Section 2.1 2 Trigonometric Functions 16. 1. C = 2π r 2. A = π r 2 17. 3. standard position 4. rθ ; 1 r 2θ 2 5. s ; θ 18. t t 6. False 7. True 8. True 9. True 10. False 11. 12. 13. 19. 20. 21. 22. 14. _ ⎛ 1 1 1 ⎞ º 23. 40º10 ' 25" = ⎜ 40 + 10 ⋅ + 25 ⋅ ⋅ ⎟ ⎝ 60 60 60 ⎠ 15. ≈ (40 + 0.1667 + 0.00694)º ≈ 40.17º ⎛ 1 1 1 ⎞ º 24. 61º 42 ' 21" = ⎜ 61 + 42 ⋅ + 21⋅ ⋅ ⎟ ⎝ 60 60 60 ⎠ ≈ (61 + 0.7000 + 0.00583)º ≈ 61.71º
0.4(60 ") 37. 240 ⋅ π radian = 4π 180 3 radians = 61º +14 '+ 24" 38. 330°= 330 ⋅ π radian = 11π radians = 61º14 ' 24 " 180 6 31. 18.255º = 18º + 0.255º = 18º + 0.255(60 ') 39. − 60° = − 60 ⋅ π radian 180 = − π radian 3 = 18º +15.3 ' 40. = −30 ⋅ π radian = − π radian = 18º +15 '+ 0.3 ' = 18º +15 '+ 0.3(60 ") = 18º +15 '+ 18" = 18º15 '18" 180 6 41. 180°= 180 ⋅ π radian = π radians 180 42. 270°= 270 ⋅ π radian = 3π radians 180 2 −3
43. −135° = −135 ⋅ π radian = − 3π radians 60. 73°= 73 ⋅ π radian 180 4 = 73π radians 44. − 225° = − 225 ⋅ π radian = − 5π radians 180 180 4 ≈ 1.27 radians 45. − 90° = − 90 ⋅ π radian = − π radians 61. − 40° = − 40 ⋅ π radian 180 2 180 2π radian 46. = −180 ⋅ π 180 radian = −π radians =− 9 ≈ − 0.70 radian 47. π = π ⋅ 180 degrees = 60° π 3 3 π 62. − 51° = − 51⋅ 180 48. 5π = 5π ⋅ 180 degrees = 150° 17π radian 6 6 π 60 ≈ − 0.89 radian 49. − 5π = − 5π ⋅ 180 degrees = − 225° 4 4 π 63. 125°= 125 ⋅ π 180 50. − 2π = − 2π ⋅ 180 degrees = −120° 25π 3 3 π = radians 36 51. π = π ⋅ 180 degrees = 90° °
s = rθ = 2 ⋅ π = π ≈ 1.047 inches 6 3 θ = 6 = 0.24 radian 25 84. r = 6 meters; A = 8 m2 78. r = 3 meters; θ = 120º = 120 ⋅ π = 2π radians 1 2 79. 180 3 s = rθ = 3 ⋅ 2π = 2π ≈ 6.283 meters 3 r = 10 meters; θ = 1 radian 2 A = 1 r 2θ = 1 (10)2 ⎛ 1 ⎞ = 100 =25 m2 A = r θ 2 8 = 1 (6)2 θ 2 8 = 18θ θ = 8 = 4 ≈ 0.444 radian 18 9 ⎜ ⎟ 2 2 ⎝ ⎠ 4 85. r = 2 θ = 30º = 30 ⋅ π = π radian 80. r = 6 feet; θ = 2 radians 180 6 A = 1 r 2θ = 1 2 2 ⎛ π ⎞ = π ≈ 1.047 in 2 A = 1 r 2θ = 1 (6)2 (2) =36 ft 2 2 2 ( ) ⎜ ⎟ 2 2 ⎝ ⎠ 3
⎜ ⎟ 2 2 ⎝ ⎠ 2 = 1250θ 90. r = 9 cm; θ = 50º = 50 ⋅ π = 5π radian θ = 100 = 2 = 0.08 radian 180 18 s = rθ = 9 ⋅ 5π ≈ 7.854 cm 1250 25 or 2 ⋅ 180 = ⎛ 72 ⎞ ≈ 4.58° 18 A = 1 r 2θ = 1 (9)2 ⎛ 5π ⎞ = 45π ≈ 35.343 cm2 25 π ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ 4 97. r = 5 cm; t = seconds; θ = 1 radian 3 91. r = 6 In 15 minutes, ω = θ t (1/ 3) = 20 = 1 ⋅ 1 = 1 radian/sec 3 20 60 θ = 15 rev = 1 ⋅ 360º = 90º = π radians s rθ 5 ⋅ (1 / 3) 5 1 1 60 4 2 v = = = = ⋅ = cm/sec s = rθ = 6 ⋅ π = 3π ≈ 9.42 inches t t 20 3 20 12 2 98. r = 2 meters; t = 20 seconds; s = 5 meters In 25 minutes, θ ( s / r ) (5 / 2) 5 1 1 ω = = = = ⋅ = radian/sec θ = 25 rev = 5 ⋅ 360º = 150º = 5π radians 60 12 6 t t 20 2 20 8 s 5 1 v = = = m/sec s = rθ = 6 ⋅ 5π = 5π ≈ 15.71 inches 6 t 20 4 0
99. d = 26 inches; r = 13 inches; v = 35 mi/hr 105. r = 2.39 ×105 miles v = 35 mi ⋅ 5280 ft ⋅ 12 in. ⋅ 1 hr hr mi ft 60 min = 36, 960 in./min ω = 1 rev/27.3 days = 2π radians/27.3 days π ω = v = 36, 960 in./min = 12 ⋅ 27.3 radians/hr r 13 in. 5 π ≈ 2843.08 radians/min ≈ 2843.08 rad ⋅ 1 rev min 2π rad ≈ 452.5 rev/min 106. v = rω = (2.39 ×10 ) ⋅ ≈ 2292 miles/hr 327.6 r = 9.29 ×107 miles ω = 1 rev/365 days = 2π radians/365 days 100. r = 15 inches; ω = 3 rev/sec = 6π rad/sec π v = rω = 15 ⋅ 6π in./sec = 90π ≈ 282.7 in/sec v = 90π in. ⋅ 1ft ⋅ 1mi ⋅ 3600 sec ≈ 16.1 mi/hr sec 12in. 5280ft 1hr = radians/hr 12 ⋅ 365 v = rω = (9.29 ×107 ) ⋅ π ≈ 66, 633 miles/hr 4380 101. r = 3960 miles 107. r = 2 inches; r = 8 inches; θ = 35º 9 '− 29º 57 ' = 5º12 ' = 5.2º 1 2 ω1 = 3 rev/min = 6π radians/min Find ω2 : v1 = v2 = 5.2 ⋅ π r ω = r ω 180 ≈ 0.09076 radian 1 1 2 2 2(6π) = 8ω2 102. s = rθ = 3960 ⋅ 0.09076 ≈ 359 miles r = 3960 miles θ = 38º 21'− 30º 20 ' ω = 12π 2 8 = 1.5π radians/min 1.5π = 8º1' ≈ 8.017º = rev/min 2π 3 = 8.017 ⋅ π 180 = rev/min 4 ≈ 0.1399 radian
s = rθ = 3960 ⋅ 0.1399 ≈ 554 miles 108. r = 30 feet ω = 1 rev = 2π = π ≈ 0.09 rad/sec 103. r = miles ω = 1 rev/day = 2π radians/day = π 12 radians/hr 70 sec 70 sec 35 v = rω = 30 feet ⋅ π rad = 6π ft ≈ 2.69 feet/sec 35 sec 7 sec 104. v = rω = 3429.5 ⋅ π ≈ 898 miles/hr 12 r = miles 109. r = 4 feet; v = rω = 4 ⋅ 20π ft ω = 10 rev/min = 20π radians/min ω = 1 rev/day = 2π radians/day = π radians/hr = 80π min 12 80π ft 1 mi 60 min v = rω = 3033.5 ⋅ π ≈ 794 miles/hr = min ⋅ ⋅ 5280 ft hr 12 ≈ 2.86 mi/hr
12. ⎛ 1 3 ⎞ 1 3 P = ⎜ , − ⎟ ⇒ x = , y = − 14. ⎛ 1 2 6 ⎞ 1 2 6 P = ⎜ − , ⎟ ⇒ x = − , y = ⎝ 2 2 ⎠ 2 2 ⎝ 5 5 ⎠ 5 5 sin t = y = − 3 2 sin t = y = 2 6 5 cos t = x = 1 2 ⎛ 3 ⎞ − cos t = x 1 =− 5 ⎛ 2 6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ tan t = y = ⎝ 2 ⎠ = − 3 ⋅ 2 = − 3 tan t = y = ⎝ 5 ⎠ = 2 6 ⎜ − 5 ⎟ = − 2 6 x ⎛ 1 ⎞ 2 1 x ⎛ 1 ⎞ 5 ⎝ 1 ⎠ ⎜ ⎟ ⎝ ⎠ csc t = 1 = 1 = − 2 = − 2 ⋅ 3 = − 2 3 ⎜ − ⎟ ⎝ 5 ⎠ csc t = 1 = 1 = 1⋅ 5 = 5 ⋅ 6 = 5 6 y ⎛ 3 ⎞ − 3 3 3 3 y ⎛ 2 6 ⎞ 2 6 2 6 6 12 ⎜ 2 ⎟ ⎜ 5 ⎟ ⎝ ⎠ sec t = 1 = 1 = 1⋅ 2 = 2 ⎝ ⎠ sec t = 1 = 1 = 1⎛ − 5 ⎞ = −5 x ⎛ 1 ⎞ 1 x 1 ⎜ 1 ⎟ ⎜ ⎟ ⎛ − ⎞ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ − ⎟ ⎛ ⎞ cot t = x = ⎝ 2 ⎠ 1 2 1 =− ⋅ 3 =− 3 cot t = x = ⎝ 5 ⎠ = − 1 ⎜ 5 ⎟ y ⎛ 3 ⎞ − 2 3 3 3 3 y ⎛ 2 6 ⎞ 5 ⎝ 2 6 ⎠ ⎜ 2 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ = − 1 ⋅ 6 = − 6 13. ⎛ 2 21 ⎞ 2 21 P = ⎜ − , ⎟ ⇒ x = − , y = 2 6 6 12 ⎝ 5 5 ⎠ 5 5 sin t = y = 21 15. ⎛ 2 2 ⎞ 2 2 P = − , ⇒ x = − , y =
5 ⎜ 2 2 ⎟ 2 2 ⎝ ⎠ cos t = x = − 2 2 5 ⎛ 21 ⎞ sin t = 2 2 y ⎜ ⎟ 21 ⎛ 5 ⎞ 21 tan t = = = − = − cos t = x = − 2 x ⎛ 2 ⎞ 5 ⎜ 2 ⎟ 2 ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ tan t = y = ⎛ 2 ⎞ ⎜ ⎟ ⎝ ⎠ = −1 csc t = 1 = 1 = 1⋅ 5 = 5 ⋅ 21 = 5 21 x ⎛ 2 ⎞ y ⎛ 21 ⎞ 21 21 21 21 ⎜ − 2 ⎟ ⎝ ⎠ ⎜ 5 ⎟ = 1 = 1 = ⋅ 2 = 2 ⋅ 2 = sec t = 1 = 1 = 1⎛ − 5 ⎞ = − 5 csc t y 1 ⎛ 2 ⎞ 2 2 2 2 x ⎛ 2 ⎞ ⎜ 2 ⎟ 2 ⎜ ⎟ ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ 2 ⎠ 1 1 ⎛ 2 ⎞ 2 2 ⎛ − 2 ⎞ sec t = = = 1 − x ⎛ ⎞ ⎟ = − ⋅ = − 2 ⎜ ⎟ ⎜ − 2 ⎟ ⎝ 2 ⎠ 2 2 cot t = x = ⎝ 5 ⎠ = − 2 ⋅ 5 y ⎛ 21 ⎞ 5 21 ⎝ 2 ⎠ ⎜ 5 ⎟ ⎛ 2 ⎞ ⎝ ⎠ − x 2 21 2 21 =− ⋅ = − cot t = = ⎝ 2 ⎠ = −1 21 21 21 y ⎛ 2 ⎞ ⎜ 2 ⎟ ⎝ ⎠
⎛ 5 ⎞ ⎜ − ⎟ ⎛ ⎞ cos t = x = 2 2 cot t = x = ⎝ 3 ⎠ = − 5 − 3 = 5 3 y ⎛ 2 ⎞ 3 ⎜ 2 ⎟ 2 ⎛ − 1 ⎞ tan t = = ⎝ ⎠ = − 1 ⋅ 3 ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ x ⎛ 2 2 ⎞ 3 2 2 19. sin ⎛ 11π ⎞ = sin ⎛ 3π + 8π ⎞ ⎜ 3 ⎟ ⎝ 2 ⎠ ⎝ 2 2 ⎠ ⎝ ⎠ = sin ⎛ 3π + 4π ⎞ = − 1 ⋅ 2 = − 2 ⎜ 2 ⎟ 2 2 2 4 ⎝ ⎠ = sin ⎛ 3π + 2 ⋅ 2π ⎞ csc t = 1 = 1 = 1⎛ − 3 ⎞ = ⎜ ⎟ y ⎛ 1 ⎞ ⎜ 1 ⎟ ⎝ 2 ⎠ ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ sec t = 1 = 1 = 1⎛ 3 ⎞ = 3 ⋅ 2 = 3 2 = sin ⎛ 3π ⎞ ⎝ 2 ⎠ = −1 x ⎛ 2 2 ⎞ ⎜ ⎟ 2 2 2 4 ⎜ 3 ⎟ ⎝ ⎠
20. 21. cos (7π) = cos (π + 6π ) = cos (π + 3 ⋅ 2π ) = cos (π) = −1 tan (6π) = tan(0 + 6π) = tan (0) = 0 31. sin 90º + tan 45º = 1 + 1 = 2 32. cos180º − sin 180º = −1 − 0 = −1 22. cot ⎛ 7π ⎞ = cot ⎛ π + 6π ⎞ ⎝ ⎠ ⎝ ⎠ = cot ⎛ π + 3π ⎞ = cot ⎛ π ⎞ = 33. 34. sin cos 45º = 2 ⋅ 2 = 2 = 1 2 2 4 2 tan 45º cos 30º = 1⋅ 3 = 3 ⎜ 2 ⎟ ⎜ 2 ⎟ 2 2 23. ⎝ ⎠ ⎝ ⎠ csc ⎛ 11π ⎞ = csc ⎛ 3π + 8π ⎞ 35. csc 45º tan 60º = 2 ⋅ 3 = 6 ⎜ 2 ⎟ ⎜ 2 2 ⎟ ⎝ ⎠ ⎝ ⎠ 2 3 2 3 = csc ⎛ 3π + 4π ⎞ 36. sec 30º cot 45º = ⋅1 = 3 3 ⎜ 2 ⎟ ⎝ ⎠ = csc ⎛ 3π + 2 ⋅ 2π ⎞ 37. 4 sin 90º − 3 tan 180º = 4 ⋅1 − 3 ⋅ 0 = 4 ⎜ 2 ⎟ ⎝ ⎠ = csc ⎛ 3π ⎞ ⎝ ⎠ 38. 5 cos 90º − 8 sin 270º = 5 ⋅ 0 − 8(−1) = 8 24. = −1 sec (8π) = sec (0 + 8π ) 39. 2sin π − 3 tan π = 2 ⋅ 3 − 3 ⋅ 3 = 3 6 2 3 3 − 3 = 0 25. = sec (0 + 4 ⋅ 2π ) = sec (0) = 1 cos ⎛ − 3π ⎞ = cos ⎛ 3π ⎞ 40. π + 3 tan π = 2 ⋅ 2 + 3 ⋅1 = 4 4 2 2 + 3 ⎜ 2 ⎟ ⎜ 2 ⎟ π π 2 2 ⎝ ⎠ ⎝ ⎠ 41. sin − cos = − = 0 = cos ⎛ π − 4π ⎞ 4 4 2 2 ⎜ 2 2 ⎟ ⎝ ⎠ = cos ⎛ π + (−1) ⋅ 2π ⎞ 42. tan π + cos π = 3 + 1 = 1 + 2 3 ⎜ 2 ⎟ 3 3 2 2 ⎝ ⎠
= cos ⎛ π ⎞ ⎜ 2 ⎟ 43. π + 4 cot π = 2 ⋅ 2 + 4 ⋅ 3 = 2 2 + 4 3 ⎝ ⎠ 4 3 3 3 = 0 26. sin = − sin (3π) = − sin (π + 2π ) = − sin (π) = 0 44. π + cot π = 3 ⋅ 2 3 + 1 = 2 3 + 1 3 4 3 27. sec (−π) = sec (π) = −1 45. tan π − cos 0 = 0 − 1 = −1 3π 28. tan (−3π) = − tan(3π) =− tan (0 + 3π ) = − tan (0) = 0 46. sin 2 + tan π = −1 + 0 = −1 29. sin 45º + cos 60º = 2 + 1 = 1 + 2 2 2 2 47. csc π + cot π = 1 + 0 = 1 2 2 π 30. sin 30º − cos 45º = 1 − 2 = 1 − 2 2 2 2 48. sec π − csc 2 = −1 − 1 = − 2
⎝ ⎠ 52. The point on the unit circle that corresponds to θ = 5π = 150º is ⎛ − 3 , 1 ⎞ . θ = 240º = 4π ⎛ 1 3 ⎞ is ⎜ − , − ⎟ . ⎜ ⎟ 3 ⎝ 2 2 ⎠ 6 sin 5π = 1 6 2 cos 5π =− 3 ⎝ 2 2 ⎠ 3 sin 240º =− 2 cos 240º = − 1 2 6 2 ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜= − ⎟ ⎛ ⎞ ⎜ ⎟ tan 240º = ⎝ 2 ⎠ = − 3 ⋅ ⎜ − 2 ⎟ = 3 tan 5π = ⎝ 2 ⎠ = 1 ⋅ ⎛ − 2 ⎞ ⋅ 3 = − 3 6 ⎛ 3 ⎞ 2 ⎜ 3 ⎟ 3 3 ⎛ − 1 ⎞ 2 ⎝ 1 ⎠ ⎜ − ⎟ ⎝ ⎠ ⎜ 2 ⎟ ⎝ 2 ⎠ ⎝ ⎠ csc 240º = 1 = 1⋅ ⎛ − 2 ⎞ ⋅ 3 = − 2 3 csc 5π = 1 = 1⋅ 2 = 2 ⎛ ⎞ ⎜ ⎟ 3 6 ⎛ 1 ⎞ 1 ⎜ − 3 ⎟ ⎝ 3 ⎠ 3 ⎜ ⎟ ⎝ ⎠ sec 5π = 1 = 1⋅ ⎛ − 2 ⎞ ⋅ 3 = − 2 3 sec 240º = 1 = 1⋅ ⎛ − 2 ⎞ = − 2 6 ⎛ 3 ⎞ ⎜ 3 ⎟ 3 3 ⎛ − 1 ⎞ ⎝ 1 ⎠ ⎜ − ⎟ ⎝ ⎠ ⎜ 2 ⎟ ⎝ 2 ⎠ ⎝ ⎠ ⎛ − 1 ⎞ ⎛ 3 ⎞ − ⎜ 2 ⎟ 1 ⎛ 2 ⎞ 3 3 π ⎜ ⎟ cot 240º = ⎝ ⎠ = − ⋅ ⎜ − ⎟ ⋅ = cot 5 6 = ⎝ 2 ⎛ 1 ⎞ ⎠ = − 3 ⋅ 2 = − 3 2 1 ⎛ 3 ⎞ ⎜ − ⎟ 2 ⎝ 3 ⎠ 3 3 ⎜ ⎟ ⎝ 2 ⎠ ⎝ ⎠
61. The point on the unit circle that corresponds to 63. The point on the unit circle that corresponds to θ = −45º π ⎛ 2 2 ⎞ is ⎜ , − ⎟ . θ = 5π = 450º is (0, 1) . 4 ⎝ sin (−45º ) = − 2 2 2 2 ⎠ 2 sin 5π = 1 2 cos 5π = 0 csc 5π = 1 = 1 2 1 sec 5π = 1 = not defined cos (−45º ) = 2 2 ⎛ 2 ⎞ ⎜ − ⎟ tan (−45º ) = ⎝ 2 ⎠ = − 2 ⋅ 2 = −1 2 tan 5π 2 5π 2 0 = 1 = not defined 0 0 ⎛ 2 ⎞ 2 2 cot 2 = = 0 1 ⎜ 2 ⎟ ⎝ ⎠ csc (−45º ) = 1 = 1⋅ ⎛ − 2 ⎞ ⋅ 2 = − 2 64. The point on the unit circle that corresponds to ⎛ 2 ⎞ ⎜ 2 ⎟ 2 θ = 5π = 900º is (−1, 0) . ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ sin 5π = 0 csc 5π = 1 = not defined 0 sec (−45º ) = 1 = 1⋅ 2 ⋅ 2 = 2 1 ⎛ 2 ⎞ 2 2 cos 5π = −1 sec 5π = = −1 ⎜ 2 ⎟ −1 ⎝ ⎠ ⎛ 2 ⎞ ⎜ ⎟ tan 5π = 0 = 0 cot 5π = −1 = not defined 0 cot (−45º ) = ⎝ 2 ⎠ ⎛ 2 ⎞ − = −1 65. The point on the unit circle that corresponds to ⎜ 2 ⎟ θ = 720º = 4π is (1, 0) . ⎝ ⎠ 62. The point on the unit circle that corresponds to sin 720º = 0 csc 720º = 1 = not defined 0 θ = − 60º = − π ⎛ 1 3 ⎞ is ⎜ , − ⎟ . cos 720º = 1 sec 720º = 1 = 1
3 sin (− 60º ) = − 3 2 ⎝ 2 2 ⎠ tan 720º = 0 = 0 1 1 cot 720º = 1 = not defined 0 cos (− 60º ) = 1 2 66. The point on the unit circle that corresponds to 7π ⎛ 3 ⎞ ⎜ − ⎟ tan (− 60º ) = ⎝ 2 ⎠ = − 3 ⋅ 2 = − 3 θ = 630º = 2 is (0, − 1) . 1 ⎛ 1 ⎞ 2 1 ⎜ ⎟ ⎝ ⎠ sin 630º = −1 cos 630º = 0 csc 630º = = −1 −1 sec 630º = 1 = not defined csc (− 60º ) = 1 = 1⋅ ⎛ − 2 ⎞ ⋅ 3 = − 2 3 0 ⎛ 3 ⎞ ⎜ 3 ⎟ 3 3 − ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ tan 630º = 1 = not defined 0 sec (− 60º ) = 1 = 1⋅ 2 = 2 cot 630º = 0 = 0 ⎛ 1 ⎞ 1 ⎜ ⎟ ⎝ ⎠ ⎛ 1 ⎞ ⎜ ⎟ cot (− 60º ) = ⎝ ⎠ = ⋅ ⎜ − ⎟ ⋅ = − −1 67. Set the calculator to degree mode: sin 28º ≈ 0.47 . ⎛ 3 ⎞ 2 ⎝ − 3 ⎠ 3 3 ⎜ 2 ⎟ ⎝ ⎠
9 + 4 = 13 5 12 sin θ = − 2 ⋅ 13 = − 2 13 85. For the point (2, −3) , x = 2 , y = −3 , 13 13 13 r = x2 + y 2 = 4 + 9 = 13 cosθ = −3 ⋅ 13 = − 3 13 sin θ = −3 ⋅ 13 3 13 13 cscθ =− 13 13 13 − 2 2 13 13 13 3 tan θ = = −3 3 cosθ = 2 ⋅ 13 = 2 13 secθ = 13 13 13 13 13 13 2 cscθ = = − − 2 2 tan θ 3 cot θ = − 2 13 13 2 3 secθ = = − − 3 3 86. For the point (−1, −2) , x = −1 , y = − 2 , cot θ = − 3 = 3 r = x2 + y2 = 1 + 4 = 5 − 2 2 sin θ = − 2 ⋅ 5 = − 2 5 cscθ = 5 = − 5 90. For the point (2, 2) , x = 2 , y = 2 , 5 5 5 − 2 2 r = x2 + y 2 = 4 + 4 = 8 = 2 2 cosθ = −1 ⋅ 5 = − 5 secθ = 5 = − 5 2 2 2 2 2 5 5 5 −1 sin θ = ⋅ = 2 2 2 cscθ = = 2 2 2 tan θ = − 2 = 2 cot θ = −1 = 1 −1 − 2 2 cosθ = 2 ⋅ 2 = 2 secθ = 2 2 = 2 2 2 2 2 2 tan θ = 2 = 1 cot θ = 2 = 1 2 2
92. For the point (−0.3, −0.4) , x = − 0.3 , y = − 0.4 , 1 1 5 r = x2 + y2 = 0.09 + 0.16 = 0.25 = 0.5 101. If sin θ = , then cscθ = = 1⋅ = 5 . 5 ⎛ 1 ⎞ 1 ⎜ ⎟ sin θ = − 0.4 = − 4 0.5 5 cscθ = 0.5 = − 5 − 0.4 4 ⎝ 5 ⎠ 2 1 3 3 cosθ = − 0.3 = − 3 0.5 5 tan θ = − 0.4 = 4 − 0.3 3 secθ = 0.5 = − 5 − 0.3 3 cot θ = − 0.3 = 3 − 0.4 4 102. If cosθ = , then secθ = = 1⋅ = . 3 ⎛ 2 ⎞ 2 2 ⎜ ⎟ ⎝ 3 ⎠ 93. sin ( 45º ) + sin (135º ) + sin ( 225º ) + sin (315º ) 103. f (60º ) = sin (60º ) = 3 2 2 2 ⎛ 2 ⎞ ⎛ 2 ⎞ 1 = + + ⎜ − ⎟ + ⎜ − ⎟ 104. g (60º ) = cos (60º ) = 2 2 ⎝ = 0 2 ⎠ ⎝ 2 ⎠ 2
115. g = 9.8 m/sec2 θ = 30º ; v = 150 m/sec : 2 (150)2 sin 30º ⋅ cos 30º R (30º ) = g ≈ 1988.32 m Use the formula H (θ ) = v0 sin θ 2 g with f (θ ) = sin θ θ approaches 1 as θ approaches 0. g = 9.8 m/sec2 ; θ = 30º ; v = 150 m/sec : 2 H (30º ) = = ≈ 286.99 m .6 8)
119. Use the formula R (θ ) = 2v0 sin θ cosθ with x (45) = cos (45º ) + 16 + 0.5 cos(2 ⋅ 45º ) g g = 9.8 m/sec2 ; θ = 25º ; v = 500 m/sec : = cos (45º ) + 16 + 0.5 cos (90º ) ≈ 4.707 cm R (25º ) = 2 (500) sin 25º ⋅ cos 25º ≈ 19, 541.95 m g 123. Note: time on road = distance on road rate on road 2 2 = 8 − 2 x Use the formula H (θ ) = v0 sin θ 2 g with 8 x g = 9.8 m/sec2 ; θ = 25º ; v = 500 m/sec : 5002 sin 2 25º H (25º ) = ≈ 2278.14 m = 1 − 4 1 2(9.8) 2 = 1 − tan θ 4 = 1 − 1 120. Use the formula R (θ ) = 2v0 sin θ cosθ g with 4 tan θ g = 32.2ft/sec2 ; θ = 50º ; v = 200 ft/sec : a. T (30º ) = 1 + 2 − 1 3 sin 30º 4 tan 30º 2 (200 )=2 sin 50º ⋅ cos 50º R (50º ) = ≈ 1223.36 ft = 1 + 2 − 1 g 3 ⋅ 1 4 ⋅ 1 Use the formula H (θ ) = v0 sin θ 2 g with 2 3 = 1 + 4 − 3 ≈ 1.9 hr 3 4 g = 32.2ft/sec2 ; θ = 50º ; v = 200 ft/sec : 2002 sin 2 50º H (50º ) = ≈ 364.49 ft 2(32.2) Sally is on the paved road for 1 − 1 ≈ 0.57 hr . 4 tan 30º 121. Use the formula (θ ) = 2a g sin θ cosθ with b. T (45º ) = 1 + 2 − 1 3sin 45º 4 tan 45º 2 1 g = 32 ft/sec2 and a = 10 feet : = 1 + 3 ⋅ −
1 4 ⋅1 2 a. t (30) = 2(10 ) ≈ 1.20 seconds 2 2 1 32 sin 30º ⋅ cos 30º = 1 + − ≈ 1.69 hr 3 4 b. t (45) = c. t (60) = 2(10 ) ≈ 1.12 seconds 32 sin 45º ⋅ cos 45º 2(10 ) ≈ 1.20 seconds 32 sin 60º ⋅ cos 60º Sally is on the paved road for 1 − 1 = 0.75 hr . 4 tan 45o c. T (60º ) = 1 + 2 − 1 3sin 60o 4 tan 60o 122. Use the formula = 1 + 2 − 1 x (θ ) = cosθ + 16 + 0.5 cos(2θ ) . 3 ⋅ 3 4 ⋅ 3 2 x (30) = cos (30º ) + = cos (30º ) + ≈ 4.897 cm 16 + 0.5 cos(2 ⋅ 30º ) 16 + 0.5 cos (60º ) = 1 + 4 − 1 ≈ 1.63 hr 3 3 4 3 Sally is on the paved road for 1 − 1 ≈ 0.86 hr . 4 tan 60o
b. When t = 4 , the coordinate on the unit circle is approximately (−0.7, −0.8) . Thus, 15. csc 450º = csc(360º + 90º ) = csc 90º = 1 sin 4 ≈ −0.8 cos 4 ≈ −0.7 tan 4 ≈ −0.8 ≈ 1.1 csc 4 ≈ 1 ≈ −1.3 −0.8 sec 4 ≈ 1 ≈ −1.4 −0.7 cot 4 ≈ −0.7 ≈ 0.9 16. sec 540º = sec(360º + 180º ) = sec180º = −1 17. cot 390º = cot(180º + 180º + 30º ) = cot 30º = 3 18. sec 420º = sec(360º + 60º ) = sec 60º = 2 −0.7 −0.8 33π ⎛ π ⎞ ⎛ π ⎞ Set the calculator on RADIAN mode: 19. cos = cos ⎜ + 8π ⎟ = cos ⎜ + 4 ⋅ 2π ⎟ 4 ⎝ 4 ⎠ ⎝ 4 ⎠ = cos π 4 129 – 131. Answers will vary. 20. = 2 2 sin 9π = sin ⎛ π + 2π ⎞ = sin π = 2 4 ⎜ 4 ⎟ 4 2 Section 2.3 1. All real numbers except − 1 ; ⎧ x | x 1 ⎫ 21. ⎝ ⎠ tan (21π) = tan(0 + 21π) = tan (0) = 0 ⎨ ≠− ⎬ 2 ⎩ 2 ⎭ 9π ⎛ π ⎞ ⎛ π ⎞ 22. csc = csc ⎜ + 4π ⎟ = csc ⎜ + 2 ⋅ 2π ⎟ 2. even 3. False 4. True 2 ⎝ 2 ⎠ ⎝ 2 ⎠ = csc π 2 = 1 5. 2π , π 23. sec 17π = sec ⎛ π + 4π ⎞ = sec ⎛ π + 2 ⋅ 2π ⎞ 4 ⎜ 4 ⎟ ⎜ 4 ⎟ 6. All real number, except odd multiples of π 2 ⎝ ⎠ ⎝ ⎠ = sec π 4 7. All real numbers between −1 and 1, inclusive. is, { y | −1 ≤ y ≤ 1} or [−1, 1] .
17π = 2 ⎛ π ⎞ ⎛ π ⎞ 24. cot = cot ⎜ + 4π ⎟ = cot ⎜ 4 + 2 ⋅ 2π ⎟ 8. True 9. False 10. False 4 ⎝ 4 ⎠ ⎝ ⎠ = cot π 4 = 1 19π = ⎛ π + π ⎞ = π = 3 11. sin 405º = sin(360º + 45º ) = sin 45º = 2 25. tan 6 tan ⎜ 6 3 ⎟ tan 6 3 12. 2 cos 420º = cos(360º + 60º ) = cos 60º = 1 26. ⎝ ⎠ sec 25π = sec ⎛ π + 4π ⎞ = sec ⎛ π + 2 ⋅ 2π ⎞ 2 6 ⎜ 6 ⎟ ⎜ 6 ⎟ 13. tan 405º = tan(180º + 180º + 45º ) = tan 45º = 1 ⎝ ⎠ ⎝ ⎠ = sec π 6 2 3 14. sin 390º = + 30º ) = sin 30º = 1 = 3 2
⎛ − 3 ⎞ ⎝ ⎠ cot θ = 1 = 1 tan θ = sin θ = ⎝ 5 ⎠ = − 3 ⋅ 5 = − 3 tan θ 2 cosθ ⎛ 4 ⎞ ⎜ ⎟ ⎝ ⎠ 5 4 4 38. 5 sin θ =− , cosθ 2 5 =− 5 5 cscθ = 1 = 1 = − 5 ⎛ 5 ⎞ sin θ ⎛ − 3 ⎞ 3 ⎜ − ⎟ ⎜ 5 ⎟ sin θ ⎝ 5 ⎠ ⎛ 5 ⎞ ⎛ 5 ⎞ 1 ⎝ ⎠ secθ = 1 = 1 = 5 tan θ = = cosθ ⎛ 2 5 ⎞ ⎜ − ⎟ = ⎜ − ⎝ ⎟ ⋅ ⎜ − 5 ⎠ ⎝ ⎟ = 2 5 ⎠ 2 cosθ ⎛ 4 ⎞ 4 ⎜ 5 ⎟ ⎝ 5 ⎠ ⎝ ⎠ cscθ = 1 = 1 = 1⋅ ⎛ − 5 ⎞ ⋅ 5 = − 5 cot θ = 1 = − 4 sin θ ⎛ 5 ⎞ ⎝ − 5 ⎠ 5 tan θ 3 ⎜ 5 ⎟ ⎝ ⎠ secθ = 1 = 1 = ⎛ − 5 ⎞ ⋅ 5 = − 5 cosθ ⎛ 2 5 ⎞ ⎜ 2 5 ⎟ 5 2 ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠
cot θ = 1 = 1 = 1⋅ 2 = 2 cscθ = 1 = 1 = 1⋅ ⎛ − 3 ⎞ = −3 tan θ ⎛ 1 ⎞ 1 sin θ ⎛ 1 ⎞ ⎜ 1 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ secθ = 1 = 1 = 3 ⋅ 2 = 3 2 39. sin θ = 1 , cosθ = 3 cosθ ⎛ 2 2 ⎞ 2 2 2 4 2 2 ⎜ 3 ⎟ tan θ = sin θ ⎛ 1 ⎞ ⎜ 2 ⎟ = 1 ⋅ 2 ⋅ 3 = 3 ⎝ ⎠ cot θ = 1 = 1 = − 4 ⋅ 2 = − 2 2 cosθ ⎛ 3 ⎞ 2 3 3 3 tan θ ⎛ 2 ⎞ 2 2 − ⎜ 2 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ cscθ = 1 = 1 = 1⋅ 2 = 2 sin θ ⎛ 1 ⎞ 1 42. sin θ = 2 2 , cosθ 1 ⎜ ⎟ ⎝ ⎠ ⎛ 2 2 ⎞ secθ = 1 = 1 = 2 ⋅ 3 = 2 3 ⎜ 3 ⎟ cosθ ⎛ 3 ⎞ 3 3 3 tan θ = sin θ = ⎝ ⎠ = 2 2 ⋅ − 3 = − 2 2 ⎜ 2 ⎟ cosθ ⎛ − 1 ⎞ 3 1 ⎝ ⎠ ⎜ 3 ⎟ cot θ = 1 = 1 = 3 ⋅ 3 = 3 ⎝ ⎠ 1 1 3 2 3 2 tan θ ⎛ 3 ⎞ 3 3 cscθ = = = 1⋅ ⋅ = sin θ ⎛ 2 2 ⎞ 2 2 2 4 ⎜ 3 ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ 3 ⎟ ⎝ ⎠ secθ = 1 = 1 = 1⋅ ⎛ − 3 ⎞ = −3 40. sin θ = 3 , cosθ = 1 cosθ ⎛ 1 ⎞ ⎜ 1 ⎟ 2 2 ⎜ − ⎟ ⎝ ⎠ ⎛ 3 ⎞ ⎝ 3 ⎠ sin θ ⎜ ⎟ ⎝ ⎠ 3 2 cot θ = 1 = 1 = − 1 ⋅ 2 = − 2 tan θ = = = ⋅ = 3 tan θ − 2 2 2 2 2 4 cosθ ⎛ 1 ⎞ 2 1 ⎜ ⎟
⎝ ⎠ 12 cscθ = 1 = 1 = 1⋅ 2 ⋅ 3 = 2 3 43. sin θ = , θ 13 in quadrant II sin θ ⎛ 3 ⎞ 3 3 3 2 2 ⎜ 2 ⎟ Solve for cosθ : sin θ + cos θ = 1 ⎝ ⎠ secθ = 1 = 1 = 1⋅ 2 = 2 cos2 θ = 1 − sin 2 θ cosθ =± 1 − sin 2 θ cosθ ⎛ 1 ⎞ 1 ⎜ ⎟ ⎝ ⎠ Since θ is in quadrant II, cosθ < 0 . cot θ = 1 = 1 = 1 ⋅ 3 = 3 cosθ =− 1 − sin 2 θ tan θ 3 3 3 3 ⎛ 12 ⎞2 = − 1 − ⎜ ⎟ = − ⎝ 13 ⎠ 1 − 144 = − 25 = − 5 169 169 13 41. sin θ 1 , cosθ = 2 2 3 3 sin θ ⎛ 12 ⎞ ⎜ 13 ⎟ 12 ⎛ 13 ⎞ 12 tan θ = = ⎝ ⎠ = ⋅ ⎜ − ⎟ = − tan θ = sin θ ⎛ − 1 ⎞ ⎜ 3 ⎟ = − 1 ⋅ 3 ⋅ 2 = − 2 cosθ ⎛ 5 ⎞ ⎜ − 13 ⎟ 13 ⎝ 5 ⎠ 5 cosθ ⎛ 2 2 ⎞ 3 2 2 2 4 1 1 13 ⎜ 3 ⎟ cscθ = = = ⎝ ⎠ sin θ ⎛ 12 ⎞ 12 ⎜ ⎟ ⎝ ⎠
secθ = 1 = 1 = − 13 cscθ = 1 = 1 = − 5 cosθ ⎛ − 5 ⎞ 5 sin θ ⎛ − 3 ⎞ 3 ⎜ 13 ⎟ ⎜ 5 ⎟ ⎝ ⎠ cot θ = 1 = 1 = − 5 ⎝ ⎠ secθ = 1 = 1 = − 5 tan θ ⎛ − 12 ⎞ 12 cosθ ⎛ − 4 ⎞ 4 ⎜ 5 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ cot θ = 1 = 1 = 4 44. cosθ = 3 , 5 θ in quadrant IV tan θ ⎛ 3 ⎞ 3 ⎜ ⎟ ⎝ ⎠ Solve for sin θ : sin 2 θ + cos2 θ = 1 sin 2 θ = 1 − cos2 θ 46. sin θ 5 =− , θ in III Since θ θ =± 1 − cos2 θ is in quadrant IV, sin θ < 0 . 2 Solve for cosθ : sin 2 θ + cos2 θ = 1 cos2 θ = 1 − sin 2 θ 2 sin θ =− 1 − cos2 θ = − ⎛ 3 ⎞ ⎜ ⎟ ⎝ 5 ⎠ 16 4 =− 25 5 Since θ 1 − sin θ is in quadrant III, cosθ < 0 . 2 cosθ =− 1 − sin 2 θ = − 1 − ⎛ − 5 ⎞ ⎛ − 4 ⎞ ⎜ 13 ⎟ tan θ = sin θ = ⎝ 5 ⎠ = − 4 ⋅ 5 = − 4 ⎝ ⎠ 144 12 cosθ ⎛ 3 ⎞ ⎜ ⎟ ⎝ ⎠ 5 3 3 =− =− 169 13 ⎛ − 5 ⎞ cscθ = 1 = 1 = − 5 sin θ ⎜ 13 ⎟ 5 ⎛ 13 ⎞ 5 sin θ ⎛ − 4 ⎞ 4 tan θ = = ±
tan θ = sin θ ⎛ 5 ⎞ ⎜ ⎟ = ⎝ 13 ⎠ = 5 ⋅ ⎛ − 13 ⎞ = − 5 sin θ = 1 − cos2 θ 2 cosθ ⎛ 12 ⎞ 13 ⎜ 12 ⎟ 12 = 1 − ⎛ − 1 ⎞ = 1 − 1 = 8 = 2 2 ⎜ − ⎟ ⎝ ⎠ ⎜ 3 ⎟ 9 9 3 ⎝ 13 ⎠ ⎝ ⎠ cscθ = 1 = 1 = 13 ⎛ 2 2 ⎞ ⎜ ⎟ sin θ ⎛ 5 ⎞ 5 sin θ ⎝ 3 ⎠ 2 2 ⎛ 3 ⎞ ⎜ 13 ⎟ tan θ = = = ⋅ ⎜ − ⎟ = − 2 2 ⎝ ⎠ cosθ ⎛ 1 ⎞ ⎜ ⎟ 3 ⎝ 1 ⎠ secθ = 1 = 1 = − 13 ⎝ 3 ⎠ cosθ ⎛ − 12 ⎞ 12 cscθ = 1 = 1 = 3 ⋅ 2 = 3 2 ⎜ 13 ⎟ ⎝ ⎠ sin θ ⎛ 2 2 ⎞ ⎜ ⎟ 2 2 2 4 cot θ = 1 = 1 = − 12 ⎝ 3 ⎠ tan θ ⎛ − 5 ⎞ 5 1 1 ⎜ 12 ⎟ secθ = = = −3 1 ⎝ ⎠ cosθ ⎛ ⎞ ⎜ − ⎟ ⎝ ⎠ 48. cosθ = 4 , 270º < θ < 360º ; 5 θ in quadrant IV cot θ = 1 = 1 ⋅ 2 = − 2 Solve for sin θ : sin 2 θ + cos2 θ = 1 tan θ − 2 2 2 4 2 3π sin θ =± 1 − cos2 θ 50. sin θ =− , π < θ < , θ in III Since θ is in quadrant IV, sin θ < 0 . 2 3 2 Solve for cosθ : sin 2 θ + cos2 θ = 1 sin θ =− 1 − cos2 θ = − 1 − ⎛ 4 ⎞ ⎝ ⎠ ⎛ − 3 ⎞ 9 3 = − = − 25 5 Since θ =± 1 − sin 2 θ is in quadrant III, cosθ < 0 . 2
tan θ = sin θ = ⎝ 5 ⎠ = − 3 ⋅ 5 = − 3 cosθ =− 1 − sin 2 θ = − 1 − ⎛ − 2 ⎞ cosθ ⎛ 4 ⎞ ⎜ ⎟ ⎝ ⎠ 5 4 4 =− 5 = − 5 ⎝ 3 ⎠ cscθ = 1 = 1 = − 5 9 3 sin θ ⎛ − 3 ⎞ 3 ⎛ − 2 ⎞ ⎜ 5 ⎟ sin θ ⎜ 3 ⎟ ⎝ ⎠ secθ = 1 = 1 = 5 tan θ = = cosθ ⎝ ⎠ ⎛ 5 ⎞ ⎜ − ⎟ cosθ ⎛ 4 ⎞ 4 ⎜ ⎟ ⎝ ⎠ ⎝ 2 ⎛ 3 3 ⎠ ⎞ 5 2 5 =− ⋅ ⎜ − ⎟ ⋅ = cot θ = 1 = 1 = − 4 3 ⎝ 5 ⎠ 5 5 tan θ ⎛ − 3 ⎞ 3 1 1 3 ⎜ 4 ⎟ cscθ = = =− 49. cosθ 1 ⎝ ⎠ π < θ < π , θ in quadrant II sin θ ⎛ 2 ⎞ 2 ⎜ ⎟ ⎝ ⎠ 3 2 1 1 3 5 3 5 Solve for sin θ : sin 2 θ + cos2 θ = 1 secθ = cosθ = = − ⎛ 5 ⎞ ⎜ ⎟ ⋅ = − 5 5 5 sin 2 θ = 1 − cos2 θ ⎝ 3 ⎠ sin θ =± 1 − cos2 θ 1 1 5 5 5 Since θ is in quadrant II, sin θ > 0 . cot θ = = = ⋅ = tan θ ⎛ 2 5 ⎞ 2 5 5 2 ⎜ 5 ⎟ ⎝ ⎠
51. sin θ = 2 , tan θ < 0, so θ is in quadrant II secθ = 1 = 1 4 = − 4 3 cosθ ⎛ 1 ⎞ 1 ⎜ ⎟ Solve for cosθ : sin2 θ + cos2 θ = 1 cos2 θ = 1 − sin 2 θ ⎝ 4 ⎠ cot θ = 1 = 1 ⋅ 15 = 15 cosθ =± 1 − sin 2 θ tan θ 15 15 15 Since θ is in quadrant II, cosθ < 0 . 53. secθ = 2, sin θ < 0, so θ is in quadrant IV cosθ =− 1 − sin 2 θ Solve for cosθ : cosθ = 1 = 1 ⎛ 2 ⎞2 4 5 5 secθ 2 =− 1 − ⎜ ⎟ ⎝ 3 ⎠ = − 1 − = − =− 9 9 3 Solve for sin θ : sin 2 θ + cos2 θ = 1 2 2 ⎛ 2 ⎞ ⎜ ⎟ sin θ = 1 − cos θ 2 tan θ = sin θ = ⎝ 3 ⎠ = 2 ⋅ ⎛ − 3 ⎞ = − 2 5 sin θ =± 1 − cos θ cosθ ⎛ 5 ⎞ 3 ⎜ 5 ⎟ 5 Since θ is in quadrant IV, sin θ < 0 . ⎝ ⎠ ⎜ − 3 ⎟ ⎝ ⎠ sin θ =− 1 − cos θ cscθ = 1 = 1 = 3 ⎛ 1 ⎞2 1 3 3 sin θ ⎛ 2 ⎞ 2 ⎜ ⎟ ⎝ ⎠ =− 1 − ⎜ ⎟ ⎝ 2 ⎠ = − 1 − =− =− 4 4 2 secθ = 1 = 1 3 3 5 =− sin θ ⎛ 3 ⎞ ⎜= − ⎟ ⎝ 2 ⎠ 3 2 cosθ ⎛ 5 ⎞ 5 5 ⎜ ⎟ tan θ = = cosθ ⎛ 1 ⎞ = − ⋅ = − 3 2 1 ⎝ 3 ⎠ ⎜ ⎟ cot θ = 1 = 1 = − 5 = − 5 cscθ = 1 = 1 = − 2 = − 2 3 tan θ ⎛ − 2 5 ⎞ 2 5 2 sin θ ⎛ 3 ⎞ 3 3 ⎜ 5 ⎟ ⎜ − ⎟
⎝ ⎠ ⎝ 2 ⎠ 52. cosθ 1 , tan θ > 0 cot θ = 1 = 1 = − 3 =− 4 Since tan θ = sin θ > 0 and cosθ < 0 , sin θ < 0 . tan θ − 3 3 cosθ Solve for sin θ : sin 2 θ + cos2 θ = 1 sin θ =± 1 − cos2 θ sin θ =− 1 − cos2 θ 54. cscθ = 3, cot θ < 0, so θ is in quadrant II Solve for sin θ : sin θ = 1 = 1 cscθ 3 Solve for cosθ : sin2 θ + cos2 θ = 1 =− 1 − ⎛ − ⎝ 1 ⎞2 ⎟ = − ⎠ 1 − 1 16 Since θ 1 − sin 2 θ is in quadrant II, cosθ < 0 . 2 =− 15 = − 15 cosθ =− 1 − sin θ 16 4 ⎛ 15 ⎞ ⎜ − ⎟ ⎛ 1 ⎞2 =− 1 − ⎜ ⎟ = − ⎝ 3 ⎠ 1 − 1 = − 8 2 2 9 9 3 tan θ = sin θ 15 ⎛ 4 ⎞ = =− ⋅ − = 15 ⎛ 1 ⎞ cosθ ⎛ 1 ⎞ 4 ⎜ 1 ⎟ ⎜ ⎟ ⎜ − ⎟ ⎝ ⎠ tan θ = sin θ = ⎝ 3 ⎠ ⎝ 4 ⎠ cosθ ⎛ 2 2 ⎞ ⎜ ⎟ cscθ = 1 = 1 = − 4 ⋅ 15 = − 4 15 ⎝ 3 ⎠ sin θ ⎛ 15 ⎞ ⎜ ⎟ 15 15 15 = 1 ⋅ ⎛ − 3 ⎞ ⋅ 2 = − 2 ⎝ 4 ⎠ 3 ⎝ 2 2 ⎠ 2 4
secθ = 1 = 1 =− ⋅ 3 sin θ =− 1 − cos2 θ cosθ ⎛ 2 2 ⎞ ⎜ ⎟ 2 2 2 4 = − 1 − ⎛ − 4 ⎞2 ⎟ = − 1 − 16 = − 9 = − 3 ⎝ 3 ⎠ ⎝ 5 ⎠ 25 25 5 cot θ = 1 = 1 = − 4 ⋅ 2 = − 2 2 cscθ = 1 = 1 = − 5 tan θ ⎛ 2 ⎞ 2 2 sin θ ⎛ − 3 ⎞ 3 ⎜ − ⎟ ⎜ 5 ⎟ ⎝ 4 ⎠ ⎝ ⎠ 1 55. tan θ = 3 , sin θ < 0, so θ is in quadrant III 4 57. tan θ , sin θ > 0, so θ is in quadrant II 3 Solve for secθ : sec2 θ = 1 + tan 2 θ secθ =± 1 + tan 2 θ Solve for secθ : sec2 θ = 1 + tan 2 θ secθ =± 1 + tan 2 θ Since θ is in quadrant III, secθ < 0 . Since θ is in quadrant II, secθ < 0 . 2 secθ =− 1 + tan 2 θ secθ =− 1 + tan θ 2 ⎛ 3 ⎞2 9 25 5 =− 1 + ⎛ − 1 ⎞ = − 1 + 1 = − 10 = − 10 =− 1 + ⎜ ⎟ = − 1 + = − =− ⎜ 3 ⎟ 9 9 3 ⎝ 4 ⎠ cosθ = 1 = − 4 16 16 4 ⎝ ⎠ 3 3 10 cosθ = = =− =− secθ 5 secθ ⎛ 10 ⎞ ⎜ − ⎟ 10 10 sin θ =− 1 − cos2 θ ⎝ 3 ⎠ =− 1 − ⎛ − 4 ⎞2 ⎟ =− 1 − 16 = − 9 = − 3 sin θ = 1 − cos2 θ ⎝ 5 ⎠ 25 25 5 ⎛ = 1 − ⎜ − 2 3 10 ⎞ ⎟ = 1 − 90 cscθ = 1 = 1 = − 5 ⎝ 10 ⎠ 100 sin θ ⎛ − 3 ⎞ 3
⎜ 5 ⎟ 10 10 ⎝ ⎠ cot θ = 1 = 1 = 4 = = 100 10 1 1 tan θ ⎛ 3 ⎞ 3 cscθ = = = 10 ⎜ ⎟ sin θ ⎝ ⎠ ⎛ 10 ⎞ ⎜ ⎟ ⎝ 10 ⎠ 56. cot θ = 4 , cosθ < 0, so θ is in quadrant III cot θ = 1 = 1 = −3 3 tan θ ⎛ 1 ⎞ ⎜ ⎟ tan θ = 1 = 1 = 3 ⎝ 3 ⎠ cot θ ⎛ 4 ⎞ 4 ⎜ ⎟ ⎝ ⎠ 58. secθ = − 2, tan θ > 0, so θ is in quadrant III Solve for secθ : sec2 θ = 1 + tan 2 θ secθ =± 1 + tan 2 θ Solve for tan θ : sec2 θ = 1 + tan 2 θ tan θ =± sec2 θ − 1 Since θ is in quadrant III, secθ < 0 . tan θ = sec2 θ − 1 = (− 2)2 − 1 = 4 − 1 = 3 secθ =− 1 + tan 2 θ cosθ = 1 1 ⎛ 3 ⎞2 9 25 5 secθ 2 =− 1 + ⎜ ⎟ ⎝ 4 ⎠ = − 1 + = − =− 16 16 4 cot θ = 1 = 1 ⋅ 3 = 3 cosθ = 1 = − 4 tan θ 3 3 3 secθ 5
sin θ =− 1 − cos2 θ 75. sec ⎛ − π ⎞ π ⎟ = sec = 2 3 =− 1 − ⎛ − ⎝ 1 ⎞2 ⎟ ⎠ =− 1 − 1 = − 3 = − 3 4 4 2 ⎝ 6 ⎠ ⎛ π ⎞ 6 3 π 2 3 cscθ = 1 = 1 = − 2 ⋅ 3 = − 2 3 76. csc ⎜ − ⎝ ⎟ = − csc = − 3 ⎠ 3 3 sin θ ⎛ 3 ⎞ 3 3 3 ⎜ ⎟ 77. sin 2 (40º ) + cos2 (40º ) = 1 ⎝ 2 ⎠ 59. sin(− 60º ) = − sin 60º = − 3 78. ) ( ) sec2 2 18º − tan 2 18º = 1 1 60. 61. 62. ) = cos 30º = 3 2 tan(−30º ) = − tan 30º = − 3 3 sin(−135º ) = − sin 135º = − 2 2 79. 80. 81. sin (80º ) csc (80º ) = sin (80º ) ⋅ = 1 sin (80º ) tan (10º ) cot (10º ) = tan (10º ) ⋅ 1 = 1 tan (10º ) tan (40º ) − sin (40º ) = tan (40º ) − tan (40º ) = 0 cos (40º ) cos (20º ) 63. sec(− 60º ) = sec 60º = 2 64. csc(−30º ) = − csc 30º = − 2 65. sin(−90º ) = − sin 90º = −1 cos(− 270º ) = cos 270º = 0 67. tan ⎛ − π ⎞ = − tan π = −1 82. 83. cot (20º ) − = cot (20º ) − cot (20º ) = 0 sin (20º )
cos (400º ) ⋅ sec (40º ) = cos (40º +360º ) ⋅ sec (40º ) = cos (40º ) ⋅ sec (40º ) = (40º ) ⋅ 1 cos (40º ) = 1 ⎜ 4 ⎟ 4 84. tan ) ⋅ cot (20º ) = tan (20º +180º ) ⋅ cot (20º ) ⎝ ⎠ 68. sin(−π) = − sin π = 0 = tan (20º ) ⋅ cot (20º ) = tan (20º ) ⋅ 1 tan (20º ) 69. cos ⎛ − π ⎞ = cos π = 2 ⎜ 4 ⎟ 4 2 = 1 ⎝ ⎠ 85. sin ⎛ − π ⎞ csc ⎛ 25π ⎞ = − sin ⎛ π ⎞ csc ⎛ 25π ⎞ 70. sin ⎛ − π ⎞ = − sin π = − 3 ⎜ 12 ⎟ ⎜ 12 ⎟ ⎜ 12 ⎟ ⎜ 12 ⎟ ⎜ 3 ⎟ 3 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = − sin ⎛ π ⎞ csc ⎛ π 24π ⎞ + ⎟ 71. tan(−π) = − tan π = 0 ⎝ 12 ⎠ ⎝ 12 12 ⎠ = − sin ⎛ π ⎞ csc ⎛ π + 2π ⎞ 72. sin ⎛ − 3π ⎞ = − sin 3π = −(−1) = 1 ⎝ 12 ⎠ ⎝ 12 ⎠ ⎜ 2 ⎟ 2 ⎛ π ⎞ ⎛ π ⎞ ⎝ ⎠ 73. csc ⎛ − π ⎞ = − csc π = − 2 = − sin ⎜ ⎟ csc ⎜ ⎟ ⎝ 12 ⎠ ⎝ 12 ⎠ = − sin ⎛ π ⎞ ⋅ 1 ⎜ 4 ⎟ 4 ⎝ 12 ⎠ sin ⎛ π ⎞ 74. ⎝ ⎠ sec (−π) = sec π = −1 = −1 ⎜ ⎟ ⎝ 12 ⎠
89. If sin θ = 0.3 , then sin θ + sin (θ + 2π) + sin (θ + 4π) = 0.3 + 0.3 + 0.3 = 0.9 90. If cosθ = 0.2 , then + 2 + = 2 cos1º +2 cos 2º +2 cos 3º +... + 2 cos(180º − 2º ) + 2 cos(180º −1º ) + cos (180º ) = 2 cos1º +2 cos 2º +2 cos 3º +... − 2 cos 2º − 2 cos1º + cos180º = cos180º = −1 95. The domain of the sine function is the set of all real numbers. 96. The domain of the cosine function is the set of all real numbers. cosθ + cos (θ + 2π) + cos (θ + 4π) = −0.2 + 0.2 + 0.2 97. f (θ ) = tan θ is not defined for numbers that are π = 0.6 odd multiples of . 2
116. a. b. 117. a. b. 118. a. b. f (−a) = f (a) = 1 4 f (a) + f (a + 2π) + f (a − 2π) = f (a) + f (a) + f (a) = 1 + 1 + 1 4 4 4 = 3 4 f (−a) =− f (a) = − 2 f (a) + f (a + π) + f (a + 2π) = f (a) + f (a) + f (a) = 2 + 2 + 2 = 6 f (−a) =− f (a) = − (−3) = 3 f (a) + f (a + π) + f (a + 4π) = f (a) + f (a) + f (a) = −3 + (−3) + (−3) = −9 f (−a) = f (a) = − 4 f (a) + f (a + 2π) + f (a + 4π) = f (a) + f (a) + f (a) = − 4 + (− 4) + (− 4) = −12 f (−a) =− f (a) = − 2 f (a) + f (a + 2π) + f (a + 4π) = f (a) + f (a) + f (a) = 2 + 2 + 2 = 6
121. Let P = ( x, y) be the point on the unit circle that then cos ⎛ π + p ⎞ = cos ⎛ π ⎞ = 0 ; so that p = π . ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ If θ = 0 , then cos (0 + p ) = cos (0) . But p = π . Thus cos (π) = −1 = cos (0) = 1 , or −1 = 1 . This is impossible. The smallest positive number p for which cos(θ + p) = cosθ corresponds to an angle t. Consider the equation tan t = y = a . Then y = ax . Now x2 + y2 = 1 , for all θ must then be p = 2π . x so x2 + a2 x2 = 1 . Thus, x =± 1 + a2 and = 1 : Since cosθ has period 2π , so cosθ does secθ . a y =± . That is, for any real number a , 1 + a2 there is a point P = ( x, y) on the unit circle for which tan t = a . In other words, −∞ < tan t < ∞ , and the range of the tangent function is the set of all real numbers. 126. cscθ = 1 : Since sin θ has period 2π , so sin θ does cscθ .
127. If P = (a, b) is the point on the unit circle Section 2.4 corresponding to θ , then Q = (−a, −b) is the point on the unit circle corresponding to θ + π . Thus, tan(θ + π) = −b = b = tan θ . If there 1. y = 3x2 Using the graph of y = x2 , vertically stretch the exists a number −a a p, 0 < p < π , for which graph by a factor of 3. tan(θ + p) = tan θ for all θ , then if θ = 0 , tan ( p ) = tan (0) = But this means that p is a multiple of π . Since no multiple of π exists in the interval (0, π) , this is impossible. Therefore, the fundamental period of f (θ ) = tan θ is π . 128. cot θ = 1 : Since tan θ has period π , so tanθ does cot θ . 2. y = − x2 Using the graph of y = x2 , reflect the graph 129. Let P = (a, b) be the point on the unit circle across the x-axis. corresponding to θ . Then cscθ = 1 = 1 b sin θ secθ = 1 = 1 a cosθ cot θ = a = 1 = 1 b ⎛ b ⎞ ⎜ ⎟ ⎝ ⎠ tan θ 130. Let P = (a, b) be the point on the unit circle π corresponding to θ . Then 3. 1; + 2π k , k is any integer 2 tan θ = b = sin θ
4. 3; π a cosθ cot θ = a = cos θ 2π π b sin θ 5. 3; = 6 3 131. (sin θ cosφ )2 + (sin θ sin φ )2 + cos2 θ = sin 2 θ cos2 φ + sin 2 θ sin 2 φ + cos2 θ = sin 2 θ (cos2 φ + sin 2 φ ) + cos2 θ = sin 2 θ + cos2 θ = 1 6. True 7. False 8. True 9. The graph of y = sin x crosses the y-axis at the 132 – 135. Answers will vary. point (0, 0), so the y-intercept is 0. 10. The graph of y = cos x crosses the at the point (0, 1), so the y-intercept is 1. y-
11. The graph of − π < x < π . y = sin x is increasing for 22. y = − sin ⎛ 1 ⎞ ⎝ 2 ⎟ 2 2 This is in the form y = A sin(ω x) where A = −1 12. The graph of y = cos x is decreasing for and ω = 1 . Thus, the amplitude is 2 A = − 1 = 1 0 < x < π . 2π 2π and the period is T = = = 4π . 13. The largest value of y = sin x is 1. 2 14. The smallest value of y = cos x is −1 . 23. y = x) is in the y = A sin(ω x) where A = 6 15. sin x = 0 when x = 0, π, 2π . and ω = π . Thus, the amplitude is A = 6 = 6 = 2π = 2π = . 16. cos x = 0 when x = π , 3π . 2 2 24. the is T 2 ω π y = − 3cos(3x) 17. sin x = 1 when x = − 3π , π ; This is in the form y = A cos(ω x) where A = − 3 sin x = −1 when x 2 2 π 3π =− 2 2 and ω = 3 . Thus, the amplitude is and the period is T = 2π = 2π . A = − 3 = 3 18. cos x = 1 when x = − 2π, 0, 2π; cos x = −1 when x = −π, π. 25. ω 3 1 ⎛ 3 ⎞ y =− ⎜ ⎟ 2 ⎝ 2 ⎠ 19. y = 2sin x This is in the form y = A cos(ω x) where This is in the form y = A sin(ω x) where A = 2 A = − 1 and ω = 3 . Thus, the amplitude is d
and ω = 1 . Thus, the amplitude is A = 2 = 2 2 2 1 1 and the period is T = 2π = 2π = 2π . A = − 2 = and the period is 2 ω 1 T = 2π = 2π = 4π . 20. y = x This is in the form y = A cos(ω x) where 3 2 A = 3 and ω = 1 . Thus, the amplitude is A = 3 = 3 26. y = 4 sin ⎛ 2 x ⎞ 3 ⎜ 3 ⎟ ⎝ ⎠ and the period is T = 2π = 2π = 2π . 4 ω 1 This is in the form y = A sin(ω x) where A = 3 21. y = − 4 cos(2x) and ω = 2 . Thus, the amplitude is A = 4 = 4 This is in the form y = A cos(ω x) where 3 3 3 A = − 4 and ω = 2 . Thus, the amplitude is and the period is T = 2π = 2π = 3π . A = − 4 2 = 4 and the period is 3 T = 2π = 2π = π . ω 2
27. y = 5 sin ⎛ − 2π x ⎞ = − 5 sin ⎛ 2π x ⎞ 43. Comparing y = 4 cos x to y = A cos (ω x ) , we 3 ⎜ 3 ⎟ 3 ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ find A = 4 and ω = 1 . Therefore, the amplitude This is in the form y = A sin(ω x) where A = − 5 3 is 4 = 4 and the period is 2π = 2π . Because 1 and ω = 2π . Thus, the amplitude is the amplitude is 4, the graph of y = 4 cos x will 3 A = − 5 = 5 and the period is lie between −4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [0, 2π ] 3 3 T = 2π = 2π = 3 . into four subintervals, each of length 2π = π by ω 2π 3 4 2 finding the following values: 0, π , π , 3π , and 2π 28. y = 9 cos ⎛ − 3π x ⎞ = 9 cos ⎛ 3π x ⎞ 2 2 5 ⎜ 2 ⎟ 5 ⎜ 2 ⎟ These values of x determine the x-coordinates of ⎝ ⎠ ⎝ ⎠ the five key points on the graph. To obtain the y- This is in the form y = A cos(ω x) where A = 9 5 of the five key points for y = 4 cos x , we multiply the y-coordinates of the and ω = 3π . Thus, the amplitude is 2 five key points for key points are y = cos x by A = 4 . The five A = 9 = 9 and the period is ( ) ⎛ π ⎞ ⎛ 3π ⎞ 5 5 0, 4 , ⎜ 2 ,0 ⎟ , (π , −4) , ⎜ ,0 ⎟ , (2π , 4) 2 ⎝ ⎠ ⎝ ⎠ T = 2π = 2π = 4 . We plot these five points and fill in the graph of ω 29. F 30. E 31. A 32. I 33. H 3π 3 2 the curve. We then extend the graph in either
direction to obtain the graph shown below. 34. B 44. Comparing y = 3sin x to y = A sin (ω x ) , we 35. C find A = 3 and ω = 1 . Therefore, the amplitude 36. G is 3 = 3 and the period is 2π = 2π . Because 1 37. J the amplitude is 3, the graph of y = 3sin x will 38. D 39. A 40. C 41. B lie between −3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [0, 2π ] into four subintervals, each of length 2π = π by 4 2 finding the following values: 42. D 0, π 2 , π , 3π , and 2π 2
These values of x determine the x-coordinates of the five key points on the graph. To obtain the coordinates of the five key points for y = 3sin x , we multiply the y-coordinates of the five key points for y = sin x by A = 3 . The five key points are (0, 0) , ⎛ π ⎞ , (π ,0) , ⎛ 3π , −3 ⎞ , ⎜ 2 ⎟ ⎜ 2 ⎟ (2π , 0) ⎝ ⎠ ⎝ ⎠ We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. 46. Comparing y = −3cos x to y = A cos (ω x ) , we find A = −3 and ω = 1 . Therefore, the amplitude is −3 = 3 and the period is 2π = 2π . Because 1 the amplitude is 3, the graph of y = −3cos x will lie between −3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [0, 2π ] into four subintervals, each of length 2π = π by 4 2 45. Comparing y = −4sin x to y = A sin (ω x ) , we finding the following values: find A = −4 and ω = 1 . Therefore, the amplitude 0, π , π , 3π , and 2π is −4 = 4 and the period is 2π = 2π . Because 2 2 1 the amplitude is 4, the graph of y = −4sin x will These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- lie between −4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and coordinates of the five key points for y = −3cos x , we multiply the y-coordinates of
48. Comparing y = sin (3x ) to y = A sin (ω x ) , we Because the amplitude is 1, the graph of y = − sin (2 x ) will lie between −1 and 1 on the y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [0, π ] into four subintervals, each of length π by finding the following values: 4 find A = 1 and ω = 3 . Therefore, the amplitude 0, π , π , 3π , and π is 1 = 1 and the period is 2π . Because the 4 2 4 amplitude is 1, the graph of 3 y = sin (3x ) will lie These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- between −1 and 1 on the y-axis. Because the coordinates of the five key points for y = − sin (2 x ) , we multiply the y-coordinates of period is 2π , one cycle will begin at x = 0 and 3 end at x = 2π . We divide the interval ⎡0, 2π ⎤ the five key points for five key points are y = sin x by A = −1 .The 3 ⎢ 3 ⎥ ⎛ π ⎞ ⎛ π ⎞ ⎛ 3π ⎞ ⎣ ⎦ (0, 0) , ⎜ , −1⎟ , ⎜ ,0 ⎟ , ⎜ ,1⎟ , (π ,0) ⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 4 ⎠ into four subintervals, each of length 2π / 3 = π 4 6 by finding the following values: We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
y ⎛ 1 ⎞ 2 ( –– –, 1) 4 (–3– –, 1) ( , 0) 51. Comparing y = 2sin ⎜ x ⎟ to ⎝ ⎠ y = A sin (ω x ) , we find A = 2 and ω = 1 . Therefore, the 2 2 (0, 0) 2 x (– – , 0) 2 amplitude is 2 = 2 and the period is 2π 1/ 2 Because the amplitude is 2, the graph of = 4π . 2 (– – , 1) 4 y = 2sin ⎛ 1 ⎝ 2 x ⎞ will lie between −2 and 2 on the ⎠ 50. Since cosine is an even function, we can plot the y-axis. Because the period is 4π , one cycle will equivalent form y = cos (2 x ) . begin at x = 0 and end at x = 4π . We divide Comparing y = cos (2 x ) to y = A cos (ω x ) , we the interval [0, 4π ] into four subintervals, each find A = 1 and ω = 2 . Therefore, the amplitude of length 4π = π 4 by finding the following is 1 = 1 and the period is 2π = π . Because the 2 values: 0, π , 2π , 3π , and 4π amplitude is 1, the graph of y = cos (2 x ) will lie These values of x determine the of between −1 and 1 on the y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [0, π ] into the five key points on the graph. To obtain the y- coordinates of the five key points for y = 2sin ⎛ 1 x ⎞ , we multiply the y-coordinates of ⎜ 2 ⎟ four subintervals, each of length π by finding 4 the following values: ⎝ ⎠ the five key points for five key points are y = sin x by A = 2 . The 0, π , π , 3π , and π (0, 0) , (π ,2) , (2π , 0) , (3π , −2) , (4π , 0) 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- coordinates of the five key points for coordinates
y = cos (2 x ) , we multiply the y-coordinates of We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. the five key points for five key points are y = cos x by A = 1 .The (0,1) , ⎛ π ⎞ , ⎛ π , −1⎞ , ⎛ 3π ⎞ , (π ,1) ⎜ 4 ⎟ ⎜ 2 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. 52. Comparing y = 2 cos ⎛ 1 x ⎞ to y = A cos (ω x ) , ⎜ 4 ⎟ we find ⎝ ⎠ A = 2 and ω = 1 . Therefore, the 4 amplitude is 2 = 2 and the period is 2π 1/ 4 Because the amplitude is 2, the graph of = 8π . y = 2 cos ⎛ 1 x ⎞ will lie between −2 and 2 on ⎜ 4 ⎟ ⎝ ⎠ the y-axis. Because the period is 8π , one cycle
3 (0, 2) (2, 2) and 2 on the y-axis. The five key points are (0, −2) , (1, 2) , (2, −2) , (3, −6) , (4, −2) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. 2 3 –– 2 2 x 3 –– 2 1 –– 2 9 (1,
⎜ 4 ⎟ ⎜ 2 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ by A = −4 and then adding 2 units. Thus, the We plot these five points and fill in the graph of the curve. We then extend the graph in either graph of y = −4 cos (3x ) + 2 will lie between −2 direction to obtain the graph shown below. and 6 on the y-axis. The five key points are ⎞ ⎛ π ⎞ ⎛ π ⎞ ⎛ 2π ⎞ y (0, −2) , ⎛ π ,2 , ⎜ ,6 ⎟ , ⎜ ,2 ⎟ , ⎜ , −2 ⎟ 10 ⎛ 3π ⎞ ⎝ 6 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎛ =π ⎞ ,5 ⎜ ,8 ⎟ ⎝ 4 ⎠ We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. ⎜ ⎟ ⎝ ⎠ (0, 5) (π ,5) ⎛ π ⎞ ⎜ ,2 ⎟ ⎝ 4 ⎠ −π π 3 3
63. Since sine is an odd function, we can plot the 64. Since cosine is an even function, we consider the equivalent form y 5 sin ⎛ 2π x ⎞ . equivalent form y = 9 cos ⎛ 3π x ⎞ . Comparing 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ Comparing y 5 sin ⎛ 2π x ⎞ to y = 9 cos ⎛ 3π x ⎞ to y = A cos (ω x ) , we find 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ y = A sin (ω x ) , we find 5 A =− 3 and ω = 2π . 3 A = 9 5 and ω = 3π . Therefore, the amplitude is 2 Therefore, the amplitude is − 5 = 5 3 3 and the 9 = 9 5 5 and the period is 2π = 4 . / 2 3 period is 2π = 3 . Because the amplitude is 2π / 3 the amplitude is 9 , the graph of 5 5 , the graph of y 5 sin ⎛ 2π x ⎞ will lie y = 9 cos ⎛ 3π x ⎞ will lie between − 9 and 9 3 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ 5 5 between − 5 3 and 5 3 on the y-axis. Because the on the y-axis. Because the period is 4 , one 3 period is 3 , one cycle will begin at x = 0 and end at x = 3 . We divide the interval [0, 3] into cycle will begin at x = 0 and end at x = 4 . We 3 four subintervals, each of length 3 by finding divide the interval ⎡0, 4 ⎤ into four subintervals, 4 ⎢ 3 ⎥ ⎣ ⎦ the following values: 0, 3 , 3 , 9 , and 3 each of length 4/3 = 1 4 3 by finding the following
( 4, 2) y –– 2 2 (4, 2) 1 –– 2 x 8 4 4 8 ( 8, 1) (0, 1) (8, 1) 2 66. We begin by considering y 1 sin ⎛ π x ⎞ . 65. We begin by considering y 3 cos ⎛ π x ⎞ . 2 1 ⎛ π ⎞ ⎝ 8 ⎠ Comparing 2 y 3 cos ⎛ π x ⎞ to ⎝ 4 ⎠ y =− 2 1 sin ⎜ ⎝ 8 x ⎟ to ⎠ π y = A sin (ω x ) , 2 ⎝ 4 ⎠ we find A = − and ω = . Therefore, the 2 8 y = A cos (ω x ) , we find A 3 and ω = π . 1 1 Therefore, the amplitude is 2 − 3 = 3 4 and the is 2π − = and the period is 2 2 1 period is 2π 2 2 = 8 . Because the amplitude is 3 , π /8 = 16 . Because the amplitude is , the 2 π /4 2 graph of y 1 sin ⎛ π x ⎞ will lie between − 1 the graph of y 3 cos ⎛ π x ⎞ will lie between 2 ⎝ 8 ⎠ 2 1 2 ⎝ 4 ⎠ and on the y-axis. Because the period is 16, 2 − 3 and 3 on the y-axis. Because the period is 2 2 8, one cycle will begin at x = 0 and end at x = 8 . We divide the interval [0, 8] into four one cycle will begin at x = 0 and end at x = 16 . We divide the interval [0,16] into four subintervals, each of length 16 = 4 by finding 4 subintervals, each of length 8 = 2 by finding the 4 following values: 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- the following values: 0, 4, 8, 12, and 16
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- coordinates of the five key points for coordinates of the five key points for y = − 1 sin ⎛ π x ⎞ + 3 , we the y- y 3 cos ⎛ π x ⎞ + 1 , we multiply the y- 2 ⎝ 8 ⎠ ⎜ ⎟ 2 ⎝ 4 ⎠ 2 coordinates of the five key points for y = sin x coordinates of the five key points for y = cos x 1 3 by A 3 and then add 1 unit. Thus, the by A = − and then add 2 units. Thus, the 2 2 2 1 ⎛ π ⎞ 3 graph of y 3 cos ⎛ π x ⎞ + 1 will lie between graph of y = − sin ⎜ x ⎟ + 2 8 2 will lie between =− ⎜ ⎟ ⎝ ⎠ 2 ⎝ 4 ⎠ 2 1 and 2 on the y-axis. The five key points are −1 and 2 on the y-axis. The five key points are ⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ (0, −1) , ⎛ 2, 1 ⎞ , (4, 2) , ⎛ 6, 1 ⎞ , (8, −1) ⎜ 0, 2 ⎟ , (4,1) , ⎜ 8, 2 ⎟ , (12, 2) , ⎜16, 2 ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
( 4, 2) y 2.5 2) graph is a reflected cosine graph with amplitude 3 and period 4π. 2π 3 –– 2 (8, 3 ) Find ω : 4π= ω 2 3 (4, 1) –– x 4πω = 2π ω = 2π = 1 4π 2 16 12 8 4 0.5 4 8 12 16 The equation is: y = −3cos ⎛ =1 x ⎞ . ⎜ 2 ⎟ 67. A = 3; T = π; ω = 2π = 2π = 2 ⎝ ⎠ 74. The graph is a reflected sine graph with T π y = ±3sin(2 x) amplitude 2 and period 4. 2π Find ω : 4 = ω 68. A = 2; T = 4π; ω = 2π = 2π = 1 4ω = 2π T 4π 2 2π π y = ±2sin ⎛ 1 x ⎞ ω = = 4 2 ⎜ 2 ⎟ ⎝ ⎠ The equation is: y = − 2sin ⎛ π x ⎞ . ⎜ 2 ⎟ 69. ⎝ ⎠ A = 3; T = 2; ω = 2π = 2π = π T 2 y = ±3sin(πx) 75. The graph is a sine graph with amplitude period 1. 3 and 4 70. A = 4; T = 1; ω = 2π = 2π = 2π 2π T 1 y = ±4sin(2π x) Find ω : 1 = ω ω = 2π 71. The graph is a cosine graph with amplitude 5 and period 8. The equation is: y = 3 sin (2π x ) . 4 Find ω : 7 T
8 = 2π ω 76. The graph is a reflected cosine graph with 5 8ω = 2π ω = 2π = π 8 4 amplitude Find ω : and period 2. 2 2 = 2π ω The equation is: y = 5cos ⎛ π x ⎞ . 2ω = 2π ⎜ 4 ⎟ ⎝ ⎠ 72. The graph is a sine graph with amplitude 4 and ω = 2π =π 2 period 8π. Find ω : 8π= 2π ω 8πω = 2π ω = 2π = 1 The equation is: y 5 cos (πx ) . 2 8π 4 The equation is: y = 4sin ⎛ 1 x ⎞ . ⎜ 4 ⎟ ⎝ ⎠
77. The graph is a reflected sine graph with amplitude 1 and period 4π . 81. The graph is a sine graph with amplitude 3 and period 4. 3 Find ω : 4π = 2π Find ω : 4 = 2π ω 3 ω 4ω = 2π 4πω = 6π ω = 6π = 3 ω = 2π = π 4 2 4π 2 The equation is: y = 3sin ⎛ π x ⎞ . The equation is: y = − sin ⎛ 3 x ⎞ . ⎝ 2 ⎠ ⎜ 2 ⎟ ⎝ ⎠ 78. The graph is a reflected cosine graph with 82. The graph is a reflected cosine graph with amplitude 2 and period 2. amplitude π and period 2π. Find ω : 2 = 2π ω Find ω : 2π= 2π ω 2πω = 2π ω = 2π = 1 2ω = 2π ω = 2π =π 2 2π The equation is: y = − 2 cos(π x) . The equation is: y = −π cos x . 83. The graph is a reflected cosine graph with 79. The graph is a reflected cosine graph, shifted up 1 unit, with amplitude 1 and period 3 . 2 amplitude 4 and period 2π . 3 Find ω : 2π = 2π Find ω : 3 = 2π 3 ω 2 ω 2πω = 6π 3ω = 4π ω = 4π ω = 6π = 3 2π The equation is: 3 y = − cos ⎛ 4π x ⎞ + 1 . The equation is: y = − 4 cos (3x ) . ⎜ 3 ⎟ 84. The graph is a sine graph with amplitude 4 and ⎝ ⎠
80. The graph is a reflected sine graph, shifted down period π. Find ω : π= 2π ω 1 unit, with amplitude 1 2 Find ω : 4π = 2π and period 4π . 3 πω = 2π ω = 2π = 2 3 ω π 4πω = 6π ω = 6π = 3 The equation is: y = 4sin (2 x ) . 4π 2 85. f (π /2) − f (0 ) sin (π / 2 ) − sin (0 ) = The equation is: y 1 sin ⎛ 3 x ⎞ − 1 . π /2 − 0 π / 2 =− ⎜ ⎟ 1 − 0 2 2 ⎝ 2 ⎠ = = π /2 π The average rate of change is 2 . π
86. f (π /2) − f (0 ) = cos (π /2) − cos (0 ) 90. ( f D g ) ( x ) = cos ⎛ 1 x ⎞ π /2 − 0 π / 2 = 0 −1 = − 2 ⎝ 2 ⎠ π /2 π The average rate of change is − 2 . π sin ⎛ 1 ⋅ π ⎞ − sin ⎛ 1 ⋅ 0 ⎞ 87. f (π /2) − f (0 ) π /2 − 0 ⎜ 2 2 ⎟ ⎜ 2 ⎟ = π / 2 sin (π / 4) − sin (0) = π /2 2 ( g D f ) ( x ) = 1 (cos x ) = 1 cos x 2 2 = 2 = 2 ⋅ 2 = 2 π /2 2 π π The average rate of change is 2 . π ⎛ π ⎞ 88. f (π /2) − f (0 ) = π /2 − 0 ⎜ 2 ⋅ ⎟ − cos (2 ⋅ 0 ) ⎝ ⎠ π / 2 cos (π ) − cos (0) = = π /2 = −2 ⋅ 2 = − 4 π π The average rate of change is − 4 . π −1 − 1 π /2 91. ( f D g ) ( x ) = −2 (cos x ) = −2 cos x 89. ( f D g ) ( x ) = sin (4 x ) ( g D f ) ( x ) = 4 (sin x ) = 4 sin x ( g D f ) ( x ) = cos (−2 x )
92. ( f D g ) ( x ) = −3 (sin x ) = −3 sin x 95. V (t ) = 220 sin (120π t ) a. Amplitude: A = 220 = 220 Period: T = 2π = 2π = 1 b, e. ω 120π 60 ( g D f ) ( x ) = sin (−3x ) c. V = IR 220 sin(120π t ) = 10I 22 sin(120π t) = I → I (t ) = 22 sin (120π t ) d. Amplitude: A = 22 = 22 Period: T = 2π = 2π = 1 93. I (t ) = 220 sin(60π t ), t ≥ 0 96. ω V (t ) = 120 sin (120π t ) 120π 60 Period: T = 2π = 2π = 1 ω 60π 30 a. Amplitude: A = 120 = 120 Amplitude: A = 220 = 220 Period: T = 2π = 2π = 1 b, e. ω 120π 60
94. I (t ) = 120 sin(30π t ), t ≥ 0 Period: T = 2π = 2π = 1 c. V = IR ω 30π 15 120 sin(120π t ) = 20I Amplitude: A = 120 = 120 t ) = I → I (t ) = 6sin (120π t ) d. Amplitude: A = 6 = 6 Period: T = 2π = 2π = 1 ω 120π 60
97. a. 2 P (t ) = ⎣ ⎦ R 99. a. Physical potential: ω = 2π ; 23 (V sin (2πft ))2 Emotional potential: ω = 2π = π ; = 0 R V 2 sin 2 (2πft ) = 0 R Intellectual potential: b. Graphing: 28 14 ω = 2π 33 V 2 = 0 sin 2 (2πft ) R b. The graph is the reflected cosine graph translated up a distance equivalent to the amplitude. The period is 1 , so ω = 4π f . 2 f 1 V 2 V 2 The amplitude is ⋅ 0 = 0 . 110 0 0 #1, #2, #3 ⎛ 2π ⎞ 2 R 2R The equation is: #1 : P (t ) = 50 sin ⎜ ⎝ 23 t ⎟ + 50 ⎠ P (t ) V0 cos (4πft ) + V0 #2 : P (t ) = 50 sin ⎛ π t ⎞ + 50 =− ⎜ ⎟ 2R 2R ⎝ ⎠ V 2 = 0 ⎡1 − cos (4πft )⎤ 2R c. Comparing the formulas: sin 2 (2πft ) = 1 ⎡⎣1 − cos (4πft ) #3 : c. No. d. P (t ) = 50sin ⎛ 2π ⎝ 33 110 t ⎞ + 50 ⎠ 2 98. a. Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus, #2 #1 #3 T = 2π = 2 (28) = 56 ω or ω = π . 28 7305 0 7335 The tunnel has a maximum height of 15 feet
101. y = sin x , − 2π ≤ x ≤ 2π y 1 14. 15. y = csc x has vertical asymptotes when x = − 2π, − π, 0, π, 2π . y = tan x has vertical asymptotes when x 3π , π , π , 3π –2π –π 0 –1 x π 2π 16. − . 2 2 2 2 y = cot x has vertical asymptotes when x = − 2π, − π, 0, π, 2π . 102 – 105. Answers will vary. Section 2.5 17. y = 3 x ; graph of vertically by a factor of 3. y 6 y = tan x is stretched (– – , 3) x 1. x = 4 ( – – , 3) 2. False; depending on the function, there can be any number of vertical asymptotes. 6 3. origin; x = odd multiples of π 2 18. y = −2 tan x ; The graph of y = tan x is stretched 4. y-axis; x = odd multiples of π 2 vertically by a factor of 2 and reflected about the x-axis. y 5. y = cos x 4 – – , 2 6. True 7. The y-intercept of y = tan x is 0. ( 4 ) x 8. y = cot x has no y-intercept. ( 4 , 2) 4 9. The y-intercept of y = sec x is 1. 19. y = 4 cot x ; The graph of y = cot x is stretched 10. y = csc x has no y-intercept. vertically by a factor of 4.
y 11. sec x = 1 when x = − 2π, 0, 2π; sec x = −1 when x = −π, π 16 (– – , 4 ) 12. csc x = 1 when x = − 3π , π ; 2 2 csc x = −1 when x = − π , 3π 2 2 16 x ( – – , 4 ) 13. y = sec x has vertical asymptotes when x 3π , π , π , 3π =− − . 2 2 2 2
y = csc x is vertically 1 ( – – , 1) 8 compressed by a factor of . 2 y 2 23. y = cot ⎛ 1 x ⎞ ; The graph of y = cot x is – – 1 ⎜ 4 ⎟ ( –3– –, ( 2 , 2 ) ⎝ ⎠ horizontally stretched by a factor of 4. 2 2 x 3 –– 2 3 ––– –1– – – 1 ( 2 , 2 ) ( , 2 ) 2 –– 2
27. y = −3csc x ; The graph of y = csc x is 1 vertically stretched by a factor of 3 and reflected about the x-axis. 30. y = csc (2x ) ; The graph of 2 y = csc x is 1 y 14 10 ( – – , 3) 6 2 3 ––– 2 6 3 (–––, 3) x ( – – , 3) by a factor of 2 vertically compressed by factor of 1 . 2 y 1.0 and 10 ( –3– – , –1–) 1 4 2 (––, –– ) 14 ( – – , –1– ) – – 2 4 2 – – x 2 28. y = −4sec x ; The graph of y = sec x is 4 2 (–3– –, –1–) vertically stretched by a factor of 4 and reflected about the x-axis. 4 2 1.0 31. y = −2 csc (π x ) ; The graph of y = csc x is horizontally compressed by a factor of 1 , π vertically stretched by a factor of 2, and reflected about the x-axis. y 6 29. y = 4sec ⎛ 1 x ⎞ ; The graph of y = sec x is ( –– , 2) 3 ⎜ 2 ⎟ 2 ⎝ ⎠ horizontally stretched by a factor of 2 and vertically stretched by a factor of 4. y 36 2.0 1.0 1.0 2.0 x 1 –– 3 2 –– 2 6 (0, 4) 32. y = −3sec ⎛ π x ⎞ ; The graph of y = sec x is ⎜ 2 ⎟ 4 2 ( 2 , 4)
36 2 3 x , 4) ⎝ ⎠ horizontally compressed by a factor of 2 , π vertically stretched by a factor of 3, and reflected about the x-axis. y 8 ( 2, 3) (2, 3) 4 2 2 4 x (0, 3) 8
33. y = tan ⎛ 1 x ⎞ + 1 ; The graph of y = tan x is 36. y = csc ⎛ 3π x ⎞ ; The graph of y = csc x is ⎜ 4 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ horizontally stretched by a factor of 4 and shifted 2 up 1 unit. y 8 horizontally compressed by a factor of . 3π ( 3 , 2) ( , 2) (0, 1) (3 , 0)x 4 4 8 34. y = 2 cot x − 1 ; The graph of y = cot x is vertically stretched by a factor of 2 and shifted down 1 unit. y 37. y = 1 tan ⎛ 1 x ⎞ − 2 ; The graph of y = tan x is 10 2 ⎜ 4 ⎟ ( –3– –, 1) ⎝ ⎠ horizontally stretched by a factor of 4, vertically 1 2 ( – – , 1) (– – , 1) – – x 2 2 compressed by a factor of units. y 8 , and shifted down 2 2 – – , 1 2 10 ( , –3– 35. y = sec ⎛ 2π x ⎞ + 2 ; The graph of y = sec x is 3 3 x ⎜ 3 ⎟ ⎝ ⎠ ( , –3– 2) horizontally compressed by a factor of 3 2π and 2 8 shifted up 2 units. y 8 38. y = 3cot ⎛ 1 x ⎞ − 2 ; The graph of y = cot x is ⎜ 2 ⎟ ( , 1)
(0, 3) ( –3– , 1) ⎝ ⎠ horizontally stretched by a factor of 2, vertically stretched by a factor of 3, and shifted down 2 units. y 2 2 x 4 ( –3– – , 1 ( , 2) 6 ( – – , 1 x ( , 2) ( – – , 6 10
39. y = 2 csc ⎛ 1 x ⎞ − 1 ; The graph of y = csc x is f ⎛ π ⎞ − f (0) ⎜ 3 ⎟ ⎜ 6 ⎟ tan (2 ⋅π / 6 ) − tan (2 ⋅ 0 ) ⎝ ⎠ horizontally stretched by a factor of 3, vertically stretched by a factor of 2, and shifted down 1 unit. y 43. ⎝ ⎠ = π − 0 6 π /6 3 − 0 6 3 6 = π /6 = π 6 3 ( 9– – , 1) (3– –, 1) 4 2 2 4 x The average rate of change is . π ( 3– – , 3) (9– – , 3) 8 44. f ⎛ π ⎞ − f (0 ) ⎝ ⎠ = π − 0 6 sec (2 ⋅ π / 6 ) − sec (2 ⋅ 0 ) π /6 = 2 −1 = 6 40. y = 3sec ⎛ 1 x ⎞ + 1 ; The graph of y = sec x is π /6 π 6 ⎜ 4 ⎟ The average rate of change is . ⎝ ⎠ horizontally stretched by a factor of 4, vertically stretched by a factor of 3, and shifted up 1 unit. y 8 π 45. ( f D g ) ( x ) = tan (4 x ) ( 8 , 5) (0, 5) (8 , 5) 8 8 x ( 4 , 1) (4 , 1) 8 ⎛ π ⎞ ( g D f ) ( x ) = 4 ( tan x ) = 4 tan x f ⎜ ⎟ − f (0 ) ( ) ( ) 3 − 0 ⎝ 6 ⎠ tan π / 6 − tan 0 3 41. π − 0 6 = = π /6 = 3 ⋅ 6 = 2 3 π /6 3 π π
46. ( f D g ) ( x ) = 2sec ⎛ 1 x ⎞ 48. ( f D g ) ( x ) = 1 (2 csc x ) = csc x ⎜ 2 ⎟ 2 ⎝ ⎠ ( g D f ) ( x ) = 2 csc ⎛ 1 x ⎞ ( g D f ) ( x ) = 1 (2sec x ) = sec x 2 ⎝ 2 ⎠ 47. ( f D g ) ( x ) = −2 (cot x ) = −2 cot x 49. a. Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide. Then, cosθ = 3 x x = 3 cosθ and sin θ = 4 y y = 4 θ Thus, L = x + y = 3 + 4 = 3secθ + 4 cscθ . cosθ 3 4 sin θ b. Let Y1 = + . ( g D f ) ( x ) = cot (−2 x ) cos x 25 0 0 sin x _π 2
c. Use MINIMUM to find the least value: 25 d. d (0.1) − d (0) = 3.2492 − 0 ≈ 32.492 0.1 − 0 0.1 − 0 (0.2) − d (0.1) = 7.2654 − 3.2492 ≈ 40.162 0.2 − 0.1 0.2 − 0.1 d (0.3) − d (0.2) = 13.764 − 7.2654 ≈ 64.986 0 _π 0 2 0.3 − 0.2 0.3 − 0.2 d (0.4) − d (0.3) 30.777 −13.764 L is least when θ ≈ 0.83 . = 0.4 − 0.3 0.4 − 0.3 ≈ 170.13 d. L ≈ 3 + 4 ≈ 9.86 feet . e. The first differences represent the average cos (0.83) sin (0.83) rate of change of the beam of light against 50. a. Note that rounding up will result in a ladder that won’t fit around the corner. Answers will vary. d (t ) = 10 tan(π t ) 51. measured in feet per second. For example, between t = 0 seconds and t = 0.1 seconds, the average rate of change of the beam of light against the wall is 32.492 feet per second. y 5 x – π π 3π 2 2 b. d (t ) = 10 tan(π t ) is undefined at t = π 2 –5 y = tan x y 5 and t = 3π , or in general at 2 ⎧ k k is an odd integer ⎫ . At these x – π π 3π ⎨ 2 ⎬ 2 2 ⎩ ⎭ instances, the length of the beam of light approaches infinity. It is at these instances in the rotation of the beacon when the beam of light being cast on the wall changes from one side of the beacon to the other. c. –5 y = –cot x + π 2 Yes, the two functions are equivalent.
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Amplitude: A = 3 = 3 Period: T = 2π = 2π Phase Shift: ω 3 φ = π ω 3 Interval defining one cycle: ⎡ φ , φ + T ⎤ = ⎡π ,π ⎤ ⎢ω ω ⎥ ⎢ 3 ⎥ ⎣ ⎦ ⎣ ⎦ Subinterval width: T = 2π / 3 = π 4 4 6
6. y = 3cos (2x + π) Amplitude: A = 3 = 3 Period: T = 2π = 2π = π ω 2 Phase Shift: φ = −π = − π ω 2 2 Interval defining one cycle: ⎡ φ , φ + T ⎤ = ⎡− π , π ⎤ ⎢ω ω ⎥ ⎢ 2 2 ⎥ ⎣ ⎦ ⎣ ⎦ Subinterval width: 8. y = − 2 cos ⎛ 2 x − π ⎞ ⎜ ⎟ ⎝ ⎠ T = π 4 4 Amplitude: A = − 2 = 2 Key points: Period: T = 2π = 2π = π ⎞ ⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞ ω 2 ⎛ − π ,3 , − ,0 , (0, −3) , ,0 , ,3 ⎜ 2 ⎟ ⎜ 4 ⎟ ⎜ 4 ⎟ ⎜ 2 ⎟ π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Phase Shift: φ = 2 = π ω 2 4 Interval defining one cycle: ⎡ φ , φ + T ⎤ = ⎡π , 5π ⎤ ⎢ω ω ⎥ ⎢ 4 4 ⎥ ⎣ ⎦ ⎣ ⎦ Subinterval width: T = π 4 4 Key points: ⎛ π , −2 ⎞ , ⎛ π ⎞ , ⎛ 3π ⎞ , (π ,0) , ⎛ 5π , −2 ⎞ ⎜ 4 ⎟ ⎜ 2 ⎟ ⎜ 4 ⎟ ⎜ 4 ⎟ 7. y = −3sin ⎛ 2x + π ⎞ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎜ 2 ⎟ ⎝ ⎠ Amplitude:
9. y = 4sin(πx + 2) − 5 Amplitude: A = 4 = 4 Period: T = 2π = 2π = 2 ω π Phase Shift: φ = ω − 2 = − 2 π π Interval defining one cycle: ⎡ φ , φ + T ⎤ = ⎡− 2 , 2 − 2 ⎤ ⎢ω ω ⎥ ⎢ π π ⎥ ⎣ ⎦ ⎣ ⎦ Subinterval width: T = 2 = 1 4 4 2 11. y = 3cos(πx − 2) + 5 Key points: ⎛ − 2 , −5 ⎞ , ⎛ 1 − 2 , −1⎞ , ⎛1 − 2 , −5 ⎞ , Amplitude: A = 3 = 3 2π 2π ⎜ π ⎟ ⎜ 2 π ⎟ ⎜ π ⎟ Period: T = = = 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ω π ⎛ 3 − 2 , −9 ⎞ , ⎛ 2 − 2 , −5 ⎞ φ 2 ⎜ 2 π ⎟ ⎜ π ⎟ Phase Shift: = ⎝ ⎠ ⎝ ⎠ ω π Interval defining one cycle: ⎡ φ , φ + T ⎤ = ⎡ 2 , 2 + 2 ⎤ ⎢ω ω ⎥ ⎢π π ⎥ ⎣ ⎦ ⎣ ⎦ Subinterval width: T = 2 = 1 4 4 2 Key points: ⎞ ⎛ 2 ,8 ⎞ , ⎛ 1 2 ,5 , ⎜1 + ⎛ 3 2 , 2 ⎟ , + ,5 , ⎜ π ⎟ 2 π ⎟ π ⎜ 2 π ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 2 + 2 , 8 ⎞ ⎜ π ⎟ 10. ⎝ ⎠ y = 2 cos(2πx + 4) + 4 Amplitude: A = 2 = 2 Period:
15. A = 2; T = π; φ = 1 19. y = 2 tan (4 x − π ) ω 2 Begin with the graph of y = tan x and apply the ω = 2π = 2π = 2 φ = φ = 1 following transformations: T π ω 2 2 φ = 1 1) Shift right π units ⎡ y = tan ( x − π )⎤ Assuming A is positive, we have that y = A sin(ω x − φ ) = 2 sin(2 x −1) = 2 sin ⎡2 ⎛ x − 1 ⎞⎤ 2) Horizontally compress by a factor of 1 4 ⎡ y = tan (4x − π )⎤ ⎢ ⎜ 2 ⎟⎥ ⎣ ⎦ 16. A = 3; T = π ; 2 ⎣ φ = 2 ω ⎝ ⎠⎦ 3) stretch by a factor of 2 ⎣ y = 2 tan (4 x − π )⎦ ω = 2π = 2π = 4 φ = φ = 2 T π ω 2 4 φ = 8 Assuming A is positive, we have that y = A sin(ω x − φ ) = 3sin(4x − 8) = 3sin ⎡⎣4 ( x − 2)⎤⎦ 17. A = 3; T = 3π; φ = − 1 ω ω = 2π = 2π = 2 3 φ = φ = − 1 20. y = 1 (2 x − π ) 2 Begin with the graph of y = cot x and apply the T 3π 3 ω 2 3 3 φ 1 ⋅ 2 = − 2 3 3 9 following transformations: 1) Shift right π units ⎡ y = cot ( x − π )⎤ 1 Assuming A is positive, we have that y = A sin(ω x − φ ) = 3 sin ⎛ 2 x + 2 ⎞ ⎝ ⎠ 2) Horizontally compress by a factor of 2 ⎡ y = cot (2 x − π )⎤ 1 = 3sin ⎡ 2 ⎛ x + 1 ⎞⎤ 3) Vertically compress by a factor of ⎢ 3 ⎜ 3 ⎟⎥ 2 Vertical
⎣ ⎝ ⎠⎦ ⎡ y = 1 cot (2x − π )⎤ 18. A = 2; T = π; φ = − 2 ⎢ 2 ⎥ ω ω = 2π = 2π = 2 ⎣ ⎦ φ = φ = − 2 T π ω 2 φ = − 4 Assuming A is positive, we have that y = A sin(ω x − φ ) = 2 sin(2 x + 4) = 2 sin ⎡⎣2 ( x + 2)⎤⎦
21. y = 3csc ⎛ 2 x − π ⎞ ⎜ 4 ⎟ ⎝ ⎠ Begin with the graph of y = csc x and apply the following transformations: 1) Shift right π units ⎡ y = csc ⎛ x − π ⎞⎤ 4 ⎢ ⎜ 4 ⎟⎥ ⎝ ⎠ 2) Horizontally compress by a factor of 1 2 ⎡ ⎛ π ⎞⎤ y = csc 2 x − ⎢ ⎜ 4 ⎟⎥ ⎛ π ⎞ ⎝ ⎠ 23. y = − cot ⎜ 2x + ⎟ 3) Vertically stretch by a factor of 3 ⎝ 2 ⎠ Begin with the graph of ⎡ ⎛ π ⎞⎤ y = 3csc 2 x − following transformations: ⎢ ⎜ 4 ⎟⎥ ⎝ ⎠ 1) Shift left π units ⎡ y = cot ⎛ x + π ⎞⎤ 2 ⎢ ⎜ 2 ⎟⎥ ⎝ ⎠ 2) Horizontally compress by a factor of 1 2 ⎡ ⎛ π ⎞⎤ y = cot 2 x + ⎢ ⎜ 2 ⎟⎥ ⎝ ⎠ 3) Reflect about the x-axis ⎡ ⎛ π ⎞⎤ y = − cot 2x + ⎢ ⎜ 2 ⎟⎥ ⎝ ⎠ 22. y = 1 sec (3x − π ) 2 Begin with the graph of y = sec x and apply the following transformations: 1) Shift right π units ⎡ y = sec ( x − π )⎤ 2) Horizontally compress by a factor of 1 3 ⎡ y = sec (3x − π )⎤
24. y =− tan ⎛ 3x + π ⎞ 26. y = − csc ⎛ − 1 π x + π ⎞ ⎜ 2 ⎟ ⎜ 2 4 ⎟ ⎝ ⎠ Begin with the graph of y = tan x and apply the ⎝ ⎠ Begin with the graph of y = csc x and apply the following transformations: 1) Shift left π units ⎡ y = tan ⎛ x + π ⎞⎤ following transformations: 1) Shift left π units ⎡ y = csc ⎛ x + π ⎞⎤ 2 ⎢ ⎜ 2 ⎟⎥ 4 ⎢ ⎜ 4 ⎟⎥ ⎝ ⎠ ⎝ ⎠ 2) Horizontally compress by a factor of 1 ⎡ ⎛ π ⎞⎤ 2) Reflect about the y-axis y = csc − x + 3 ⎢ ⎜ 4 ⎟⎥ ⎝ ⎠ ⎡ ⎛ π ⎞⎤ y = tan 3x + ⎢ ⎜ 2 ⎟⎥ 2 ⎝ ⎠ 3) Horizontally compress by a factor of π ⎡ ⎛ π ⎞⎤ 3) Reflect about the x-axis y =− tan x + ⎛ 1 π ⎞ ⎢ ⎜ 2 ⎟⎥ ⎡ y = csc − ⎤ π x + ⎝ ⎠ ⎢ ⎜ 2 4 ⎟⎥ ⎝ ⎠ 3) Reflect about the x-axis ⎡ ⎛ 1 π ⎞⎤ ⎛ π ,1⎞ ⎢ y = − csc ⎜ − π x + ⎟⎥ ⎜12 ⎟ ⎢ ⎝ 2 4 ⎠⎥ ⎝ ⎠ ⎣ ⎦ ⎛ π , −1⎞ ⎜ 4 ⎟ ⎝ ⎠ 25. y = − sec (2π x + π ) Begin with the graph of y = sec x and apply the following transformations: 1) Shift left π units ⎡ y = sec ( x + π )⎤ 27. I (t ) = 120 sin ⎛ 30π t − π ⎞ , t ≥ 0 ⎣ ⎦ ⎜ 3 ⎟
2) Horizontally compress by a factor of 1 Period: ⎝ ⎠ T = 2π = 2π = 1 2π ω 30π 15 ⎡ y = sec (2π x + π )⎤ 3) Reflect about the x-axis Amplitude: A = 120 π = 120 ⎡ y = − sec (2π x + π )⎤ Phase Shift: φ = 3 = 1 ⎣ ⎦ ω 30π 90
28. I (t ) = 220 sin ⎛ 60π t − π ⎞ , c. t ≥ 0 ⎜ 6 ⎟ Period: ⎝ ⎠ T = 2π = 2π = 1 ω 60π 30 Amplitude: A = 220 π = 220 Phase Shift: φ = 6 = 1 d. y = 15.62 sin (0.517 x − 2 .096) + 40.377 ω 60π 360 e. 60 29. a. 0 13 20 30. a. b. Amplitude: A = 56.0 − 24.2 = 31.8 = 15.9 2 2 80.0 − 34.6 45.4 Shift: 56.0+24.2 = 80.2 = 40.1 b. Amplitude: A = = = 22.7 2 2 2 2 80.0+34.6 114.6 ω = 2π = π 12 6 Phase shift (use y = 24 .2, x = 1): Vertical Shift: ω = 2π = π 12 6 = = 57.3 2 2 24.2 = 15.9 sin ⎛ π ⋅1 − φ ⎞ + 40.1 Phase shift (use y = 34.6, x = 1): ⎜ 6 ⎟ ⎝ ⎠ −15.9 = 15.9 sin ⎛ π − φ ⎞ 34.6 = 22.7 sin ⎛ π ⎝ 6
⋅1 − φ ⎞ + 57.3 ⎠ ⎜ 6 ⎟ ⎝ ⎠ −1 = sin ⎛ π − φ ⎞ − 22.7 = 22.7 sin ⎛ π ⎝ 6 − φ ⎞ ⎠ ⎜ 6 ⎟ ⎝ ⎠ ⎛ π ⎞ − π = π − φ −1 = sin ⎜ − φ ⎟ ⎝ ⎠ 2 6 φ = 2π 3 − π = π − φ 2 6 2π Thus, y = 15.9 sin ⎛ π x − 2π ⎞ + 40.1 or ⎝ ⎠ φ = 3 = ⎛ π − 2π ⎞ ⎟ + or y = 15.9 sin ⎡ π ( x − 4)⎤ + 40.1 . Thus, y 22.7 sin x ⎝ 6 57.3 3 ⎠ ⎢ 6 ⎥ y = 22.7 sin ⎡ π ( x − 4)⎤ + 57.3 . ⎢ 6 ⎥ ⎣ ⎦
c. d. y = 22.61sin (0.503x − 2 .038) + 57.17 c. d. y = 25.693 sin(0.476 x −1.814) + 49.854 e. 90 e. 80 0 13 20 31. a. 0 13 20 32. a. b. Amplitude: A = 75.4 − 25.5 = 49.9 = 24.95 2 2 b. Amplitude: A = 77.0 − 31.8 = 45.2 = 22.6 2 2 Vertical Shift: ω = 2π = π 12 6 75.4+25.5 = 100.9 = 50.45 2 2 Vertical Shift: ω = 2π = π 77.0+31.8 = 108.8 = 54.4 2 2 Phase shift (use y = 25.5, x = 1): 12 6 Phase shift (use y = 31.8, x = 1) : 25.5 = 24.95 sin ⎛ π ⋅1 − φ ⎞ + 50.45 ⎛ π ⎞ ⎜ 6 ⎟ 31.8 = 22.6 sin ⋅1 − φ + 54.4 ⎝ ⎠ ⎜ 6 ⎟ ⎝ ⎠ − 24.95 = 24.95 sin ⎛ π − φ ⎞ ⎛ π ⎞ ⎜ 6 ⎟ − 22.6 = 22.6 sin − φ ⎝ ⎠ ⎜ 6 ⎟ −1 = sin ⎛ π − φ ⎞ ⎝ ⎠ ⎛ π ⎞ ⎜ 6 ⎟ −1 = sin − φ ⎝ ⎠ ⎜ 6 ⎟ ⎝ ⎠ − π = π − φ π π 2 6 φ = 2π − = − φ 2 6 2π Thus,
c. y 80 70 60 50 20 0 1 2 3 4 5 6 7 8 9 101112 c. y 9 7 5 3 1 x 2 4 6 8 12 d. y = 22.46 sin(0.506 x − 2.060) + 54.35 d. y = 4.4 sin ⎛ 4π (16.1333) − 0.2555 ⎞ + 3.8 ⎜ 25 ⎟ ⎝ ⎠ ≈ 8.2 feet e. 80 34. a. 8.1833 + 12.5 = 20.6833 hours which is at 8:41 PM. 13.2 − 2.2 11 b. Amplitude: A = = = 5.5 2 2 Vertical Shift: 13.2 + 2.2 = 15.4 = 7.7 2 2 0 13 ω = 2π = π = 4π 20 33. a. 3.6333 + 12.5 = 16.1333 hours which is at 4:08 PM. 12.5 6.25 25 Phase shift (use y = 13.2, x = 8.1833): 13.2 = 5.5 sin ⎛ 4π ⋅ 8.1833 − φ ⎞ + 7.7 b. Amplitude: A = 8 .2 − (− 0.6) = 8 .8 = 4.4 ⎜ 25 ⎟ ⎝ ⎠ 2 2 ⎛ 4π ⎞ Vertical Shift: 8.2 + ( − 0.6) = 7.6 = 3.8 2 2 5.5 = 5.5 sin ⎜ ⋅ 8.1833 − φ ⎟ ⎝ ⎠ 1 = sin ⎛ 32.7332π − φ ⎞ ω = 2π = π = 4π ⎜ ⎟ 12.5 6.25 25 Phase shift (use y = 8.2, x = 3.6333) : 8.2 = 4.4 sin ⎛ 4π ⋅ 3.6333 − φ ⎞ + 3.8 ⎝ 25 ⎠ π = 32.7332π − φ 2 25 ⎜ 25 ⎟ φ ≈ 2.5426 ⎝ ⎠ ⎛ 4π ⎞ 4.4 = 4.4 sin ⎛ 4π ⋅ 3.6333 − φ ⎞ Thus, y = 5.5 sin ⎜ 25 x − 2.5426 ⎟ + 7.7 or ⎜ 25 ⎟ ⎝ ⎠
⎝ ⎠ y = 5.5 sin ⎡ 4π x − 5.0583 ⎤ + 7.7 . 1 = sin ⎛ 14.5332π − φ ⎞ ⎢ 25 ( )⎥ ⎜ 25 ⎟ ⎣ ⎦ ⎝ ⎠ π = 40.5332π − φ 2 25 φ ≈ 0.2555 c. y 14 12 10 Thus, y = 4.4 sin ⎛ 4π x − 0.2555 ⎞ + 3.8 or 8 ⎜ 25 ⎟ 6 ⎝ ⎠ y = 4.4 sin ⎡ 4π ( x − 0.5083)⎤ + 3.8 . ⎣ ⎦ 4 2 1 3 5 x 7 9 11 13 d. y = 5.5 sin ⎛ 4π (20.6833) − 2.5426 ⎞ + 7.7 ⎜ 25 ⎟ ⎝ ⎠ ≈ 13.2 feet
35. a. Amplitude: A = 13.75 −10.53 = 1.61 2 36. a. Amplitude: A = 15.30 − 9.08 = 3.11 2 Vertical Shift: 13.75 + 10.53 = 12.14 2 ω = 2π 365 Phase shift (use y = 13.75, x = 172): 13.75 = 1.61sin ⎛ 2π ⋅172 − φ ⎞ + 12.14 Vertical Shift: 15.30 + 9.08 = 12.19 2 ω = 2π 365 Phase shift (use y = 15.30, x = 172): 15.30 = 3.11sin ⎛ 2π ⋅172 − φ ⎞ + 12.19 ⎜ 365 ⎟ ⎜ 365 ⎟ ⎝ ⎠ 1.61 = 1.61sin ⎛ 2π ⋅172 − φ ⎞ ⎝ ⎠ 3.11 = 3.11sin ⎛ 2π ⋅172 − φ ⎞ ⎜ 365 ⎟ ⎜ 365 ⎟ ⎝ ⎠ 1 = sin ⎛ 344π − φ ⎞ ⎝ ⎠ 1 = sin ⎛ 344π − φ ⎞ ⎜ 365 ⎟ ⎜ 365 ⎟ ⎝ ⎠ π = 344π − φ 2 365 φ ≈ 1.3900 ⎝ ⎠ π = 344π − φ 2 365 φ ≈ 1.39 Thus, y = 1.61sin ⎛ 2π x −1.39 ⎞ + 12.14 or Thus, y = ⎛ 2π x − 1.39 ⎞ + 12.19 or ⎜ 365 ⎟ ⎜ 365 ⎟ ⎝ ⎠ y = 1.61sin ⎡ 2π ( x − 80.75)⎤ + 12.14 . ⎝ ⎠ y = 3.11sin ⎡ 2π ( x − 80.75)⎤ + 12.19 . ⎢ 365 ⎥ ⎢ 365 ⎥ ⎣ ⎦ b. y = 1.61sin ⎡ 2π (91 − 80.75)⎤ + 12.14 ⎣ ⎦ b. y = 3.11sin ⎛ 2π (91) −1.3900 ⎞ + 12.19 ⎢ 365 ⎥ ⎜ 365 ⎟ ⎣ ⎦ ≈ 12.42 hours c. y 20 10 0 140 280 420 x d. The actual hours of sunlight on April 1, 2005 was 12.43 hours. This is very close to the predicted amount of 12.42 hours. ⎝ ⎠ ≈ 12.74 hours c. y 20
cos π + sin π = 1 + 1 = 3 3 2 2 2 = sin 50º sin (50°+ 360°) 11. 3sin 45º − 4 tan π = 3 ⋅ 2 − 4 ⋅ 3 = 3 2 − 4 3 6 2 3 2 3 = sin 50º sin 50º = 1 12. 4 cos 60º +3 tan π = 4 ⋅ 1 + 3 ⋅ 3 = 2 + 3 3 26. tan 20º ⋅ cos (−20°) ⋅ csc 20° 13. 3 2 6 cos 3π + 2 tan ⎛ − π ⎞ = 6 ⎛ − 2 ⎞ + 2 (− 3 ) = sin 20° ⋅ cos 20° ⋅ cos 20° = 1 1 sin 20° 4 ⎝ 3 ⎠ ⎝ 2 ⎠ 14. = −3 2 − 2 3 3sin 2π − 4 cos 5π = 3⎛ 3 ⎞ − 4 (0) = 3 3 27. sin (−40º ) ⋅ csc 40° = − sin 40° ⋅ csc 40° = − sin 40° ⋅ 1 sin 40º = −1 3 2 ⎝ 2 ⎠ 2 tan −20º cot 20º =− tan 20º ⋅ 1 = −1 15. sec ⎛ − π ⎞ − cot ⎛ − 5π ⎞ = sec π + cot 5π = 2 + 1 = 3 28. ( ) tan 20º ⎜ 3 ⎟ ⎜ 4 ⎟ 3 4 ⎝ ⎠ ⎝ ⎠
cscθ = 1 = 1 = 1⋅ 5 = 5 quadrant III. Thus, secθ 13 sin θ 4 4 4 5 cosθ = 1 = 1 = − 5 secθ = 1 = 1 = 1⋅ 5 = 5 cosθ 3 3 3 secθ 13 13 1 1 3 3 tan θ = sin θ , so cot θ = = tan θ 4 = 1⋅ = 4 4 sin θ = ( tan θ ) (cosθ ) = 12 ⎛ − 5 ⎞ = − 12 32. tan θ = 1 4 and 0 < θ < π , so θ lies in quadrant I. 2 5 ⎝ cscθ = 1 = 1 = − 13 13 ⎠ 13 Using the Pythagorean Identities: sec2 θ = tan 2 θ + 1 sin θ 12 12 cot θ = 1 = 1 = 5 sec2 θ = ⎛ 1 ⎞ ⎝ ⎠ + 1 = 1 + 1 = 17 16 16 tan θ 12 12 secθ =± 17 =± 17 16 4 Note that secθ must be positive since θ lies in quadrant I. Thus, secθ = 17 . 4
⎝ ⎠ ⎝ ⎠ tan θ =± 9 3 cosθ =± 25 5 16 4 Note that tan θ < 0 , so tan θ 3 4 cosθ = 1 = 1 = − 4 169 13 Note that cosθ must be negative because θ lies in quadrant II. Thus, cosθ = − 5 . 13 secθ − 5 5 sin θ 12 12 ⎛ 13 ⎞ 12 4 tan θ = = 13 = ⎜ − ⎟ = − tan θ = sin θ , so cosθ − 13 13 ⎝ 5 ⎠ 5 cosθ cscθ = 1 = 1 = 13 sin θ = ( tan θ ) (cosθ ) = − 3 ⎛ − 4 ⎞ = 3 . sin θ 12 4 ⎜ 5 ⎟ 5 ⎝ ⎠ secθ = 1 = 1 = − 13 cscθ = 1 = 1 = 5 cosθ 5 13 sin θ 3 cot θ = 1 = 1 = − 5 cot θ = 1 = 1 = − 4 tan θ 12 12 tan θ 3 3
38. cosθ 3 θ lies in quadrant III. 5 40. = 12 13 and 3π < θ < 2π (quadrant IV) 2 Using the Pythagorean Identities: sin 2 θ = 1 − cos2 θ 2 Using the Pythagorean Identities: sin 2 θ = 1 − cos2 θ 2 sin 2 θ = 1 − ⎛ − 3 ⎞ = 1 − 9 = 16 sin 2 θ = 1 − ⎛ 12 ⎞ = 1 − 144 = 25 ⎜ 5 ⎟ 25 25 ⎜ 13 ⎟ 169 169 ⎝ ⎠ sin θ =± 16 4 25 5 Note that sin θ must be negative because θ lies in quadrant III. Thus, sin θ 4 5 ⎝ ⎠ sin θ =± 25 = 5 169 13 Note that sin θ must be negative because θ lies in quadrant IV. Thus, sin θ 5 . =− 13 sin θ − 4 4 ⎛ 5 ⎞ 4 sin θ − 5 5 ⎛ 13 ⎞ 5 tan θ = = 5 = − ⎜ − ⎟ = tan θ = = 13 = − ⎜ ⎟ = − cosθ 3 5 ⎝ 3 ⎠ 3 cosθ 12 13 ⎝ 12 ⎠ 12 cscθ = 1 = 1 = − 5 cscθ = 1 = 1 = − 13 sin θ 4 4 sin θ 5 13 secθ = 1 = 1 = − 5 secθ = 1 = 1 = 13 cosθ 3 3 cosθ 12 cot θ = 1 = 1 = 3 cot θ = 1 = 1 = − 12 tan θ 4 tan θ 5 12 39. 5 sin θ =− 13 and 3π < θ < 2π (quadrant IV) 2 41. tan θ = 1 3 and 180º < θ < 270º (quadrant III) Using the Pythagorean Identities: cos2 θ = 1 − sin 2 θ 2 Using the Pythagorean Identities: sec2 θ = tan 2 θ + 1 2 cos2 θ = 1 − ⎛ − 5 ⎞
42. tan θ 2 90º < θ < 180º 3 (quadrant II) 44. cscθ = − 4 and π< θ < 3π 2 (quadrant III) Using the Pythagorean Identities: sec2 θ = tan 2 θ + 1 Using the Pythagorean Identities: cot 2 θ = csc2 θ − 1 sec2 θ = ⎛ − 2 ⎞ + 1 = 4 + 1 = 13 cot 2 θ = (−4)2 − 1 = 16 −1 = 15 ⎜ 3 ⎟ 9 9 ⎝ ⎠ 13 secθ =± =± cot θ =± 15 Note that cot θ must be positive since θ lies in 9 3 Note that secθ must be negative since θ lies in quadrant III. Thus, cot θ = sin θ = 1 = 1 = − 1 15 . quadrant II. Thus, secθ = − 13 . cscθ −4 4 3 cosθ = 1 = 1 = − 3 ⋅ 13 = − 3 13 cot θ = cosθ , so sin θ secθ − 13 13 13 13 cosθ = (cot θ ) (sin θ ) = 15 ⎛ − 1 ⎞ = − 15 3 ⎜ 4 ⎟ 4 tan θ = sin θ , so ⎝ ⎠ tan θ = 1 = 1 ⋅ 15 = 15 cosθ cot θ 15 15 15 sin θ = ( tan θ ) (cosθ ) = − 2 ⎛ − 3 13 ⎞ = 2 13 1 1 4 15 4 15 ⎜ ⎟ secθ = = cosθ − = − ⋅ 15 = − 15 15 3 ⎝ 13 ⎠ 13 15 cscθ = 1 = 1 = 13 = 13 π sin θ 2 13 2 13 2 13 45. cot θ = − 2 and < θ < π (quadrant II) 2 cot θ = 1 = 1 = − 3 Using the Pythagorean Identities: csc2 θ = 1 + cot 2 θ tan θ 2 2 43. secθ = 3 and 3π < θ < 2π (quadrant IV) 2 Using the Pythagorean Identities: tan 2 θ = sec2 θ −1
tan 2 θ = 32 −1 = 9 −1 = 8 tan θ =± 8 = ±2 2 Note that tan θ must be negative since θ lies in quadrant IV. Thus,. tan θ = −2 2 . csc2 θ = 1 + (−2)2 = 1 + 4 = 5 cscθ =± 5 Note that cscθ must be positive because θ lies in quadrant II. Thus, cscθ = 5 . sin θ = 1 = 1 ⋅ 5 = 5 cscθ 5 5 5 cot θ = cosθ , so sin θ cosθ = cot θ sin θ ⎛ 5 ⎞ 2 5 = −2 = − . cosθ = 1 = 1 ( ) ( ) ⎜ ⎟ secθ tan θ = sin θ 3 , so tan θ = 1 = 1 = − 1 ⎝ 5 ⎠ 5 cosθ cot θ −2 2 sin θ = ( tan θ ) (cosθ ) = −2 2 ⎛ 1 ⎞ = − 2 2 . secθ = 1 = 1 =− =− ⎜ ⎟ cosθ 2 5 2 5 2 cscθ = 1 = 1 = − 3 ⋅ 2 = − 3 2 sin θ 2 2 3 2 2 2 4 cot θ = 1 = 1 ⋅ 2 = − 2 tan θ −2 2 2 4
51. y = tan( x + π) 54. y = 4 tan(2x) The graph of y = tan x is shifted π units to the The graph of y = tan x is stretched vertically by 52. left. y =− tan ⎛ x − π ⎞ a factor of 4 and compressed horizontally by a factor of 1 . 2 ⎜ 2 ⎟ ⎝ ⎠ The graph of y = tan x is shifted π units to the 55. y = cot ⎛ x + π ⎞ 2 ⎜ 4 ⎟ ⎝ ⎠ right and reflected across the x-axis. The graph of left. y = cot x is shifted π units to the 4 53. y = − 2 tan(3x) The graph of y = tan x is stretched vertically by 56. y = − 4 cot(2 x) a factor of 2, reflected across the x-axis, and The graph of y = cot x is stretched vertically by compressed horizontally by a factor of 1 . 3 a factor of 4, reflected across the x-axis and compressed horizontally by a factor of 1 . 2
63. 64. 65. y = 4 cos x Amplitude = 4 y = sin(2 x) Amplitude = 1 y = − 8sin ⎛ π x ⎞ = 4 ; Period = 2π = 1 ; Period = 2π = π 2 ⎜ 2 ⎟ ⎝ ⎠ Amplitude = −8 = 8 ; Period = 2π = 4 π 69. y = 2sin (2x − π ) 2 Amplitude: A = 2 = 2 Period: T = 2π = 2π = π 66. y = − 2 cos(3π x) ω 2 Amplitude = −2 = 2 ; Period = 2π = 2 Phase Shift: φ = π = π 67. y = 4sin(3x) Amplitude: A = 4 = 4 3π 3 ω 2 2 Period: T = 2π = 2π Phase Shift: ω 3 φ = 0 = 0 ω 3 70. y = − cos ⎛ 1 x + π ⎞ ⎝ ⎠ Amplitude: A = −1 = 1 Period: T = 2π = 2π = 4π ω 1 2 − π Phase Shift: φ = 2 = −π
68. y = 2 cos ⎛ 1 x ⎞ ω 1 ⎜ 3 ⎟ 2 ⎝ ⎠ Amplitude: A = 2 = 2 Period: T = 2π = 2π = 6π ω 1 3 Phase Shift: φ = 0 = 0 ω 1 3
71. y = 1 sin ⎛ 3 x − π ⎞ 2 ⎜ 2 ⎟ ⎝ ⎠ Amplitude: A = 1 = 1 Period: 2 2 T = 2π = 2π = 4π ω 3 3 2 Phase Shift: φ = π = 2π ω 3 3 2 74. y = −7 sin ⎛ π x + 4 ⎞ ⎝ ⎠ Amplitude: A = − 7 = 7 Period: T = 2π = 2π = 6 ω π 3 − 4 Phase Shift: φ = 3 = − 4 ω π π 3 72. y = 3 cos (6 x + 3π) 2 Amplitude: A = 3 = 3 Period: 2 2 T = 2π = 2π = π Phase Shift: ω 6 3 φ = −3π − π = ω 6 2 75. The graph is a cosine graph with amplitude 5 and period 8π. Find ω : 8π= 2π ω 8πω = 2π ω = 2π = 1
8π 4 The equation is: y = 5cos ⎛ 1 x ⎞ . ⎜ 4 ⎟ ⎝ ⎠ 73. y 2 cos (πx − 6) 3 76. The graph is a sine graph with amplitude 4 and period 8π. π Amplitude: A = − 2 = 2 Find ω : 8π= 2 ω Period: 3 3 T = 2π = 2π = 2 8πω = 2π ω = 2π = 1 ω π 8π 4 Phase Shift: φ = 6 The is: y = 4sin ⎛ 1 x ⎞ . ω π ⎜ 4 ⎟ ⎝ ⎠
77. The graph is a reflected cosine graph with amplitude 6 and period 8. cos t = − 1 ; 3 t = 1 = −3 ⎛ − 1 ⎞ Find ω : 8 = 2π ⎜ 3 ⎟ ⎝ ⎠ ω ⎛ 2 2 ⎞ 8ω = 2π ⎜ 3 ⎟ 2 2 ⎛ 3 ⎞ tan t = ⎝ ⎠ = ⋅ ⎜ − ⎟ = −2 2 ; ω = 2π = π ⎛ − 1 ⎞ 3 ⎝ 1 ⎠ 8 4 ⎜ 3 ⎟ The equation is: y = − 6 cos ⎛ π x ⎞ . ⎝ ⎠ 1 2 2 ⎜ 4 ⎟ cot t = ⋅ = − 4 ⎝ ⎠ 78. The graph is a reflected sine graph with amplitude 7 and period 8. −2 2 2 84. The point P = (−2, 5) is on a circle of radius Find ω : 8 = 2π r = (−2)2 + 52 = 4 + 25 = 29 with the center ω at the origin. So, we have x = −2 , y = 5 , and 8ω = 2π y 5 5 29 r = 29 . Thus, sin t = = = ; ω = 2π = π 8 4 r 29 29 x −2 2 29 y 5 The equation is: y = −7 sin ⎛ π x ⎞ . cos t = = r = − 29 29 ; tan t = x =− . 2 ⎜ 4 ⎟ ⎝ ⎠ 79. Set the calculator to radian mode: sin π ≈ 0.38 . 85. The domain of ⎧ y = sec x is π ⎫ 8 ⎨ x x ≠ odd multiple of ⎬ . ⎩ 2 ⎭ The range of y = sec x is {y y ≤ −1 or y ≥ 1}. The period is 2π . 80. Set the calculator to degree mode: = 1 ≈ 1.02 . cos10o
86. a. b. 32o 20 ' 35" = 32 + 20 + 35 ≈ 32.34o 60 63.18o 0.18o = (0.18)(60 ') = 10.8 ' 0.8 ' = (0.8)(60 ") = 48" Thus, 63.18o = 63o10 ' 48" 81. Terminal side of θ in Quadrant III implies sin θ < 0 cscθ < 0 87. r = 2 θ = 30º or θ = π 6 cosθ < 0 secθ < 0 tan θ > 0 cot θ > 0 82. cosθ > 0, tan θ < 0 ; θ lies in quadrant IV. s = rθ = 2 ⋅ π = π ≈ 1.047 feet 6 3 A = 1 ⋅ r 2θ = 1 ⋅ (2)2 ⋅ π = π ≈ 1.047 square feet 2 2 6 3 83. ⎛ 1 2 2 ⎞ P = − , 88. In 30 minutes: r = 8 inches, θ = 180º or θ = π ⎜ 3 3 ⎟ s = rθ = 8 ⋅ π= 8π≈ 25.13 inches ⎝ ⎠ 2π sin t = 2 2 ; csc t = 1 = 3 ⋅ 2 = 3 2 In 20 minutes: r = 8 inches, θ = 120º or θ = 3 3 ⎛ 2 2 ⎞ 2 2 2 4 ⎜ 3 ⎟ s = rθ = 8 ⋅ 2π = 16π ≈ 16.76 inches ⎝ ⎠ 3 3