Chapter 4 Trigonometric Functions

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<strong>Trigonometric</strong> <strong>Functions</strong> 29. Because the tangent is negative and the sine is positive, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus, 2 y 2 tan θ = − = = , x = − 3, y = 2. Furthermore, 3 x −3 r x y 2 2 2 2 = + = − + = + = ( 3) 2 9 4 13 Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . y 2 2 13 2 13 sinθ = = = ⋅ = r 13 13 13 13 x −3 −3 13 3 13 cosθ = = = ⋅ = − r 13 13 13 13 r 13 cscθ = = y 2 r 13 13 secθ = = = − x −3 3 x −3 3 cotθ = = = − y 2 2 30. Because the tangent is negative and the sine is positive, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus, 1 y 1 tan θ =− = = , y = 1, x =−3 . Furthermore, 3 x −3 r x y 2 2 2 2 = + = − + = + = ( 3) 1 9 1 10 Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . y 1 1 10 10 sinθ = = = ⋅ = r 10 10 10 10 x −3 −3 10 3 10 cosθ = = = ⋅ = − r 10 10 10 10 r 10 cscθ = = = 10 y 1 r 10 10 secθ = = = − x −3 3 x −3 cotθ = = =−3 y 1 31. Because the tangent is positive and the cosine is negative, θ lies in quadrant III. In quadrant III, x is 4 y −4 negative and y is negative. Thus, tan θ = = = , 3 x − 3 x = –3, y = –4. Furthermore, 2 2 2 2 r = x + y = ( − 3) + ( − 4) = 9+ 16 = 25 = 5 Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . y −4 4 sinθ = = = − r 5 5 x −3 3 cosθ = = =− r 5 5 r 5 5 cscθ = = = − y −4 4 r 5 5 secθ = = = − x −3 3 x −3 3 cotθ = = = y −4 4 32. Because the tangent is positive and the cosine is negative, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, 5 y −5 tan θ = = = , x =− 12, y =−5 . Furthermore, 12 x −12 2 2 2 2 r = x + y = ( − 12) + ( − 5) = 144+ 25 = 169 = 13 Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . y −5 5 sinθ = = =− r 13 13 x −12 12 cosθ = = = − r 13 13 r 13 13 cscθ = = = − y −5 5 r 13 13 secθ = = = − x −12 12 x −12 12 cotθ = = = y −5 5 524 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
PreCalculus 4E Section 4.4 33. Because the secant is negative and the tangent is positive, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, r 3 secθ =− 3 = = , x =− 1, r = 3 . Furthermore, x −1 2 2 2 x + y = r 2 2 2 ( − 1) + y = 3 2 y = 9− 1= 8 y =− 8 =−2 2 Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . y −2 2 2 2 sinθ = = =− r 3 3 x −1 1 cosθ = = =− r 3 3 y −2 2 tanθ = = = 2 2 x −1 r 3 3 2 3 2 cscθ = = = ⋅ = − y −2 2 −2 2 2 4 x −1 1 2 2 cotθ = = = ⋅ = y − 2 2 2 2 2 4 34. Because the cosecant is negative and the tangent is positive, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, r 4 cscθ =− 4 = = , y =− 1, r = 4 . Furthermore, y −1 2 2 2 x + y = r 2 2 2 x + ( − 1) = 4 2 x = 16 − 1 = 15 x =− 15 Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . y −1 1 sinθ = = = − r 4 4 x − 15 15 cosθ = = = − r 4 4 y −1 1 15 15 tanθ = = = ⋅ = x − 15 15 15 15 r 4 4 15 4 15 secθ = = = − ⋅ = − x − 15 15 15 15 x − 15 cotθ = = = 15 y − 1 35. Because 160° lies between 90° and 180°, it is in quadrant II. The reference angle is θ ′ = 180°− 160°= 20°. 36. Because 170° lies between 90° and 180°, it is in quadrant II. The reference angle is θ ′ = 180°− 170°= 10°. 37. Because 205° lies between 180° and 270°, it is in quadrant III. The reference angle is θ ′ = 205°− 180°= 25°. 38. Because 210° lies between 180° and 270°, it is in quadrant III. The reference angle is θ ′ = 210°− 180°= 30°. 39. Because 355° lies between 270° and 360°, it is in quadrant IV. The reference angle is θ ′ = 360°− 355°= 5°. 40. Because 351° lies between 270° and 360°, it is in quadrant IV. The reference angle is θ ′ = 360°− 351°= 9°. 41. Because 7 π 3π 6π 8π lies between = and 2π = , it 4 2 4 4 is in quadrant IV. The reference angle is 7π 8π 7π π θ′ = 2π − = − = . 4 4 4 4 42. Because 5π 4 lies between π = 4π 4 and 3π 2 = 6π 4 , it is in quadrant III. The reference angle is θ ′ = 5π 4 − π = 5π 4 − 4π 4 = π 4 . 43. Because 5 π π 3π lies between = and π = 6 2 6 in quadrant II. The reference angle is 5π 6π 5π π θ′ = π − = − = . 6 6 6 6 6π 6 , it is 44. Because 5 π 10 π π 7π = lies between = and 7 14 2 14 14π π = , it is in quadrant II. The reference angle is 14 5π 7π 5π 2π θ′ = π − = − = . 7 7 7 7 45. − 150°+ 360°= 210° Because the angle is in quadrant III, the reference angle is θ ′ = 210°− 180°= 30°. 525 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: PreCalculus 4E Section 4.4 25. In q
Page 39 and 40: PreCalculus 4E Section 4.4 64. 240

Chapter 4 Trigonometric Functions

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