Chapter 4 Trigonometric Functions

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<strong>Trigonometric</strong> <strong>Functions</strong> 46. − 250°+ 360°= 110° Because the angle is in quadrant II, the reference angle is θ′ = 180°− 110°= 70°. 47. − 335°+ 360°= 25° Because the angle is in quadrant I, the reference angle is θ′ = 25°. 48. − 359°+ 360°= 1° Because the angle is in quadrant I, the reference angle is θ′ = 1°. 57. 58. 11 11 16 5 − π + 4π = − π + π = π 4 4 4 4 Because the angle is in quadrant III, the reference 5π π angle is θ′ = − π = . 4 4 17 17 24 7 − π + 4π =− π + π = π 6 6 6 6 Because the angle is in quadrant III, the reference 7π π angle is θ′ = − π = . 6 6 49. Because 4.7 lies between π ≈ 3.14 and 3 π ≈ 4.71 , it 2 is in quadrant III. The reference angle is θ′ = 4.7 −π ≈ 1.56 . 50. Because 5.5 lies between 3 π ≈ 4.71 and 2π ≈ 6.28 , 2 it is in quadrant IV. The reference angle is θ′ = 2π −5.5 ≈ 0.78 . 51. 565°− 360°= 205° Because the angle is in quadrant III, the reference angle is θ ′ = 205°− 180°= 25°. 52. 553°− 360°= 193° Because the angle is in quadrant III, the reference angle is θ′ = 193°− 180°= 13°. 53. 54. 55. 56. 17 π 17 12 5 − 2π = π − π = π 6 6 6 6 Because the angle is in quadrant II, the reference 5π π angle is θ′ = π − = . 6 6 11 π 11 8 3 − 2π = π − π = π 4 4 4 4 Because the angle is in quadrant II, the reference 3π π angle is θ′ = π − = . 4 4 23 π 23 16 7 − 4π = π − π = π 4 4 4 4 Because the angle is in quadrant IV, the reference 7π π angle is θ′ = 2π − = . 4 4 17 π 17 12 5 − 4π = π − π = π 3 3 3 3 Because the angle is in quadrant IV, the reference 5π π angle is θ′ = 2π − = . 3 3 59. 60. 25 25 36 11 − π + 6π =− π + π = π 6 6 6 6 Because the angle is in quadrant IV, the reference 11π π angle is θ′ = 2π − = . 6 6 13 13 18 5 − π + 6π = − π + π = π 3 3 3 3 Because the angle is in quadrant IV, the reference 5π π angle is θ′ = 2π − = . 3 3 61. 225° lies in quadrant III. The reference angle is θ ′ = 225°− 180°= 45°. 2 cos 45°= 2 Because the cosine is negative in quadrant III, 2 cos 225 ° = − cos 45 ° = − . 2 62. 300° lies in quadrant IV. The reference angle is θ ′ = 360°− 300°= 60° . 3 sin 60°= 2 Because the sine is negative in quadrant IV, 3 sin 300°=− sin 60°=− . 2 63. 210° lies in quadrant III. The reference angle is 210 180 30 θ ′ = °− °= °. 3 tan 30°= 3 Because the tangent is positive in quadrant III, 3 tan 210 ° = tan 30°= . 3 526 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
PreCalculus 4E Section 4.4 64. 240° lies in quadrant III. The reference angle is θ ′ = 240°− 180°= 60° . sec60°= 2 Because the secant is negative in quadrant III, sec 240°=− sec60°− 2 . 65. 420° lies in quadrant I. The reference angle is 420 360 60 θ ′ = °− °= °. tan 60°= 3 Because the tangent is positive in quadrant I, tan 420 ° = tan 60 ° = 3 . 66. 405° lies in quadrant I. The reference angle is θ ′ = 405°− 360°= 45° . tan 45°= 1 Because the tangent is positive in quadrant I, tan 405 ° =tan45 ° =1. 67. 2π lies in quadrant II. The reference angle is 3 2π 3π 2π π θ′ = π − = − = . 3 3 3 3 π 3 sin = 3 2 Because the sine is positive in quadrant II, 2π π 3 sin =sin = . 3 3 2 70. 71. 72. 7π lies in quadrant IV. The reference angle is 4 7π 8π 7π π θ′ = 2 π − = − = . 4 4 4 4 π cot = 1 4 Because the cotangent is negative in quadrant IV, 7π π cot = − cot = − 1 . 4 4 9π lies in quadrant I. The reference angle is 4 9 9 8 θ′ = π − 2π = π − π = π . 4 4 4 4 π tan = 1 4 Because the tangent is positive in quadrant I, 9π π tan =tan = 1 4 4 9π lies on the positive y-axis. The reference angle is 2 9 9 8 θ′ = π − 4π = π − π = π . 2 2 2 2 π 9π Because tan is undefined, tan is also 2 2 undefined. 68. 69. 3π lies in quadrant II. The reference angle is 4 3π 4π 3π π θ′ = π − = − = . 4 4 4 4 π 2 cos = 4 2 Because the cosine is negative in quadrant II, 3π π 2 cos =– cos =− . 4 4 2 7π lies in quadrant III. The reference angle is 6 7 7 6 θ′ = π − π = π − π = π . 6 6 6 6 π csc = 2 6 Because the cosecant is negative in quadrant III, 7π π csc = − csc =− 2 . 6 6 73. –240° lies in quadrant II. The reference angle is 240 180 60 θ ′ = °− °= °. 3 sin 60°= 2 Because the sine is positive in quadrant II, 3 sin( − 240 ° )= sin 60 ° = . 2 74. –225° lies in quadrant II. The reference angle is θ ′ = 225°− 180°= 45 ° . 2 sin 45°= 2 Because the sine is positive in quadrant II, 2 sin( − 225 ° ) = sin 45°= . 2 527 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Page 1 and 2: Chapter 4 Trigonometric Functions S
Page 3 and 4: PreCalculus 4E Section 4.1 16. 17.
Page 7 and 8: PreCalculus 4E Section 4.1 88. Firs
Page 9 and 10: PreCalculus 4E Section 4.2 117. doe
Page 11 and 12: PreCalculus 4E Section 4.2 2. The p
Page 15 and 16: PreCalculus 4E Section 4.2 0 π 57.
Page 17 and 18: PreCalculus 4E Section 4.2 π H = 1
Page 19 and 20: PreCalculus 4E Section 4.3 3. Apply
Page 21 and 22: PreCalculus 4E Section 4.3 3. 2 2 2
Page 23 and 24: PreCalculus 4E Section 4.3 π π π
Page 25 and 26: PreCalculus 4E Section 4.3 43. π t
Page 29 and 30: PreCalculus 4E Section 4.4 80. a. L
Page 31 and 32: PreCalculus 4E Section 4.4 c. 8. a.
Page 33 and 34: PreCalculus 4E Section 4.4 7. We ne
Page 35 and 36: PreCalculus 4E Section 4.4 25. In q
Page 37: PreCalculus 4E Section 4.4 33. Beca

Chapter 4 Trigonometric Functions

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