fi 95 90 85 80 75 70 65 60 2 8 12 10 9 3 0 1 8. 9. Find the percentile rank of 2 for the data in Exercise 3. 10. Find the percentile rank of 20 for the data in Exercise 4. 11. Find the percentile rank of 8 for the data in Exercise 5. 12. Find the percentile rank of 6 for the data in Exercise 6. In 13–18, find the mean and the median for each set of data to the nearest tenth. 13. xi fi 21–25 16–20 11–15 6–10 1–5 2 3 12 6 1 14. xi fi 91–100 5 81–90 8 71–80 10 61–70 6 51–60 0 41–50 1 15. xi fi $1.51–$1.60 $1.41–$1.50 $1.31–$1.40 $1.21–$1.30 $1.11–$1.20 $1.01–$1.10 2 5 14 4 2 3 14411C15.pgs 8/14/08 10:37 AM Page 613 Measures of Central Tendency for Grouped Data 16. xi fi 17–19 14–16 11–13 8–10 5–7 20 27 32 39 32 17. 18. xi fi $60–$69 $50–$59 $40–$49 $30–$39 $20–$29 $10–$19 16 5 16 2 5 37 613 xi fi 0.151–0.160 0.141–0.150 0.131–0.140 0.121–0.130 0.111–0.120 16 5 0 6 0 Applying Skills 19. The table shows the number of correct answers on a test consisting of 15 questions. Find the mean, the median, and the mode for the number of correct answers. Correct Answers 6 7 8 9 10 11 12 13 14 15 Frequency 1 0 1 3 5 8 9 6 5 2 20. The ages of students in a calculus class at a high school are shown in the table. Find the mean and median age. Age Frequency 19 18 17 16 15 2 8 9 1 1 21. Each time Mrs. Taggart fills the tank of her car, she estimates, from the number of miles driven and the number of gallons of gasoline needed to fill the tank, the fuel efficiency of her car, that is, the number of miles per gallon. The table shows the result of the last 20 times that she filled the car. a. Find the mean and the median fuel efficiency (miles per gallon) for her car. b. Find the percentile rank of 34 miles per gallon. Miles per Gallon 32 33 34 35 36 37 38 39 40 Frequency 1 3 2 5 3 3 2 0 1 22. The table shows the initial weights of people enrolled in a weight-loss program. Find the mean and median weight. Weight Frequency Weight Frequency 191–200 201–210 211–220 221–230 231–240 241–250 1 1 5 7 10 12 251–260 261–270 271–280 281–290 291–300 13 16 8 5 2 14411C15.pgs 8/14/08 614 10:37 AM Page 614 Statistics 23. In order to improve customer relations, an auto-insurance company surveyed 100 people to determine the length of time needed to complete a report form following an auto accident. The result of the survey is summarized in the following table showing the number of minutes needed to complete the form. Find the mean and median amount of time needed to complete the form. Minutes Frequency 26–30 31–35 36–40 41–45 46–50 51–55 56–60 61–65 66–70 2 8 12 15 10 24 26 1 2 Hands-On Activity Organize the data from the survey in the Hands-On Activity of Section 15-1 using intervals of five seconds. Use the table to find the mean number of seconds. Compare this result with the mean found using the individual data values. 15-4 MEASURES OF DISPERSION The mean and the median of a set of data help us to describe a set of data. However, the mean and the median do not always give us enough information to draw meaningful conclusions about the data. For example, consider the following sets of data. Ages of students on a middle-school basketball team Ages of students in a community center tutoring program 11 12 12 13 13 13 13 13 14 14 14 14 6 8 9 9 10 13 13 15 17 18 19 19 The mean of both sets of data is 13 and the median of both sets of data is 13, but the two sets of data are quite different. We need a measure that indicates how the individual data values are scattered or spread on either side of the mean. A number that indicates the variation of the data values about the mean is called a measure of dispersion. Range The simplest of the measures of dispersion is called the range. The range is the difference between the highest value and the lowest value of a set of data. In the sets of data given above, the range of ages of students on the middle-school basketball team is 14 2 11 or 3 and the range of the ages of the students in the community center tutoring program is 19 2 6 or 13. The difference in the ranges indicates that the ages of the students on the basketball team are more closely grouped about the mean than the ages of the students in the tutoring program. 14411C15.pgs 8/14/08 10:37 AM Page 615 Measures of Dispersion 615 The range is dependent on only the largest data value and the smallest data value. Therefore, the range can be very misleading. For example consider the following sets of data: Ages of members of the chess club: 11, 11, 11, 11, 15, 19, 19, 19, 19 Ages of the members of the math club: 11, 12, 13, 14, 15, 16, 17, 18, 19 For each of these sets of data, the mean is 15, the median is 15, and the range is 8. But the sets of data are very different. The range often does not tell us critical information about a set of data. Interquartile Range Another measure of dispersion depends on the first and third quartiles of a distribution. The difference between the first and third quartile values is the interquartile range. The interquartile range tells us the range of at least 50% of the data. The largest and smallest values of a set of data are often not representative of the rest of the data. The interquartile range better represents the spread of the data. It also gives us a measure for identifying extreme data values, that is, those that differ significantly from the rest of the data. For example, consider the ages of the members of a book club. 21, 24, 25, 27, 28, 31, 35, 35, 37, 39, 40, 41, 69 ↑ ↑ ↑ 26 35 39.5 median Q3 Q1 For these 13 data values, the median is the 7th value. For the six values below the median, the first quartile is the average of the 3rd and 4th values from 27 the lower end: 25 1 5 26. The third quartile is the average of the 3rd and 2 40 4th values from the upper end: 39 1 5 39.5. The interquartile range of 2 the ages of the members of the book club is 39.5 2 26 or 13.5. The age of the oldest member of the club differs significantly from the ages of the others. It is more than 1.5 times the interquartile range above the upper quartile: 69 . 39.5 1 1.5(13.5) We call this data value an outlier. DEFINITION An outlier is a data value that is greater than the upper quartile plus 1.5 times the interquartile range or less than the lower quartile minus 1.5 times the interquartile range. When we draw the box-and-whisker plot for a set of data, the outlier is indicated by a ✸ and the whisker is drawn to the largest or smallest data value that is not an outlier. The box-and-whisker plot for the ages of the members of the book club is shown on page 616. 14411C15.pgs 8/14/08 616 10:37 AM Page 616 Statistics * 21 26 39.5 40 35 69 We can use the graphing calculator to graph a box-and-whisker plot with outliers. From the STAT PLOT menu, choose the ı-▫-ı -- option. For example, with the book club data entered into L1, the following keystrokes will graph a box-and-whisker plot with outliers. ENTER: 2nd STAT PLOT 1 䉴 ENTER 䉲 䉴 ENTER 䉲 2nd 䉲 ALPHA DISPLAY: 䉴 L1 1 ZOOM 9 EXAMPLE 1 The table shows the number of minutes, rounded to the nearest 5 minutes, needed for each of 100 people to complete a survey. a. Find the range and the interquartile range for this set of data. b. Does this data set include outliers? Minutes 30 35 40 45 50 55 60 65 70 85 Frequency 3 8 12 15 10 24 17 8 2 1 Cumulative Frequency 3 11 23 38 48 72 89 97 99 100 Solution a. The range is the difference between the largest and smallest data value. Range 5 85 2 30 5 55 To find the interquartile range, we must first find the median and the lower and upper quartiles. Since there are 100 values, the median is the average of the 50th and the 51st values. Both of these values lie in the interval 55. Therefore, the median is 55. There are 50 values above the mean and 50 values below the mean. Of the lower 50 values, the lower quartile is the average of the 25th and 26th values. Both of these values lie in the interval 45. The lower quartile is 45. The upper quartile is the average of the 25th and the 26th from the upper end of the distribution (or the 75th and 76th from the lower end). These values lie in the interval 60. The upper quartile is 60. Interquartile range 5 60 2 45 5 15 14411C15.pgs 8/14/08 10:37 AM Page 617 Measures of Dispersion 617 b. An outlier is a data value that is 1.5 times the interquartile range below the first quartile or above the third quartile. 45 2 1.5(15) 5 22.5 60 1 1.5(15) 5 82.5 The data value 85 is an outlier. Answers a. Range 5 55, interquartile range 5 15 b. The data value 85 is an outlier. Exercises Writing About Mathematics 1. In any set of data, is it always true that xi 5 i? For example, in a set of data with more than three data values, does x4 5 4? Justify your answer. 2. In a set of data, Q1 5 12 and Q3 5 18. Is a data value equal to 2 an outlier? Explain why or why not. Developing Skills In 3–6, find the range and the interquartile range for each set of data. 3. 3, 5, 7, 9, 11, 13, 15, 17, 19 4. 12, 12, 14, 14, 16, 18, 20, 22, 28, 34 5. 12, 17, 23, 31, 46, 54, 67, 76, 81, 93 6. 2, 14, 33, 34, 34, 34, 35, 36, 37, 37, 38, 40, 42 In 7–9, find the mean, median, range, and interquartile range for each set of data to the nearest tenth. 7. xi fi 50 45 40 35 30 25 20 3 8 12 15 11 7 4 8. xi fi 10 9 8 7 6 5 4 3 2 2 4 6 9 3 3 2 0 1 9. xi fi 11 16 19 31 37 32 35 5 8 9 6 5 5 6 14411C15.pgs 8/14/08 618 10:37 AM Page 618 Statistics Applying Skills 10. The following data represents the yearly salaries, in thousands of dollars, of 10 basketball players. 533 427 800 687 264 264 125 602 249 19,014 a. Find the mean and median salaries of the 10 players. b. Which measure of central tendency is more representative of the data? Explain. c. Find the outlier for the set of data. d. Remove the outlier from the set of data and recalculate the mean and median salaries. e. After removing the outlier from the set of data, is the mean more or less representative of the data? 11. The grades on a math test are shown in the stem-and-leaf diagram to the right. Stem 9 8 7 6 5 4 a. Find the mean grade. b. Find the median grade. c. Find the first and third quartiles. d. Find the range. Leaf 001269 0235557 88 46679 067 68 e. Find the interquartile range. 12. The ages of students in a Spanish class are shown in the table. Find the range and the interquartile range. Age Frequency 19 18 17 16 15 1 8 8 6 2 13. The table shows the number of hours that 40 third graders reported studying a week. Find the range and the interquartile range. Hours 3 4 5 6 7 8 9 10 11 12 Frequency 2 1 3 3 5 8 8 5 4 1 14. The table shows the number of pounds lost during the first month by people enrolled in a weight-loss program. a. Find the range. b. Find the interquartile range. c. Which of the data values is an outlier? Pounds Lost 1 2 3 4 5 6 7 8 9 11 15 Frequency 1 1 2 2 6 10 7 7 2 1 1 14411C15.pgs 8/14/08 10:37 AM Page 619 Variance and Standard Deviation 619 15. The 14 students on the track team recorded the following number of seconds as their best time for the 100-yard dash: 13.5 13.7 13.1 13.0 13.3 13.2 13.0 12.8 13.4 13.3 13.1 12.7 13.2 13.5 Find the range and the interquartile range. 16. The following data represent the waiting times, in minutes, at Post Office A and Post Office B at noon for a period of several days. A: 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 9, 10 B: 1, 2, 2, 3, 3, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10 a. Find the range of each set of data. Are the ranges the same? b. Graph the box-and-whisker plot of each set of data. c. Find the interquartile range of each set of data. d. If the data values are representative of the waiting times at each post office, which post office should you go to at noon if you are in a hurry? Explain. Hands-On Activity Find the range and the interquartile range of the data from the survey in the Hands-On Activity of Section 15-1 estimating the length of a minute. Does your data contain an outlier? 15-5 VARIANCE AND STANDARD DEVIATION Variance Let us consider a more significant measure of dispersion than either the range or the interquartile range. Let xi represent a student’s grades on eight tests. – )2 x i 2 x– (xi 2 x Grade (xi) 95 92 88 87 86 82 80 78 8 9 6 2 1 0 24 26 28 8 81 36 4 1 0 16 36 64 8 a xi 5 688 – 2 a (xi 2 x ) 5 0 – 2 a (xi 2 x ) 5 238 i51 i51 i51 8 a xii Mean 5 i51 8 5 688 8 5 86 The table shows the deviation, or difference, of each grade from the mean. Grades above the mean are positive and grades below the mean are negative. For any set of data, the sum of these differences is always 0. 14411C15.pgs 8/14/08 620 10:37 AM Page 620 Statistics In order to find a meaningful sum, we can use the squares of the differences so that each value will be positive. The square of the deviation of each data value from the mean is used to find another measure of dispersion called the variance. To find the variance, find the sum of the squares of the deviations from the mean and divide that sum by the number of data values. In symbols, the variance for a set of data that represents the entire population is given by the formula: n Variance 5 n1 a (xi 2 x) 2 i51 For the set of data given above, the variance is 18 (238) or 29.75. Note that since the square of the differences is involved, this method of finding a measure of dispersion gives greater weight to measures that are farther from the mean. EXAMPLE 1 A student received the following grades on five math tests: 84, 97, 92, 88, 79. Find the variance for the set of grades of the five tests. Solution The mean of this set of grades is: 84 1 97 1 92 1 88 1 79 5 5 440 5 5 88 xi x i 2 x– – )2 (xi 2 x 84 97 92 88 79 24 9 4 0 29 16 81 16 0 81 5 5 a xi 5 440 – 2 a (xi 2 x ) 5 194 i51 i51 – 2 5 A 1 B 194 5 384 5 38.8 Answer Variance 5 n1 a (xi 2 x) 5 5 5 i51 When the data representing a population is listed in a frequency distribution table, we can use the following formula to find the variance: n 2 a fi (xi 2 x) Variance 5 i51 n a fi i51 14411C15.pgs 8/14/08 10:37 AM Page 621 Variance and Standard Deviation 621 EXAMPLE 2 In a city, there are 50 math teachers who are under the age of 30. The table below shows the number of years of experience of these teachers. Find the variance of this set of data. Let xi represent the number of years of experience and fi represent the frequency for that number of years. xi 0 1 2 3 4 5 6 7 8 fi 1 2 7 8 6 9 8 4 5 50 xifi 0 2 14 24 24 45 48 28 40 225 Solution 8 a xifi Mean 5 i50 8 a fi 5 225 50 5 4.5 i50 For this set of data, the data value is equal to i for each xi. The table below shows the deviation from the mean, the square of the deviation from the mean, and the square of the deviation from the mean multiplied by the frequency. xi fi (xi 2 x– ) (xi 2 x– ) 2 fi(xi 2 x– ) 2 8 7 6 5 4 3 2 1 0 5 4 8 9 6 8 7 2 1 3.5 2.5 1.5 0.5 20.5 21.5 22.5 23.5 24.5 12.25 6.25 2.25 0.25 0.25 2.25 6.25 12.25 20.25 61.25 25.00 18.00 2.25 1.50 18.00 43.75 24.50 20.25 8 8 a fi 5 50 – 2 a fi (xi 2 x ) 5 214.50 i51 i50 8 a Variance 5 fi (xi 2 x–) 2 i50 8 a fi 5 214.50 50 5 4.29 Answer i50 Note that the data given in this example represents a population, that is, data for all of the teachers under consideration. 14411C15.pgs 8/14/08 622 10:37 AM Page 622 Statistics Standard Deviation Based on the Population Although the variance is a useful measure of dispersion, it is in square units. For example, if the data were a set of measures in centimeters, the variance would be in square centimeters. In order to have a measure that is in the same unit of measure as the given data, we find the square root of the variance. The square root of the variance is called the standard deviation. When the data represents a population, that is, all members of the group being studied: n Standard deviation based on a population 5 É 1 n a fi (xi 2 x) 2 i51 If the data is grouped in terms of the frequency of a given value: n 2 a fi (xi 2 x) i51 Standard deviation based on a population 5 é n a fi i51 The symbol for the standard deviation for a set of data that represents a population is s (lowercase Greek sigma). Many calculators use the symbol sx. EXAMPLE 3 Find the standard deviation for the number of years of experience for the 50 teachers given in Example 2. Solution The standard deviation is the square root of the variance. Therefore: Standard deviation 5 !4.29 5 2.071231518 Calculator (1) Clear lists 1 and 2. Solution (2) Enter the number of years of experience in L and the frequency in L . 1 2 (3) Locate the standard deviation for a population (sx) under 1-Var Stats. ENTER: STAT 䉴 2nd L1 2nd L2 ENTER DISPLAY: , ENTER > 1–Var Stats – x=4.5 x=225 x2=1227 Sx=2.092259788 x=2.071231518 n=50 (4) sx 5 2.071231518 Answer The standard deviation is approximately 2.07. 14411C15.pgs 8/14/08 10:37 AM Page 623 Variance and Standard Deviation 623 Standard Deviation Based on a Sample When the given data is information obtained from a sample of the population, the formula for standard deviation is obtained by dividing the sum of the squares of the deviation from the mean by 1 less than the number of data values. If each data value is listed separately: n Standard deviation for a sample 5 É 1 n 2 1 a (xi 2 x–) 2 i51 If the data is grouped in terms of the frequency of a given value: n 2 a fi (xi 2 x) Standard deviation for a sample 5 i51 é n a a fi b 2 1 i51 The symbol for the standard deviation for a set of data that represents a sample is s. Many calculators use the symbol sx. EXAMPLE 4 From a high school, ten students are chosen at random to report their number of online friends. The data is as follows: 15, 13, 12, 10, 9, 7, 5, 4, 3, and 2. Solution The total number of online friends for these 10 students is 80 or a mean of 8 online friends (x– 5 8). Online Friends (xi) x i 2 x– – )2 (xi 2 x 15 13 12 10 9 7 5 4 3 2 7 5 4 2 1 21 23 24 25 26 49 25 16 4 1 1 9 16 25 36 10 – 2 a (xi 2 x ) 5 182 i51 10 Standard deviation 5 É 1 10 2 1 a (xi i51 2 x) 2 5 #19 (182) 4.5 Answer 14411C15.pgs 8/14/08 624 10:37 AM Page 624 Statistics EXAMPLE 5 In each of the following, tell whether the population or sample standard deviation should be used. a. In a study of the land areas of the states of the United States, the area of each of the 50 states is used. b. In a study of the heights of high school students in a school of 1,200 students, the heights of 100 students chosen at random were recorded. c. In a study of the heights of high school students in the United States, the heights of 100 students from each of the 50 states were recorded. Solution a. Use the population standard deviation since every state is included. Answer b. Use the sample standard deviation since only a portion of the total school population was included in the study. Answer c. Use the sample standard deviation since only a portion of the total school population was included in the study. Answer EXAMPLE 6 A telephone survey conducted in Monroe County obtained information about the size of the households. Telephone numbers were selected at random until a sample of 130 responses were obtained. The frequency chart at the right shows the result of the survey. No. of People per Household Frequency 1 2 3 4 5 6 7 8 9 28 37 45 8 7 3 1 0 1 Solution The table below can be used to find the mean and the standard deviation. xi fi fixi (xi 2 x– ) (xi 2 x– ) 2 fi(xi 2 x– ) 2 1 2 3 4 5 6 7 8 9 28 37 45 8 7 3 1 0 1 28 74 135 32 35 18 7 0 9 21.6 20.6 0.4 1.4 2.4 3.4 4.4 5.4 6.4 2.56 0.36 0.16 1.96 5.76 11.56 19.36 29.16 40.96 71.68 13.32 7.20 15.68 40.32 34.68 19.36 0 40.96 9 9 9 a fi 5 130 a fi xi 5 338 – 2 a fi (xi 2 x ) 5 243.20 i51 i51 i51 14411C15.pgs 8/14/08 10:37 AM Page 625 Variance and Standard Deviation 625 9 a fixi Mean 5 i51 9 a fi 5 338 130 5 2.6 i51 9 a Standard deviation 5 fi (xi 2 x) 2 i51 9 a fi b 2 1 è ia 51 243.20 243.20 5 #130 2 1 5 # 129 5 1.373051826 Calculator Enter the number of members of the households in L1 and the frequency for Solution each data value in L2. ENTER: 䉴 STAT L1 , ENTER 2nd L2 2nd ENTER DISPLAY: > 1–Var Stats – x=2.6 x=338 x2=1122 Sx=1.373051826 x=1.367760663 n=130 The standard deviation based on the data from a sample is sx 5 1.373051826 or approximately 1.37. Answer Exercises Writing About Mathematics 1. The sets of data for two different statistical studies are identical. The first set of data represents the data for all of the cases being studied and the second represents the data for a sample of the cases being studied. Which set of data has the larger standard deviation? Explain your answer. 2. Elaine said that the variance is the square of the standard deviation. Do you agree with Elaine? Explain why or why not. Developing Skills In 3–9, the given values represent data for a population. Find the variance and the standard deviation for each set of data. 3. 9, 9, 10, 11, 5, 10, 12, 9, 10, 12, 6, 11, 11, 11 4. 11, 6, 7, 13, 5, 8, 7, 10, 9, 11, 13, 12, 9, 16, 10 5. 20, 19, 20, 17, 18, 19, 42, 41, 41, 39, 39, 40 6. 20, 101, 48, 25, 63, 31, 20, 50, 16, 14, 245, 9 14411C15.pgs 8/14/08 626 7. 10:37 AM Page 626 Statistics xi fi 30 35 40 45 50 55 60 1 7 10 9 11 8 6 8. xi fi 2 4 6 8 10 12 14 16 0 1 6 10 13 21 7 1 9. xi fi 20 25 30 35 40 45 50 55 60 21 10 1 2 2 4 5 3 7 In 10–16, the given values represent data for a sample. Find the variance and the standard deviation based on this sample. 10. 6, 4, 9, 11, 4, 3, 22, 3, 7, 10 11. 12.1, 33.3, 45.5, 60.1, 94.2, 22.2 12. 15, 10, 16, 19, 10, 19, 14, 17 13. 1, 3, 5, 22, 30, 45, 50, 55, 60, 70 14. xi fi 55 50 45 40 35 30 25 11 15 4 1 14 12 4 15. xi fi 33 34 35 36 37 38 39 3 1 4 6 5 11 6 16. xi fi 1 2 3 4 5 6 7 3 3 3 3 3 3 3 Applying Skills 17. To commute to the high school in which Mr. Fedora teaches, he can take either the Line A or the Line B train. Both train stations are the same distance from his house and both stations report that, on average, they run 10 minutes late from the scheduled arrival time. However, the standard deviation for Line A is 1 minute and the standard deviation for Line B is 5 minutes. To arrive at approximately the same time on a regular basis, which train line should Mr. Fedora use? Explain. Stem Leaf 18. A hospital conducts a study to determine if nurses need extra staffing at night. A random sample of 25 nights was used. The number of calls to the nurses’ station each night is shown in the stem-and-leaf diagram to the right. a. Find the variance. b. Find the standard deviation. 9 8 7 6 5 4 0234466 68 01234 446789 037 19 14411C15.pgs 8/14/08 10:37 AM Page 627 Variance and Standard Deviation 19. The ages of all of the students in a science class are shown in the table. Find the variance and the standard deviation. Age Frequency 18 17 16 15 1 2 9 9 627 20. The table shows the number of correct answers on a test consisting of 15 questions. The table represents correct answers for a sample of the students who took the test. Find the standard deviation based on this sample. Correct Answers 6 7 8 9 10 11 12 13 14 15 Frequency 2 1 3 3 5 8 8 5 4 1 21. The table shows the number of robberies during a given month in 40 different towns of a state. Find the standard deviation based on this sample Robberies 0 1 2 3 4 5 6 7 8 9 10 Frequency 1 1 1 2 2 6 10 7 7 2 1 22. Products often come with registration forms. One of the questions usually found on the registration form is household income. For a given product, the data below represents a random sample of the income (in thousands of dollars) reported on the registration form. Find the standard deviation based on this sample. 38 40 26 42 39 25 40 40 39 36 46 41 43 47 49 43 39 35 43 37 Hands-On Activity The people in your survey from the Hands-On Activity of Section 15-1 represent a random sample of all people. Find the standard deviation based on your sample. 14411C15.pgs 8/14/08 628 10:37 AM Page 628 Statistics 15-6 NORMAL DISTRIBUTION The Normal Curve .... ...... Imagine that we were able to determine ...... ........ ........ .......... the height in centimeters of all 10-year.......... .......... ............ ............ old children in the United States. With a ............. .............. .............. ............... scale along the horizontal axis that ................ ................ .................. ................... includes all of these heights, we will place .................... ..................... ...................... ........................ a dot above each height for each child of ......................... . ........................... . ............................. . ................................ that height. For example, we will place . .................................. ........................................ .......................................... above 140 on the horizontal scale a dot for each child who is 140 centimeters tall. 123 128 133 138 143 148 153 Do this for 139, 138, 137, and so on for Height in centimeters each height in our data. The result would be a type of frequency histogram. If we draw a smooth curve joining the top dot for each height, we will draw a bellshaped curve called the normal curve. As the average height of 10-year-olds is approximately 138 centimeters, the data values are concentrated at 138 centimeters and the normal curve has a peak at 138 centimeters. Since for each height that is less than or greater than 138 centimeters there are fewer 10-year-olds, the normal curve progressively gets shorter as you go farther from the mean. Scientists have found that large sets of data that occur naturally such as heights, weights, or shoe sizes have a bell-shaped or a normal curve. The highest point of the normal curve is at the mean of the data. The normal curve is symmetric with respect to a vertical line through the mean of the distribution. Standard Deviation and the Normal Curve A normal distribution is a set of data that can be represented by a normal curve. For a normal distribution, the following relationships exist. 1. The mean and the median of the data values lie on the line of symmetry of the curve. 2. Approximately 68% of the data values lie within one standard deviation from the mean. 99.7% of data values 95% of data values 68% of data values 3. Approximately 95% of the data values lie within two standard deviations from the mean. 4. Approximately 99.7% of the data values lie within three standard deviations from the mean. 13.5% 34% 34% 13.5% x–23s x–22s x–2s x– x–1s x–12s x–13s Normal distribution 14411C15.pgs 8/14/08 10:37 AM Page 629 Normal Distribution 629 EXAMPLE 1 A set of data is normally distributed with a mean of 50 and a standard deviation of 2. a. What percent of the data values are less than 50? b. What percent of the data values are between 48 and 52? c. What percent of the data values are between 46 and 54? d. What percent of the data values are less than or equal to 46? Solution a. In a normal distribution, 50% of the data values are to the left and 50% to the right of the mean. b. 48 and 52 are each 1 standard deviation away from the mean. 48 5 50 2 2 2.5% 13.5% 34% 34% 13.5% 52 5 50 1 2 46 48 50 50% Therefore, 68% of the data values are between 48 and 52. 52 54 50% c. 46 and 53 are each 2 standard deviations away from the mean. 46 5 50 2 2(2) 54 5 50 1 2(2) Therefore, 95% of the data values are between 48 and 52. d. 50% of the data values are less than 50. 47.5% of the data values are more than 46 and less than 50. Therefore, 50% 2 47.5% or 2.5% of the data values are less than or equal to 46. Answers a. 50% b. 68% c. 95% d. 2.5% Z-Scores The z-score for a data value is the deviation from the mean divided by the standard deviation. Let x be a data value of a normal distribution. x2x x2x z-score 5 standard deviation 5 s The z-score of x, a value from a normal distribution, is positive when x is above the mean and negative when x is below the mean. The z-score tells us how many standard deviations x is above or below the mean. 14411C15.pgs 8/14/08 630 10:37 AM Page 630 Statistics 1. The z-score of the mean is 0. 99.7% 95% 2. Of the data values, 34% have a z-score between 0 and 1 and 34% have a z-score between 21 and 0. Therefore, 68% have a z-score between 21 and 1. 68% 3. Of the data values, 13.5% have a z-score between 1 and 2 and 13.5% have a z-score between 22 and 21. 23 13.5% 34% 34% 13.5% 22 21 0 1 2 3 4. Of the data values, (34 1 13.5)% or 47.5% have a z-score between 0 and 2 and 47.5% have a z-score between 22 and 0. Therefore, 95% have a z-score between 22 and 2. 5. Of the data values, 99.7% have a z-score between 23 and 3. For example, the mean height of 10-year-old children is 138 centimeters with a standard deviation of 5. Casey is 143 centimeters tall. 138 5 55 5 1 z-score for Casey 5 143 2 5 Casey’s height is 1 standard deviation above the mean. For a normal distribution, 34% of the data is between the mean and 1 standard deviation above the mean and 50% of the data is below the mean. Therefore, Casey is as tall as or taller than (34 1 50)% or 84% of 10-year-old children. A calculator will give us this same answer. The second entry of the DISTR menu is normalcdf(. When we use this function, we must supply a minimum value, a maximum value, the mean, and the standard deviation separated by commas: > DISTR DRAW 1: normalpdf( 2:normalcdf( 3:invNorm( 4:invT( 5:tpdf( 6:tcdf( 7 x2 p d f ( normalcdf(minimum, maximum, mean, standard deviation) For the minimum value we can use 0. To find the proportion of 10-year-old children whose height is 143 centimeters or less, use the following entries. ENTER: 2nd 143 , 5 DISTR ) , ENTER 0 2 138 , DISPLAY: normalcdf(0,143, 138,5) .8413447404 The calculator returns the number 0.8413447404, which can be rounded to 0.84 or 84%. 14411C15.pgs 8/14/08 10:37 AM Page 631 Normal Distribution 631 If we wanted to find the proportion of the 10-year-old children who are between 134.6 and 141.4 centimeters tall, we could make the following entry: ENTER: 2nd 141.4 , 5 DISTR ) , 2 138 134.6 DISPLAY: , ENTER normalcdf(134.6, 141.4,138,5) .5034956838 The calculator returns the number 0.5034956838, which can be rounded to 0.50 or 50%. Since 134.6 and 141.4 are equidistant from the mean, 25% of the data is below the mean and 25% is above the mean. Therefore, for this distribution, 134.6 centimeters is the first quartile, 138 centimeters is the median or second quartile, and 141.4 centimeters is the third quartile. Note: 134.6 is 0.68 standard deviation below the mean and 141.4 is 0.68 standard deviation above the mean. For any normal distribution, data values with z-scores of 20.68 are approximately equal to the first quartile and data values with z-scores of 0.68 are approximately equal to the third quartile. 25% 25% 20.68 25% 0 25% 0.68 EXAMPLE 1 On a standardized test, the test scores are normally distributed with a mean of 60 and a standard deviation of 6. a. Of the data, 84% of the scores are at or below what score? b. Of the data, 16% of the scores are at or below what score? c. What is the z-score of a score of 48? d. If 2,000 students took the test, how many would be expected to score at or below 48? Solution a. Since 50% scored at or below the mean and 34% scored within 1 standard deviation above the mean, (50 1 34)% or 84% scored at or below 1 deviation above the mean: x– 1 s 5 60 1 6 5 66 b. Since 50% scored at or below the mean and 34% scored within 1 standard deviation below the mean, (50 2 34)% or 16% scored at or below 1 deviation below the mean: x– 1 s 5 60 2 6 5 54 – x 48 2 60 5 212 c. z-score 5 x 2 s 5 6 6 5 22 14411C15.pgs 8/14/08 632 10:37 AM Page 632 Statistics d. A test score of 48 has a z-score of 22. Since 47.5% of the scores are between 22 and 0 and 50% of the scores are less than 0, 50% 2 47.5% or 2.5% scored at or below 48. 2.5% 3 2,000 5 0.025 3 2,000 5 50 students Answers a. 66 b. 54 c. 22 d. 50 EXAMPLE 2 For a normal distribution of weights, the mean weight is 160 pounds and a weight of 186 pounds has a z-score of 2. a. What is the standard deviation of the set of data? b. What percent of the weights are between 155 and 165? 2 x– s 186 2 160 s Solution a. z-score 5 x 25 2s 5 26 s 5 13 Answer b. ENTER: 2nd 155 13 DISTR , ) 165 DISPLAY: 2 , ENTER 160 , normalcdf(155,16 5,160,13) .2994775047 About 30% of the weights are between 155 and 165. Answer Exercises Writing About Mathematics 1. A student’s scores on five tests were 98, 97, 95, 93, and 67. Explain why this set of scores does not represent a normal distribution. 2. If 34% of the data for a normal distribution lies between the mean and 1 standard deviation above the mean, does 17% of the data lie between the mean and one-half standard deviation above the mean? Justify your answer. Developing Skills In 3–9, for a normal distribution, determine what percent of the data values are in each given range. 3. Between 1 standard deviation below the mean and 1 standard deviation above the mean 4. Between 1 standard deviation below the mean and 2 standard deviations above the mean 14411C15.pgs 8/14/08 10:37 AM Page 633 Normal Distribution 633 5. Between 2 standard deviations below the mean and 1 standard deviation above the mean 6. Above 1 standard deviation below the mean 7. Below 1 standard deviation above the mean 8. Above the mean 9. Below the mean 10. A set of data is normally distributed with a mean of 40 and a standard deviation of 5. Find a data value that is: a. 1 standard deviation above the mean b. 2.4 standard deviations above the mean c. 1 standard deviation below the mean d. 2.4 standard deviations below the mean Applying Skills In 11–14, select the numeral that precedes the choice that best completes the statement or answers the question. 11. The playing life of a Euclid mp3 player is normally distributed with a mean of 30,000 hours and a standard deviation of 500 hours. Matt’s mp3 player lasted for 31,500 hours. His mp3 player lasted longer than what percent of other Euclid mp3 players? (1) 68% (2) 95% (3) 99.7% (4) more than 99.8% 12. The scores of a test are normally distributed. If the mean is 50 and the standard deviation is 8, then a student who scored 38 had a z-score of (1) 1.5 (2) 21.5 (3) 12 (4) 212 13. The heights of 10-year-old children are normally distributed with a mean of 138 centimeters with a standard deviation of 5 centimeters. The height of a 10-year-old child who is as tall as or taller than 95.6% of all 10-year-old children is (1) between 138 and 140 cm. (3) between 145 and 148 cm. (2) between 140 and 145 cm. (4) taller than 148 cm. 14. The heights of 200 women are normally distributed. The mean height is 170 centimeters with a standard deviation of 10 centimeters. What is the best estimate of the number of women in this group who are between 160 and 170 centimeters tall? (1) 20 (2) 34 (3) 68 (4) 136 15. When coffee is packed by machine into 16-ounce cans, the amount can vary. The mean weight is 16.1 ounces and the standard deviation is 0.04 ounce. The weight of the coffee approximates a normal distribution. a. What percent of the cans of coffee can be expected to contain less than 16 ounces of coffee? b. What percent of the cans of coffee can be expected to contain between 16.0 and 16.2 ounces of coffee? 8/14/08 634 10:37 AM Page 634 Statistics 16. The length of time that it takes Ken to drive to work represents a normal distribution with a mean of 25 minutes and a standard deviation of 4.5 minutes. If Ken allows 35 minutes to get to work, what percent of the time can he expect to be late? 17. A librarian estimates that the average number of books checked out by a library patron is 4 with a standard deviation of 2 books. If the number of books checked out each day approximates a normal distribution, what percent of the library patrons checked out more than 7 books yesterday? 18. The heights of a group of women are normally distributed with a mean of 170 centimeters and a standard deviation of 10 centimeters. What is the z-score of a member of the group who is 165 centimeters tall? 19. The test grades for a standardized test are normally distributed with a mean of 50. A grade of 60 represents a z-score of 1.25. What is the standard deviation of the data? 20. Nora scored 88 on a math test that had a mean of 80 and a standard deviation of 5. She also scored 80 on a science test that had a mean of 70 and a standard deviation of 3. On which test did Nora perform better compared with other students who took the tests? 15-7 BIVARIATE STATISTICS Statistics are often used to compare two sets of data. For example, a pediatrician may compare the height and weight of a child in order to monitor growth. Or the owner of a gift shop may record the number of people who enter the store with the revenue each day. Each of these sets of data is a pair of numbers and is an example of bivariate statistics. Representing bivariate sta550 tistics on a two-dimensional graph or scatter plot can help 540 us to observe the relationship 530 between the variables. For example the mean value for the critical 520 reading and for the math sections of the SAT examination for nine 510 schools in Ontario County are 500 listed in the table and shown on the graph on the right. 510 520 530 540 550 560 570 Critical Reading 14411C15.pgs Math Math 530 551 521 522 537 511 516 537 566 Critical Reading 530 529 512 518 526 500 504 515 543 14411C15.pgs 8/14/08 10:37 AM Page 635 Bivariate Statistics 635 The graph shows that there appears to be a linear relationship between the critical reading scores and the math scores. As the math scores increase, the critical reading scores also increase. We say that there is a correlation between the two scores. The points of the graph approximate a line. These data can also be shown on a calculator. Enter the math scores as L1 and the corresponding critical reading scores as L2. Then turn on Plot 1 and use ZoomStat from the ZOOM menu to construct a window that will include all values of x and y: ENTER: 2nd STAT PLOT 䉲 ENTER 䉲 2nd DISPLAY: 2nd ENTER: 1 ZOOM 9 ENTER L1 䉲 L2 DISPLAY: Plot1 Plot2 Plot3 On Off . Ty p e : ...... Xlist:L1 Xlist:L2 Mark: + . The calculator will display a graph similar to that shown above. To draw a line that approximates the data, use the regression line on the calculator. A regression line is a special line of best fit that minimizes the square of the vertical distances to each data point. In this course, you do not have to know the formula to find the regression line. The calculator can be used to determine the regression line: ENTER: STAT 䉴 4 VARS 䉴 1 1 The calculator displays values for a and b for the linear equation y = ax + b and stores the regression equation into Y1 in the Y menu. 9 Press ZOOM to display the scatter plot and the line that best approximates the data. If we round the given values of a and b to three decimal places, the linear regression equation is: y 5 0.693x 1 151.013 LinReg y=ax+b a=.6925241158 b=151.0129957 r2=.7986740639 r=.8936856628 ENTER 14411C15.pgs 8/14/08 636 10:37 AM Page 636 Statistics We will study other data that can be approximated by a curve rather than a line in later sections. The scatter plots below show possible linear correlation between elements of the pairs of bivariate data. The correlation is positive when the values of the second element of the pairs tend to increase when the values of the first elements of the pairs increase. The correlation is negative when the values of the second element of the pairs tend to decrease when the values of the first elements of the pairs increase. Strong positive correlation Moderate positive correlation No linear correlation Moderate negative correlation Strong negative correlation EXAMPLE 1 Jacob joined an exercise program to try to lose weight. Each month he records the number of months in the program and his weight at the end of that month. His record for the first twelve months is shown below: Month Weight 1 2 3 4 5 6 7 248 242 237 228 222 216 213 8 9 206 197 10 11 12 193 185 178 8/14/08 10:37 AM Page 637 Bivariate Statistics 637 a. Draw a scatter plot and describe the correlation between the data (if any). b. Draw a line that appears to represent the data. c. Write an equation of a line that best represents the data. Solution a. Use a scale of 0 to 13 along a horizontal line for the number of months in the program and a scale from 170 to 250 along the vertical axis for his weight at the end of that month. The scatter plot 250 is shown here. There appears to be a strong neg240 ative correlation between the number of months 230 and Jacob’s weight. Weight 14411C15.pgs 220 210 200 190 180 170 b. The line on the graph that appears to approximate the data intersects the points (1, 248) and (11, 185). We can use these two points to write an equation of the line. 2 185 y 2 248 5 248 1 2 11 (x 2 1) y 2 248 5 26.3x 1 6.3 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Month y 5 26.3x 1 254.3 c. On a calculator, enter the number of the month in L1 and the weight in L2. To use the calculator to determine a line that approximates the data, use the following sequence: 䉴 ENTER: STAT 1 䉴 VARS 4 1 LinReg y=ax+b a=-6.283216783 b=254.5909091 r2=.9966845178 r=-.9983408825 ENTER The calculator displays values for a and b for the linear equation y = ax 1 b and stores the equation as Y1 in the Y menu. If we round the given values of a and b to three decimal places, the linear regression equation is: y 5 26.283x 1 254.591 Turn on Plot 1, then use ZoomStat to graph the data: ENTER: 2nd 䉲 STAT PLOT ENTER ENTER 2nd ZOOM 9 䉲 ENTER 2nd L2 ENTER L1 ENTER The calculator will display the scatter plot of the data along with the regression equation. 14411C15.pgs 8/14/08 638 10:37 AM Page 638 Statistics EXAMPLE 2 In order to assist travelers in planning a trip, a travel guide lists the average high temperature of most major cities. The listing for Albany, New York is given in the following table. Month 1 2 3 4 5 6 7 8 9 10 11 12 Temperature 31 33 42 51 70 79 84 81 73 62 48 35 Can these data be represented by a regression line? Solution Draw a scatter plot the data. The graph shows that the data can be characterized by a curve rather than a line. Finding the regression line for this data would not be appropriate. SUMMARY • The slope of the regression line gives the direction of the correlation: A positive slope shows a positive correlation. A negative slope shows a negative correlation. • The regression line is appropriate only for data that appears to be linearly related. Do not calculate a regression line for data with a scatter plot showing a non-linear relationship. • The regression equation is sensitive to rounding. Round the coefficients to at least three decimal places. Exercises Writing About Mathematics 1. Explain the difference between univariate and bivariate data and give an example of each. 2. What is the relationship between slope and correlation? Can slope be used to measure the strength of a correlation? Explain. Developing Skills In 3–6, is the set of data to be collected univariate or bivariate? 3. The science and math grades of all students in a school 4. The weights of the 56 first-grade students in a school 14411C15.pgs 8/14/08 10:37 AM Page 639 Bivariate Statistics 639 5. The weights and heights of the 56 first-grade students in a school 6. The number of siblings for each student in the first grade In 7–10, look at the scatter plots and determine if the data sets have high linear correlation, moderate linear correlation, or no linear correlation. 7. 8. 9. 10. Applying Skills In 11–17: a. Draw a scatter plot. b. Does the data set show strong positive linear correlation, moderate positive linear correlation, no linear correlation, moderate negative linear correlation, or strong negative linear correlation? c. If there is strong or moderate correlation, write the equation of the regression line that approximates the data. 11. The following table shows the number of gallons of gasoline needed to fill the tank of a car and the number of miles driven since the previous time the tank was filled. Gallons 8.5 7.6 9.4 8.3 10.5 8.7 9.6 4.3 6.1 7.8 Miles 255 230 295 250 315 260 290 130 180 235 12. A business manager conducted a study to examine the relationship between number of ads placed for each month and sales. The results are shown below where sales are in the thousands. Number of Ads 10 12 14 Sales 20 26.5 32 16 18 34.8 40 20 22 24 26 28 30 47.2 49.1 56.9 57.9 65.8 66.4 13. Jack Sheehan looked through some of his favorite recipes to compare the number of calories per serving to the number of grams of fat. The table below shows the results. Calories 310 210 260 330 290 320 245 293 Fat 11 5 11 12 14 16 7 10 220 260 350 8 8 15 14411C15.pgs 8/14/08 640 10:37 AM Page 640 Statistics 14. Greg did a survey to support his theory that the size of a family is related to the size of the family in which the mother of the family grew up. He asked 20 randomly selected people to list the number of their siblings and the number of their mother’s siblings. Greg made the following table. Family 2 2 3 1 0 3 5 2 1 2 Mother’s Family 4 0 3 7 4 2 2 6 4 7 Family 3 6 0 2 1 4 1 0 4 1 Mother’s Family 5 4 6 1 0 2 1 3 3 2 15. When Marie bakes, it takes about five and a half minutes for the temperature of the oven to reach 350°. One day, while waiting for the oven to heat, Marie recorded the temperature every 20 seconds. Her record is shown below. Seconds 0 20 40 60 80 100 120 140 160 Temperature 100 114 126 145 160 174 193 207 222 Seconds 180 200 220 240 260 280 300 320 340 Temperature 240 255 268 287 301 318 331 342 350 16. An insurance agent is studying the records of his insurance company looking for a relationship between age of a driver and the percentage of accidents due to speeding. The table shown below summarizes the findings of the insurance agent. Age 17 18 21 25 30 35 40 45 50 55 60 65 % of Speeding Accidents 49 49 48 39 31 33 24 25 16 10 5 6 17. A sociologist is interested in the relationship between body weight and performance on the SAT. A random sample of 10 high school students from across the country provided the following information: Weight Score 197 193 194 157 159 170 149 169 157 185 1,485 1,061 1,564 1,729 1,668 1,405 1,544 1,752 1,395 1,214 14411C15.pgs 8/14/08 10:37 AM Page 641 Correlation Coefficient 641 15-8 CORRELATION COEFFICIENT We would like to measure the strength of the linear relationship between the variables in a set of bivariate data. The slope of the regression equation tells us the direction of the relationship but it does not tell us the strength of the relationship. The number that we use to measure both the strength and direction of the linear relationship is called the correlation coefficient, r. The value of the correlation coefficient does not depend on the units of measurement. In more advanced statistics courses, you will learn a formula to derive the correlation coefficient. In this course, we can use the graphing calculator to calculate the value of r. EXAMPLE 1 The coach of the basketball team made the following table of attempted and successful baskets for eight players. Attempted Baskets (xi) 10 12 12 13 14 15 17 19 Successful Baskets (yi) 6 7 9 8 10 11 14 15 Find the value of the correlation coefficient. Solution Enter the given xi in L1 and yi in L2. Then choose LinReg(ax+b) from the CALC STAT menu: ENTER: STAT 䉴 4 ENTER LinReg y=ax+b a=1.066666667 b=-4.933333333 r2=.9481481481 r=.9737289911 The calculator will list both the regression equation and r, the correlation coefficient. Answer r 5 0.97 Note: If the correlation coefficient does not appear on your calculator, enter 2nd CATALOG D , scroll down to DiagnosticOn, press ENTER , and press ENTER again. When the absolute value of the correlation coefficient is close to 1, the data have a strong linear correlation. When the absolute value of the correlation coefficient is close to 0, there is little or no linear correlation. Values between 0 and 1 indicate various degrees of positive moderate correlation and values between 0 and 21 indicate various degrees of negative moderate correlation. 14411C15.pgs 8/14/08 642 10:37 AM Page 642 Statistics Properties of the Correlation Coefficient, r 21 r 1.The correlation coefficient is a number between 21 and 1. When r 5 1 or 21, there is a perfect linear relationship between the data values. r 5 1 r 5 21 When r 5 0, no linear relationship exists between the data values. When |r| is close to 1, the data have a strong linear relationship.Values between 0 and 1 indicate various degrees of moderate correlation. The sign of r matches the sign of the slope of the regression line. EXAMPLE 2 An automotive engineer is studying the fuel efficiency of a new prototype. From a fleet of eight prototypes, he records the number of miles driven and the number of gallons of gasoline used for each trip. Miles Driven 310 270 350 275 380 320 290 405 Gallons of Gasoline Used 10.0 9.0 11.2 8.7 12.3 10.2 9.5 12.7 14411C15.pgs 8/14/08 10:37 AM Page 643 Correlation Coefficient 643 a. Based on the context of the problem, do you think the correlation coefficient will be positive, negative, or close to 0? b. Based on the scatter plot of the data, do you expect the correlation coefficient to be close to 21, 0, or 1? c. Use a calculator to find the equation of the regression line and determine the correlation coefficient. Solution a. As gallons of gasoline used tend to increase with miles driven, we expect the correlation coefficient to be positive. Enter the given data as L1 and L2 on a calculator. b. c. There appears to be a strong positive correlation, so r will be close to 1. y 5 0.030x 1 0.748, r 5 0.99 LinReg y=ax+b a=.0298533724 b=.7476539589 r2=.9879952255 r=.9939794895 EXAMPLE 3 The produce manager of a food store noted the relationship between the amount the store charged for a pound of fresh broccoli and the number of pounds sold in one week. His record for 11 weeks is shown in the following table. Cost per Pound Pounds Purchased $0.65 $0.85 $0.90 $1.00 $1.25 $1.50 $1.75 $1.99 $2.25 $2.50 $2.65 58 43 49 23 39 16 56 32 12 35 What conclusion could the product manager draw from this information? 11 14411C15.pgs 8/14/08 644 10:37 AM Page 644 Statistics Solution Enter the given data as L1 and L2 on the calculator, graph the scatter plot, and find the value of the correlation coefficient. LinReg y=ax+b a=-13.82881393 b=55.73638117 r2=.331571737 r=-.5758226611 From the scatter plot, we can see that there is a moderate negative correlation. Since r 5 20.58, the product manager might conclude that a lower price does explain some of the increase in sales but other factors also influence the number of sales. A Warning About Cause-and-Effect The correlation coefficient is a number that measures the strength of the linear relationship between two data sets. However, simply because there appears to be a strong linear correlation between two variables does not mean that one causes the other. There may be other variables that are the cause of the observed pattern. For example, consider a study on the population growth of a city. Although a statistician may find a linear pattern over time, this does not mean that time causes the population to grow. Other factors cause the city grow, for example, a booming economy. SUMMARY • 21 r 1. The correlation coefficient is a number between 21 and 1. • When r 5 1 or r 5 21, there is a perfect linear relationship between the data values. • When r 5 0, no linear relationship exists between the data values. • When r is close to 1, the data have a strong linear relationship. Values between 0 and 1 indicate various degrees of moderate correlation. • The sign of r matches the sign of the slope of the regression line. • A high correlation coefficient does not necessarily mean that one variable causes the other. 14411C15.pgs 8/14/08 10:37 AM Page 645 Correlation Coefficient 645 Exercises Writing About Mathematics 1. Does a correlation coefficient of 21 indicate a lower degree of correlation than a correlation coefficient of 0? Explain why or why not. 2. If you keep a record of the temperature in degrees Fahrenheit and in degrees Celsius for a month, what would you expect the correlation coefficient to be? Justify your answer. Developing Skills In 3–6, for each of the given scatter plots, determine whether the correlation coefficient would be close to 21, 0, or 1. 3. 4. 5. 6. In 7–14, for each of the given correlation coefficients, describe the linear correlation as strong positive, moderate positive, none, moderate negative, or strong negative. 7. r 5 0.9 11. r 5 1 8. r 5 21 9. r 5 20.1 12. r 5 20.5 10. r 5 0.3 13. r 5 0 14. r 5 20.95 Applying Skills In 15–19: a. Draw a scatter plot for each data set. b. Based on the scatter plot, would the correlation coefficient be close to 21, 0, or 1? Explain. c. Use a calculator to find the correlation coefficient for each set of data. 15. The following table shows the number of gallons of gasoline needed to fill the tank of a car and the number of miles driven since the previous time the tank was filled. Gallons 12.5 3.4 Miles 7.9 9.0 15.7 7.0 5.1 11.9 13.0 10.7 392 137 249 308 504 204 182 377 407 304 16. A man on a weight-loss program tracks the number of pounds that he lost over the course of 10 months. A negative number indicates that he actually gained weight for that month. Month No. of Pounds Lost 1 2 3 4 5 6 7 8 9.2 9.1 4.8 4.5 2.8 1.8 1.2 0 9 10 0.8 22.6 14411C15.pgs 8/14/08 646 10:37 AM Page 646 Statistics 17. An economist is studying the job market in a large city conducts of survey on the number of jobs in a given neighborhood and the number of jobs paying $100,000 or more a year. A sample of 10 randomly selected neighborhood yields the following data: Total Number of Jobs 24 28 17 39 32 21 39 39 24 29 No. of HighPaying Jobs 3 3 4 5 7 3 4 7 7 4 18. The table below shows the same-day forecast and the actual high temperature for the day over the course of 18 days. The temperature is given in degrees Fahrenheit. Same-Day Forecast 56 52 67 55 58 56 59 57 53 Actual Temperature 53 54 63 49 66 54 54 56 59 Same-Day Forecast 45 55 45 58 59 55 48 53 54 Actual Temperature 48 60 36 59 59 47 46 52 48 19. The table below shows the five-day forecast and the actual high temperature for the fifth day over the course of 18 days. The temperature is given in degrees Fahrenheit. Five-Day Forecast 56 52 67 55 58 56 59 57 53 Actual Temperature 50 50 84 54 57 40 70 79 48 Five-Day Forecast 45 55 45 58 59 55 48 53 54 Actual Temperature 40 61 40 70 46 75 49 46 88 20. a. In Exercises 18 and 19, if the forecasts were 100% accurate, what should the value of r be? b. Is the value of r for Exercise 18 greater than, equal to, or less than the value of r for Exercise 19? Is this what you would expect? Explain. 8/14/08 10:37 AM Page 647 Non-Linear Regression 647 15-9 NON-LINEAR REGRESSION Not all bivariate data can be represented by a linear function. Some data can be better approximated by a curve. For example, on the right is a scatter plot of the file size of a computer program called Super Type over the course of 6 different versions. The relationship does not appear to be linear. For this set of data, a linear regression would not be appropriate. There are a variety of non-linear functions that can be applied to non-linear data. In a statistics course, you will learn more rigorous methods of determining the regression model. In this course, we will use the scatter plot of the data to choose the regression model: Regression to Use Description of Scatter Plot Exponential • An exponential curve that does not pass through (0, 0) • y-intercept is positive • Data constraint: y . 0 Logarithmic • A logarithmic curve that does not pass through (0, 0) • y-intercept is positive or negative • Data constraint: x . 0 Power • Positive half of power curve passing through (0, 0) • Data constraints: x . 0, y . 0 1,800 Size (megabytes) 14411C15.pgs 1,600 1,400 1,200 1,000 800 600 400 200 0 0 1 2 3 4 5 6 7 Version Examples (continued on next page) 14411C15.pgs 8/14/08 648 10:37 AM Page 648 Statistics Regression to Use Description of Scatter Plot Examples Specific types of power regression: Quadratic • A quadratic curve Cubic • A cubic curve The different non-linear regression models can be found in the STAT CALC menu of the graphing calculator. • 5:QuadReg is quadratic regression. • 6:CubicReg is cubic regression. • 9:LnReg is logarithmic regression. • 0:ExpReg is exponential regression. • A:PwrReg is power regression. In the example of the file size of Super Type, the scatter plot appears to be exponential or power. The table below shows the data of the scatter plot: Version Size (megabytes) 1 2 3 4 5 6 155 240 387 630 960 1,612 To find the exponential regression model, enter the data into L1 and L2. Choose ExpReg from the STAT CALC menu: ENTER: STAT 䉴 0 VARS 䉴 1 1 ENTER The calculator will display the regression equation and store the equation into Y1 of the Y menu. To the nearest thousandth, the regression equation is y 5 95.699(1.596x). Press ZOOM sion equation. 9 to graph the scatter plot and the regres- 14411C15.pgs 8/14/08 10:37 AM Page 649 Non-Linear Regression 649 ExpReg y=a* b^x a=95.69902108 b=1.595666689 r2=.9994598905 r=.9997299088 To find the power regression model, with the data in L1 and L2, choose PwrReg from the STAT CALC menu: ENTER: STAT 䉴 ALPHA 䉴 VARS A 1 ENTER 1 The calculator will display the regression equation and store the equation into Y1 of the Y menu. To the nearest thousandth, the regression equation is y 5 121.591x1.273. Press ZOOM to graph the scatter plot and the regres- 9 sion equation. PwrReg y=a* x^b a=121.5914377 b=1.273149963 r2=.9307655743 r=.9647619263 From the scatter plots, we see that the exponential regression equation is a better fit for the data. EXAMPLE 1 A stone is dropped from a height of 1,000 feet. The trajectory of the stone is recorded by a high-speed video camera in intervals of half a second. The recorded distance that the stone has fallen in the first 5 seconds in given below: Seconds 1 1.5 2 2.5 3 3.5 4 4.5 Distance 16 23 63 105 149 191 260 321 a. Determine which regression model is most appropriate. b. Find the regression equation. Round all values to the nearest thousandth. 14411C15.pgs 8/14/08 650 10:37 AM Page 650 Statistics Solution a. Draw a scatter plot of the data. The data appears to approximate an exponential function or a power function. Enter the data in L1 and L2. Find and graph the exponential and power models. From the displays, it appears that the power model is the better fit. exponential power b. To the nearest thousandth, the calculator will display the power equation y = axb for a 5 13.619, b 5 2.122. Answers a. Power regression b. y 5 13.619(x2.122) EXAMPLE 2 A pediatrician has the following table that lists the head circumferences for a group of 12 baby girls from the same extended family. The circumference is given in centimeters. Age in Months Circumference 2 2 5 4 1 17 11 14 7 11 10 19 36.8 37.2 38.6 38.2 35.9 40.4 39.7 39.9 39.2 41.1 39.3 40.5 a. Make a scatter plot of the data. b. Choose what appears to be the curve that best fits the data. c. Find the regression equation for this model. 14411C15.pgs 8/14/08 10:37 AM Page 651 Non-Linear Regression 651 Solution a. b. The data appears to approximate a log function. Answer c. With the ages in L1 and the circumferences in L2, choose LnReg from the STAT CALC menu: ENTER: STAT 䉴 9 ENTER ` The equation of the regression equation, to the nearest thousandth, is LnReg y=a+blnx a=35.9381563 b=1.627161495 r2=.9281619959 r=.9634116441 y 5 35.938 1 1.627 ln x Answer Exercises Writing About Mathematics 1. At birth, the average circumference of a child’s head is 35 centimeters. If the pair (0, 35) is added to the data in Example 2, the calculator returns an error message. Explain why. 2. Explain when the power function, y = axb, has only positive or only negative y-values and when it has both positive and negative y-values. Developing Skills In 3–8, determine the regression model that appears to be appropriate for the data. 3. 4. 5. 14411C15.pgs 8/14/08 652 10:37 AM Page 652 Statistics 6. 7. 8. In 9–13: a. Create a scatter plot for the data. b. Determine which regression model is the most appropriate for the data. Justify your answer. c. Find the regression equation. Round the coefficient of the regression equation to three decimal places. 9. 10. 11. 12. x 4 7 3 8 6 5 6 3 9 4.5 y 10 7 15 9 5 6 6 14 14 8 x –2.3 1.8 21.0 4.6 1.4 3.7 5.3 1.9 0.7 4.2 y 2.2 18.3 8.2 x 3.3 23.8 22.1 0.4 3.5 23.8 21.8 20.4 2.4 1.2 y 12.5 217.1 23.6 0.4 15.0 218.9 22.1 20.4 4.2 3.4 x y 13. 1 2 63.7 15.3 43.0 89.5 22.7 12.1 54.7 3 4 5 6 7 8 9 10 27 25.6 24.8 24.2 23.8 23.4 23.1 22.8 22.6 22.4 x 1 2 3 4 5 6 7 8 9 10 y 2.7 2.3 2.0 1.7 1.5 1.2 1.1 0.9 0.8 0.7 Applying Skills 14. Mrs. Vroman bought $1,000 worth of shares in the Acme Growth Company. The table below shows the value of the investment over 10 years. Year Value ($) 1 2 3 4 1,045 1,092 1,141 1,192 5 6 7 8 9 10 1,246 1,302 1,361 1,422 1,486 1,553 a. Find the exponential regression equation for the data with the coefficient and base rounded to three decimal places. b. Predict, to the nearest dollar, the value of the Vromans’ investment after 11 years. 14411C15.pgs 8/14/08 10:37 AM Page 653 Non-Linear Regression 653 15. The growth chart below shows the average height in inches of a group of 100 children from 2 months to 36 months. Month 2 Height in Inches 4 8 10 12 14 16 18 22.7 26.1 27.5 28.9 32.1 31.7 33.1 32.7 34.0 Month 20 Height in inches 6 22 24 26 28 30 32 34 36 34.4 34.6 36.0 34.6 35.2 36.6 35.6 37.2 37.6 a. Find the logarithmic regression equation for the data with the coefficients rounded to three decimal places. b. Predict, to the nearest tenth of an inch, the average height of a child at 38 months. 16. The orbital speed in kilometers per second and the distance from the sun in millions of kilometers of each of six planets is given in the table. Planet Venus Earth Mars Jupiter Saturn Uranus Orbital Speed 34.8 29.6 23.9 12.9 9.6 6.6 Distance from the Sun 108.2 149.6 227.9 778.0 1,427 2,871 a. Find the regression equation that appears to be the best fit for the data with the coefficient rounded to three decimal places. b. Neptune has an orbital speed of 5.45 km/sec and is 4,504 million kilometers from the sun. Does the equation found for the six planets given in the table fit the data for Neptune? 17. A mail order company has shipping boxes that have square bases and varying heights from 1 to 5 feet. The relationship between the height of the box and the volume is shown in the table. Height (ft) 3 Volume (ft ) 1 1.5 2 2.5 3 3.5 2 7 16 31 54 86 4 4.5 128 182 5 250 a. Create a scatter plot for the data. Let the horizontal axis represent the height of the box and the vertical axis represent the volume. b. Determine which regression model is most appropriate for the data. Justify your answer. c. Find the regression equation. Round the coefficient of the regression equation to three decimal places. 14411C15.pgs 8/14/08 654 10:37 AM Page 654 Statistics 18. In an office building the thermostats have six settings. The table below shows the average temperature in degrees Fahrenheit for a month that each setting produced. Setting 1 2 3 4 5 6 Temperature (°F) 61 64 66 67 69 70 a. Create a scatter plot for the data. Let the horizontal axis represent the setting and the vertical axis represent temperature. b. Find the equation of best fit using a power regression. Round the coefficient of the regression equation to three decimal places. 19. The following table shows the speed in megahertz of Intel computer chips over the course of 36 years. The time is given as the number of years since 1971. Year 0 1 3 7 11 14 18 22 Speed 0.108 0.8 2 5 6 16 25 66 Year 24 26 28 29 31 34 35 36 Speed 200 300 500 1,500 1,700 3,200 2,900 3,000 One application of Moore’s Law is that the speed of a computer processor should double approximately every two years. Use this information to determine the regression model. Does Moore’s Law hold for Intel computer chips? Explain. Hands-On Activity: Sine Regression If we make a scatter plot of the following set of data on a graphing calculator, we may observe that the data points appear to form a sine curve. x 13 14 15 16 17 y 211.6 217.2 218.0 215.0 28.4 18 19 20 21 22 23 24 6.2 12.0 20.1 18.4 11.9 3.4 26.9 A sine function should be used to model the data. We can use a graphing calculator to find the sinusoidal regression equation. Enter the x-values into L1 and the y-values into L2. Then with the calculator in radian mode, choose SinReg from the STAT CALC menu: ENTER: 䉴 STAT 1 1 ALPHA ENTER C VARS 䉴 25 20 15 10 5 0 25 210 215 220 10 12 14 16 18 20 22 24 26 14411C15.pgs 8/14/08 10:37 AM Page 655 Interpolation and Extrapolation 655 To the nearest thousandth, the sinusoidal regression equation is: y 5 19.251 sin (0.551x 1 2.871) 2 0.029 Press ZOOM 9 to graph the scatter plot and the regression equation. Note: The sinusoidal regression model on the graphing calculator assumes that the x-values are equally spaced and in increasing order. For arbitrary data, you need to give the calculator an estimate of the period. See your calculator manual for details. The average high temperature of a city is recorded for 14 months. The table below shows this data. Month 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Temp (°F) 40 48 61 71 81 85 83 77 67 54 41 39 42 49 a. Create a scatter plot for the data. b. Find the sinusoidal regression equation for the data with the coefficient and base rounded to three decimal places. c. Predict the average temperature of the city at 15 months. Round to the nearest degree. d. Predict the average temperature of the city at 16 months. Round to the nearest degree. 15-10 INTERPOLATION AND EXTRAPOLATION Data are usually found for specific values of one of the variables. Often we wish to approximate values not included in the data. Interpolation The process of finding a function value between given values is called interpolation. EXAMPLE 1 Each time Jen fills the tank of her car, she records the number of gallons of gas needed to fill the tank and the number of miles driven since the last time that she filled the tank. Her record is shown in the table. Gallons of Gas 7.5 8.8 5.3 9.0 8.1 4.7 6.9 8.3 Miles 240 280 170 290 260 150 220 270 a. If Jen needs 8.0 gallons the next time she fills the tank, to the nearest mile, how many miles will she have driven? b. If Jen has driven 200 miles, to the nearest tenth, how many gallons of gasoline can she expect to need? 14411C15.pgs 8/14/08 656 10:37 AM Page 656 Statistics Solution Graph the scatter plot of the data. There appears to be positive linear correlation. With the data in L1 and L2, use the calculator to find the regression equation: ENTER: STAT 䉴 4 ENTER When rounded to the nearest thousandth, the linear equation returned is LinReg y=ax+b a=32.36537919 b=-2.076402594 r2=.9988025622 r=.9994011018 y 5 32.365x 2 2.076 a. Substitute 8.0 for x in the equation given by the calculator. y 5 32.365x 2 2.076 5 32.365(8.0) 2 2.076 5 256.844 Jen will have driven approximately 257 miles. Answer b. Substitute 200 for y in the equation given by the calculator. y 5 32.365x 2 2.076 200 5 32.365x 2 2.076 202.076 5 32.365x 6.244 x Jen will need approximately 6.2 gallons of gasoline. Answer Extrapolation Often we want to use data collected about past events to predict the future. The process of using pairs of values within a given range to approximate values outside of the given range of values is called extrapolation. 14411C15.pgs 8/14/08 10:37 AM Page 657 Interpolation and Extrapolation 657 EXAMPLE 2 The following table shows the number of high school graduates in the U.S. in the thousands from 1992 to 2004. Year 1992 1993 1994 1995 1996 1997 1998 No. of Graduates 2,478 2,481 2,464 2,519 2,518 2,612 2,704 Year 1999 2000 2001 2002 2003 2004 No. of Graduates 2,759 2,833 2,848 2,906 3,016 3,081 a. Write a linear regression equation for this data. b. If the number of high school graduates continued to grow at this rate, how many graduates would there have been in 2006? c. If the number of high school graduates continues to grow at this rate, when is the number of high school graduates expected to exceed 3.5 million? Solution a. Enter the year using the number of years since 1990, that is, the difference between the year and 1990, in L1. Enter the corresponding number of high school graduates in L2. The regression equation is: y 5 53.984x 1 2,277.286 Answer b. Use the equation y 5 53.984x 1 2,277.286 and let x 5 16: y 5 53.984(16) 1 2,277.286 3,141 If the increase continued at the same rate, the expected number of graduates in 2006 would have been approximately 3,141,000. c. Use the equation y 5 53.984x 1 2,277.286 and let y 5 3,500: 3,500 5 53.984x 1 2,277.286 1,222.714 5 53.984x 22.650 x If the rate of increase continues, the number of high school graduates can be expected to exceed 3.5 million in the 23rd year after 1990 or in the year 2013. 14411C15.pgs 8/14/08 658 10:37 AM Page 658 Statistics Unlike interpolation, extrapolation is not usually accurate. Extrapolation is valid provided we are sure that the regression model continues to hold outside of the given range of values. Unfortunately, this is not usually the case. For instance, consider the data given in Exercise 15 of Section 15-9. The growth chart below shows the average height in inches of a group of 100 children from 2 months to 36 months. Month 2 Height in Inches 4 6 8 10 12 14 16 18 22.7 26.1 27.5 28.9 32.1 31.7 33.1 32.7 34.0 Month 20 Height in inches 22 24 26 28 30 32 34 36 34.4 34.6 36.0 34.6 35.2 36.6 35.6 37.2 37.6 The data appears logarithmic. When the coefficient and the exponent are rounded to three decimal places, an equation that best fits the data is y 5 19.165 1 5.026 ln x. If we use this equation to find the height of child who is 16 years old (192 months), the result is approximately 45.6 inches or less than 4 feet. The average 16-year-old is taller than this. The chart is intended to give average growth for very young children and extrapolation beyond the given range of ages leads to errors. Exercises Writing About Mathematics 1. Explain the difference between interpolation and extrapolation. 2. What are the possible sources of error when using extrapolation based on the line of best fit? Developing Skills In 3–5: a. Determine the appropriate linear regression model to use based on the scatter plot of the given data. b. Find an approximate value for y for the given value of x. c. Find an approximate value for x for the given value of y. 3. b. x 5 5.7 c. y 5 1.25 x y 1 2 3 4 5 6 7 8 9 10 1.05 1.10 1.16 1.22 1.28 1.34 1.41 1.48 1.55 1.62 14411C15.pgs 8/14/08 10:37 AM Page 659 Interpolation and Extrapolation 659 4. b. x 5 12 c. y 5 140 x 1 2 3 4 5 6 7 8 9 10 y 3.1 3.6 4.0 4.5 5.1 5.6 6.0 6.5 6.9 7.5 6 7 8 9 10 5. b. x 5 0.5 c. y 5 0.5 x 1 2 3 4 5 y 1 2.3 3.7 5.3 6.9 8.6 10.3 12.1 14.0 15.9 In 6–9: a. Determine the appropriate non-linear regression model to use based on the scatter plot of the given data. b. Find an approximate value for y for the given value of x. c. Find an approximate value for x for the given value of y. 6. b. x 5 1.4 c. y 5 1.50 x y 1 2 3 4 5 6 7 8 9 10 0.80 1.09 1.25 1.37 1.46 1.54 1.60 1.66 1.71 1.75 7. a. x 5 12 b. y 5 80 x y 23 26 13 14 11.6 33.3 52.5 43.5 20 17 29 18 18 17 4.0 18.7 84.0 8.0 12.4 11.5 8. a. x 5 10.5 b. y 5 100.0 x –2.0 21.0 20.5 0.1 0.5 0.8 y 1.0 7.7 9.3 12.0 16.5 21.6 28.0 2.3 3.3 5.3 1.1 1.5 1.8 2.1 9. a. x 5 12 b. y 5 215 x 0.5 1.7 2.7 3.9 4.9 5.7 7.0 8.2 9.2 10.0 y 0.1 2.1 7.8 23.1 43.4 65.1 114.9 183.8 248.3 311.3 14411C15.pgs 8/14/08 660 10:37 AM Page 660 Statistics Applying Skills In 10–12, determine the appropriate linear regression model to use based on the scatter plot of the given data. 10. The following table represents the percentage of the Gross Domestic Product (GDP) that a country spent on education. Year 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 Percent 2.71 3.17 5.20 5.61 7.20 8.14 8.79 10.21 10.72 11.77 a. Estimate the percentage of the GDP spent on education in 1998. b. Assuming this model continues to hold into the future, predict the percentage of the GDP that will be spent on education in 2015. 11. The following chart gives the average time in seconds that a group of 10 F1 racing cars went from zero to the given miles per hour. Speed 75 100 125 150 175 200 225 250 275 300 Time 1.2 2.2 2.9 4.0 5.6 6.8 7.3 8.7 9.3 9.9 a. What was the average time it took the 10 racing cars to reach 180 miles per hour? b. Estimate the average time it will take the 10 racing cars to reach 325 miles per hour. 12. The relationship between degrees Celsius and degrees Fahrenheit is shown in the table at intervals of 10° Fahrenheit. Celsius 0 10 20 30 40 50 60 70 80 90 100 Fahrenheit 32 50 68 86 104 122 140 158 176 194 212 a. Find the Fahrenheit temperature when the Celsius temperature is 25°. b. Find the Celsius temperature when the Fahrenheit temperature is 24°. 13. The following table gives the number of compact cars produced in a country over the course of several years. Year 1981 1984 1987 1990 1993 1996 1999 2002 2005 2008 No. of Cars 100 168 471 603 124 1,780 1,768 4,195 6,680 10,910 a. Estimate the number of cars produced by the country in 2000 using an exponential model. b. Estimate the number of cars produced by the country in 1978 using the model from part a. 14411C15.pgs 8/14/08 10:37 AM Page 661 Interpolation and Extrapolation 661 14. In an office building the thermostats have six settings. The table below shows the average temperature in degrees Fahrenheit for a month that each setting produced. Setting 1 2 3 4 5 6 Temperature (°F) 61 64 66 67 69 70 Using a power model and assuming that it is possible to choose a setting between the given settings: a. what temperature would result from a setting halfway between 2 and 3? b. where should the setting be placed to produce a temperature of 68 degrees? In 15 and 16, determine the appropriate non-linear regression model to use based on the scatter plot of the given data. 15. A mail order company has shipping boxes that have square bases and varying height from 1 to 5 feet. The relationship between the height of the box and the volume in cubic feet is shown in the table. Height 1 1.5 2 2.5 3 3.5 4 4.5 5 Volume 2 6.75 16 31.25 54 85.75 128 182.25 250 a. If the company introduces a box with a height of 1.25 feet, what would be the volume to the nearest hundredth cubic foot? b. If the company needs a box with a volume of at least 100 cubic feet, what would be the smallest height to the nearest tenth of a foot? c. If the company needs a box with a volume of 800 square feet, what would be the height to the nearest foot? 16. Steve kept a record of the height of a tree that he planted. The heights are shown in the table. Age of Tree in Years 1 3 5 Height in Inches 7 12 15 7 9 11 13 16.5 17.8 19 20 a. Write an equation that best fits the data. b. What was the height of the tree after 2 years? c. If the height of the tree continues in this same pattern, how tall will the tree be after 20 years?