![January 27, 2005 11:58 l24-appa-sv Sheet number 4 Page number 527 black
Exercise Set A 527
43. From the figure, h = x − y but x = d tan β,
y = d tan α so h = d(tan β − tan α).
h
x
y
β α
d
44. From the figure, d = x − y but x = h cot α,
y = h cot β so d = h(cot α − cot β),
d h
h= .
cot α − cot β β
α
d y
x
√ √
45. (a) sin 2θ = 2 sin θ cos θ = 2( 5/3)(2/3) = 4 5/9
(b) cos 2θ = 2 cos2 θ − 1 = 2(2/3)2 − 1 = −1/9
√ √ √
46. (a) sin(α − β) = sin α cos β − cos α sin β = (3/5)(1/ 5) − (4/5)(2/ 5) = −1/ 5
√ √ √
(b) cos(α + β) = cos α cos β − sin α sin β = (4/5)(1/ 5) − (3/5)(2/ 5) = −2/(5 5)
47. sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = (2 sin θ cos θ) cos θ + (cos2 θ − sin2 θ) sin θ
= 2 sin θ cos2 θ + sin θ cos2 θ − sin3 θ = 3 sin θ cos2 θ − sin3 θ; similarly, cos 3θ = cos3 θ − 3 sin2 θ cos θ
cos θ sec θ cos θ sec θ cos θ cos θ
48. = = = = cos2 θ
1 + tan2 θ sec2 θ sec θ (1/ cos θ)
cos θ tan θ + sin θ cos θ(sin θ/ cos θ) + sin θ
49. = = 2 cos θ
tan θ sin θ/ cos θ
2 2 1 1
50. 2 csc 2θ = = = = csc θ sec θ
sin 2θ 2 sin θ cos θ sin θ cos θ
sin θ cos θ sin2 θ + cos2 θ 1 2 2
51. tan θ + cot θ = + = = = = = 2 csc 2θ
cos θ sin θ sin θ cos θ sin θ cos θ 2 sin θ cos θ sin 2θ
sin 2θ cos 2θ sin 2θ cos θ − cos 2θ sin θ sin θ
52. − = = = sec θ
sin θ cos θ sin θ cos θ sin θ cos θ
sin θ + cos 2θ − 1 sin θ + (1 − 2 sin2 θ) − 1 sin θ(1 − 2 sin θ)
53. = = = tan θ
cos θ − sin 2θ cos θ − 2 sin θ cos θ cos θ(1 − 2 sin θ)
54. Using (47), 2 sin 2θ cos θ = 2(1/2)(sin θ + sin 3θ) = sin θ + sin 3θ
55. Using (47), 2 cos 2θ sin θ = 2(1/2)[sin(−θ) + sin 3θ] = sin 3θ − sin θ
sin(θ/2) 2 sin2 (θ/2) 1 − cos θ
56. tan(θ/2) = = =
cos(θ/2) 2 sin(θ/2) cos(θ/2) sin θ](https://image.slidesharecdn.com/appendixapage524-121017115444-phpapp02/85/Appendix-a-page_524-4-320.jpg)
![January 27, 2005 11:58 l24-appa-sv Sheet number 5 Page number 528 black
528 Appendix A
sin(θ/2) 2 sin(θ/2) cos(θ/2) sin θ
57. tan(θ/2) = = 2 (θ/2)
=
cos(θ/2) 2 cos 1 + cos θ
58. From (52), cos(π/3 + θ) + cos(π/3 − θ) = 2 cos(π/3) cos θ = 2(1/2) cos θ = cos θ
C
1
59. From the figures, area = hc but h = b sin A
2
1 a
so area = bc sin A. The formulas b h
2
1 1
area = ac sin B and area = ab sin C
2 2 A
c
B
follow by drawing altitudes from vertices B and C, respectively.
60. From right triangles ADC and BDC, C
h1 = b sin A = a sin B so a/ sin A = b/ sin B.
From right triangles AEB and CEB,
h1 a
h2 = c sin A = a sin C so a/ sin A = c/ sin C b
thus a/ sin A = b/ sin B = c/ sin C. E h2
D
A B
c
61. (a) sin(π/2 + θ) = sin(π/2) cos θ + cos(π/2) sin θ = (1) cos θ + (0) sin θ = cos θ
(b) cos(π/2 + θ) = cos(π/2) cos θ − sin(π/2) sin θ = (0) cos θ − (1) sin θ = − sin θ
(c) sin(3π/2 − θ) = sin(3π/2) cos θ − cos(3π/2) sin θ = (−1) cos θ − (0) sin θ = − cos θ
(d) cos(3π/2 + θ) = cos(3π/2) cos θ − sin(3π/2) sin θ = (0) cos θ − (−1) sin θ = sin θ
sin(α + β) sin α cos β + cos α sin β
62. tan(α + β) = = , divide numerator and denominator by
cos(α + β) cos α cos β − sin α sin β
sin α sin β
cos α cos β and use tan α = and tan β = to get (38);
cos α cos β
tan α + tan(−β) tan α − tan β
tan(α − β) = tan(α + (−β)) = = because
1 − tan α tan(−β) 1 + tan α tan β
tan(−β) = − tan β.
63. (a) Add (34) and (36) to get sin(α − β) + sin(α + β) = 2 sin α cos β so
sin α cos β = (1/2)[sin(α − β) + sin(α + β)].
(b) Subtract (35) from (37). (c) Add (35) and (37).
A+B A−B 1
64. (a) From (47), sin cos = (sin B + sin A) so
2 2 2
A+B A−B
sin A + sin B = 2 sin cos .
2 2
(b) Use (49) (c) Use (48)
α−β α+β
65. sin α + sin(−β) = 2 sin cos , but sin(−β) = − sin β so
2 2
α+β α−β
sin α − sin β = 2 cos sin .
2 2](https://image.slidesharecdn.com/appendixapage524-121017115444-phpapp02/85/Appendix-a-page_524-5-320.jpg)
Appendix a page_524
- 1. January 27, 2005 11:58 l24-appa-sv Sheet number 1 Page number 524 black APPENDIX A Trigonometry Review EXERCISE SET A 1. (a) 5π/12 (b) 13π/6 (c) π/9 (d) 23π/30 2. (a) 7π/3 (b) π/12 (c) 5π/4 (d) 11π/12 3. (a) 12◦ (b) (270/π)◦ (c) 288◦ (d) 540◦ 4. (a) 18◦ (b) (360/π)◦ (c) 72◦ (d) 210◦ 5. sin θ cos θ tan θ csc θ sec θ cot θ √ √ √ √ (a) 21/5 2/5 21/2 5/ 21 5/2 2/ 21 √ √ √ √ (b) 3/4 7/4 3/ 7 4/3 4/ 7 7/3 √ √ √ √ (c) 3/ 10 1/ 10 3 10/3 10 1/3 6. sin θ cos θ tan θ csc θ sec θ cot θ √ √ √ √ (a) 1/ 2 1/ 2 1 2 2 1 (b) 3/5 4/5 3/4 5/3 5/4 4/3 √ √ √ √ (c) 1/4 15/4 1/ 15 4 4/ 15 15 √ √ √ √ 7. sin θ = 3/ 10, cos θ = 1/ 10 8. sin θ = 5/3, tan θ = 5/2 √ √ √ √ 9. tan θ = 21/2, csc θ = 5/ 21 10. cot θ = 15, sec θ = 4/ 15 11. Let x be the length of the side adjacent to θ, then cos θ = x/6 = 0.3, x = 1.8. 12. Let x be the length of the hypotenuse, then sin θ = 2.4/x = 0.8, x = 2.4/0.8 = 3. 13. θ sin θ cos θ tan θ csc θ sec θ cot θ √ √ √ √ (a) 225◦ −1/ 2 −1/ 2 1 − 2 − 2 1 √ √ √ √ (b) −210◦ 1/2 − 3/2 −1/ 3 2 −2/ 3 − 3 √ √ √ √ (c) 5π/3 − 3/2 1/2 − 3 −2/ 3 2 −1/ 3 (d) −3π/2 1 0 — 1 — 0 14. θ sin θ cos θ tan θ csc θ sec θ cot θ √ √ √ √ (a) 330◦ −1/2 3/2 −1/ 3 −2 2/ 3 − 3 √ √ √ √ (b) −120◦ − 3/2 −1/2 3 −2/ 3 −2 1/ 3 √ √ √ √ (c) 9π/4 1/ 2 1/ 2 1 2 2 1 (d) −3π 0 −1 0 — −1 — 524
- 2. January 27, 2005 11:58 l24-appa-sv Sheet number 2 Page number 525 black Exercise Set A 525 15. sin θ cos θ tan θ csc θ sec θ cot θ (a) 4/5 3/5 4/3 5/4 5/3 3/4 (b) −4/5 3/5 −4/3 −5/4 5/3 −3/4 √ √ √ √ (c) 1/2 − 3/2 −1/ 3 2 −2 3 − 3 √ √ √ √ (d) −1/2 3/2 −1/ 3 −2 2/ 3 − 3 √ √ √ √ (e) 1/ 2 1/ 2 1 2 2 1 √ √ √ √ (f ) 1/ 2 −1/ 2 −1 2 − 2 −1 16. sin θcos θ tan θ csc θ sec θ cot θ √ √ √ √ (a) 1/4 15/4 1/ 15 4 4/ 15 15 √ √ √ √ (b) 1/4 − 15/4 −1/ 15 4 −4/ 15 − 15 √ √ √ √ (c) 3/ 10 1/ 10 3 10/3 10 1/3 √ √ √ √ (d) −3/ 10 −1/ 10 3 − 10/3 − 10 1/3 √ √ √ √ (e) 21/5 −2/5 − 21/2 5/ 21 −5/2 −2/ 21 √ √ √ √ (f ) − 21/5 −2/5 21/2 −5/ 21 −5/2 2/ 21 17. (a) x = 3 sin 25◦ ≈ 1.2679 (b) x = 3/ tan(2π/9) ≈ 3.5753 18. (a) x = 2/ sin 20◦ ≈ 5.8476 (b) x = 3/ cos(3π/11) ≈ 4.5811 19. sin θ cos θ tan θ csc θ sec θ cot θ √ √ √ √ (a) a/3 9 − a2 /3 a/ 9 − a2 3/a 3/ 9 − a2 9 − a2 /a √ √ √ √ (b) a/ a2 + 25 5/ a2 + 25 a/5 a2 + 25/a a2 + 25/5 5/a √ √ √ √ (c) a2 − 1/a 1/a a2 − 1 a/ a2 − 1 a 1/ a2 − 1 20. (a) θ = 3π/4 ± 2nπ and θ = 5π/4 ± 2nπ, n = 0, 1, 2, . . . (b) θ = 5π/4 ± 2nπ and θ = 7π/4 ± 2nπ, n = 0, 1, 2, . . . 21. (a) θ = 3π/4 ± nπ, n = 0, 1, 2, . . . (b) θ = π/3 ± 2nπ and θ = 5π/3 ± 2nπ, n = 0, 1, 2, . . . 22. (a) θ = 7π/6 ± 2nπ and θ = 11π/6 ± 2nπ, n = 0, 1, 2, . . . (b) θ = π/3 ± nπ, n = 0, 1, 2, . . . 23. (a) θ = π/6 ± nπ, n = 0, 1, 2, . . . (b) θ = 4π/3 ± 2nπ and θ = 5π/3 ± 2nπ, n = 0, 1, 2, . . . 24. (a) θ = 3π/2 ± 2nπ, n = 0, 1, 2, . . . (b) θ = π ± 2nπ, n = 0, 1, 2, . . . 25. (a) θ = 3π/4 ± nπ, n = 0, 1, 2, . . . (b) θ = π/6 ± nπ, n = 0, 1, 2, . . . 26. (a) θ = 2π/3 ± 2nπ and θ = 4π/3 ± 2nπ, n = 0, 1, 2, . . . (b) θ = 7π/6 ± 2nπ and θ = 11π/6 ± 2nπ, n = 0, 1, 2, . . .
- 3. January 27, 2005 11:58 l24-appa-sv Sheet number 3 Page number 526 black 526 Appendix A 27. (a) θ = π/3 ± 2nπ and θ = 2π/3 ± 2nπ, n = 0, 1, 2, . . . (b) θ = π/6 ± 2nπ and θ = 11π/6 ± 2nπ, n = 0, 1, 2, . . . 28. sin θ = −3/5, cos θ = −4/5, tan θ = 3/4, csc θ = −5/3, sec θ = −5/4, cot θ = 4/3 √ √ √ √ 29. sin θ = 2/5, cos θ = − 21/5, tan θ = −2/ 21, csc θ = 5/2, sec θ = −5/ 21, cot θ = − 21/2 30. (a) θ = π/2 ± 2nπ, n = 0, 1, 2, . . . (b) θ = ±2nπ, n = 0, 1, 2, . . . (c) θ = π/4 ± nπ, n = 0, 1, 2, . . . (d) θ = π/2 ± 2nπ, n = 0, 1, 2, . . . (e) θ = ±2nπ, n = 0, 1, 2, . . . (f ) θ = π/4 ± nπ, n = 0, 1, 2, . . . 31. (a) θ = ±nπ, n = 0, 1, 2, . . . (b) θ = π/2 ± nπ, n = 0, 1, 2, . . . (c) θ = ±nπ, n = 0, 1, 2, . . . (d) θ = ±nπ, n = 0, 1, 2, . . . (e) θ = π/2 ± nπ, n = 0, 1, 2, . . . (f ) θ = ±nπ, n = 0, 1, 2, . . . 32. Construct a right triangle with one angle equal to 17◦ , measure the lengths of the sides and hypotenuse and use formula (6) for sin θ and cos θ to approximate sin 17◦ and cos 17◦ . 33. (a) s = rθ = 4(π/6) = 2π/3 cm (b) s = rθ = 4(5π/6) = 10π/3 cm 34. r = s/θ = 7/(π/3) = 21/π 35. θ = s/r = 2/5 1 2 1 1 36. θ = s/r so A = r θ = r2 (s/r) = rs 2 2 2 2π − θ 37. (a) 2πr = R(2π − θ), r = R 2π √ 4πθ − θ2 (b) h = R2 − r 2 = R2 − (2π − θ)2 R2 /(4π 2 ) = R 2π 38. The circumference of the circular base is 2πr. When cut and flattened, the cone becomes a circular sector of radius L. If θ is the central angle that subtends the arc of length 2πr, then θ = (2πr)/L so the area S of the sector is S = (1/2)L2 (2πr/L) = πrL which is the lateral surface area of the cone. 39. Let h be the altitude as shown in the figure, then √ 1 √ √ h = 3 sin 60◦ = 3 3/2 so A = (3 3/2)(7) = 21 3/4. 3 h 2 60° 7 40. Draw the perpendicular from vertex C as shown in the figure, C then √ h = 9 sin 30◦ = 9/2, a = h/ sin 45◦ = 9 2/2, √ 9 h a c1 = 9 cos 30√ = 9 3/2, c2 = a cos 45◦ = 9/2, ◦ 45° 30° c1 + c2 = 9( 3 + 1)/2, angle C = 180◦ − (30◦ + 45◦ ) = 105◦ A c1 c2 B 41. Let x be the distance above the ground, then x = 10 sin 67◦ ≈ 9.2 ft. 42. Let x be the height of the building, then x = 120 tan 76◦ ≈ 481 ft.
- 4. January 27, 2005 11:58 l24-appa-sv Sheet number 4 Page number 527 black Exercise Set A 527 43. From the figure, h = x − y but x = d tan β, y = d tan α so h = d(tan β − tan α). h x y β α d 44. From the figure, d = x − y but x = h cot α, y = h cot β so d = h(cot α − cot β), d h h= . cot α − cot β β α d y x √ √ 45. (a) sin 2θ = 2 sin θ cos θ = 2( 5/3)(2/3) = 4 5/9 (b) cos 2θ = 2 cos2 θ − 1 = 2(2/3)2 − 1 = −1/9 √ √ √ 46. (a) sin(α − β) = sin α cos β − cos α sin β = (3/5)(1/ 5) − (4/5)(2/ 5) = −1/ 5 √ √ √ (b) cos(α + β) = cos α cos β − sin α sin β = (4/5)(1/ 5) − (3/5)(2/ 5) = −2/(5 5) 47. sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = (2 sin θ cos θ) cos θ + (cos2 θ − sin2 θ) sin θ = 2 sin θ cos2 θ + sin θ cos2 θ − sin3 θ = 3 sin θ cos2 θ − sin3 θ; similarly, cos 3θ = cos3 θ − 3 sin2 θ cos θ cos θ sec θ cos θ sec θ cos θ cos θ 48. = = = = cos2 θ 1 + tan2 θ sec2 θ sec θ (1/ cos θ) cos θ tan θ + sin θ cos θ(sin θ/ cos θ) + sin θ 49. = = 2 cos θ tan θ sin θ/ cos θ 2 2 1 1 50. 2 csc 2θ = = = = csc θ sec θ sin 2θ 2 sin θ cos θ sin θ cos θ sin θ cos θ sin2 θ + cos2 θ 1 2 2 51. tan θ + cot θ = + = = = = = 2 csc 2θ cos θ sin θ sin θ cos θ sin θ cos θ 2 sin θ cos θ sin 2θ sin 2θ cos 2θ sin 2θ cos θ − cos 2θ sin θ sin θ 52. − = = = sec θ sin θ cos θ sin θ cos θ sin θ cos θ sin θ + cos 2θ − 1 sin θ + (1 − 2 sin2 θ) − 1 sin θ(1 − 2 sin θ) 53. = = = tan θ cos θ − sin 2θ cos θ − 2 sin θ cos θ cos θ(1 − 2 sin θ) 54. Using (47), 2 sin 2θ cos θ = 2(1/2)(sin θ + sin 3θ) = sin θ + sin 3θ 55. Using (47), 2 cos 2θ sin θ = 2(1/2)[sin(−θ) + sin 3θ] = sin 3θ − sin θ sin(θ/2) 2 sin2 (θ/2) 1 − cos θ 56. tan(θ/2) = = = cos(θ/2) 2 sin(θ/2) cos(θ/2) sin θ
- 5. January 27, 2005 11:58 l24-appa-sv Sheet number 5 Page number 528 black 528 Appendix A sin(θ/2) 2 sin(θ/2) cos(θ/2) sin θ 57. tan(θ/2) = = 2 (θ/2) = cos(θ/2) 2 cos 1 + cos θ 58. From (52), cos(π/3 + θ) + cos(π/3 − θ) = 2 cos(π/3) cos θ = 2(1/2) cos θ = cos θ C 1 59. From the figures, area = hc but h = b sin A 2 1 a so area = bc sin A. The formulas b h 2 1 1 area = ac sin B and area = ab sin C 2 2 A c B follow by drawing altitudes from vertices B and C, respectively. 60. From right triangles ADC and BDC, C h1 = b sin A = a sin B so a/ sin A = b/ sin B. From right triangles AEB and CEB, h1 a h2 = c sin A = a sin C so a/ sin A = c/ sin C b thus a/ sin A = b/ sin B = c/ sin C. E h2 D A B c 61. (a) sin(π/2 + θ) = sin(π/2) cos θ + cos(π/2) sin θ = (1) cos θ + (0) sin θ = cos θ (b) cos(π/2 + θ) = cos(π/2) cos θ − sin(π/2) sin θ = (0) cos θ − (1) sin θ = − sin θ (c) sin(3π/2 − θ) = sin(3π/2) cos θ − cos(3π/2) sin θ = (−1) cos θ − (0) sin θ = − cos θ (d) cos(3π/2 + θ) = cos(3π/2) cos θ − sin(3π/2) sin θ = (0) cos θ − (−1) sin θ = sin θ sin(α + β) sin α cos β + cos α sin β 62. tan(α + β) = = , divide numerator and denominator by cos(α + β) cos α cos β − sin α sin β sin α sin β cos α cos β and use tan α = and tan β = to get (38); cos α cos β tan α + tan(−β) tan α − tan β tan(α − β) = tan(α + (−β)) = = because 1 − tan α tan(−β) 1 + tan α tan β tan(−β) = − tan β. 63. (a) Add (34) and (36) to get sin(α − β) + sin(α + β) = 2 sin α cos β so sin α cos β = (1/2)[sin(α − β) + sin(α + β)]. (b) Subtract (35) from (37). (c) Add (35) and (37). A+B A−B 1 64. (a) From (47), sin cos = (sin B + sin A) so 2 2 2 A+B A−B sin A + sin B = 2 sin cos . 2 2 (b) Use (49) (c) Use (48) α−β α+β 65. sin α + sin(−β) = 2 sin cos , but sin(−β) = − sin β so 2 2 α+β α−β sin α − sin β = 2 cos sin . 2 2
- 6. January 27, 2005 11:58 l24-appa-sv Sheet number 6 Page number 529 black Exercise Set A 529 66. (a) From (34), C sin(α + φ) = C sin α cos φ + C cos α sin φ so C cos φ = 3 and C sin φ = 5, √ √ square and add to get C 2 (cos2 φ + sin2 φ) = 9 + 25, C 2 = 34. If C = 34 then cos φ = 3/ 34 √ and sin φ = 5/ 34√ φ is the first-quadrant angle for which tan φ = 5/3. so 3 sin α + 5 cos α = 34 sin(α + φ). √ (b) Follow the procedure of part (a) to get C cos φ = A and C sin φ = B, C = A2 + B 2 , tan φ = B/A where the quadrant in which φ lies is determined by the signs of A and B because √ cos φ = A/C and sin φ = B/C, so A sin α + B cos α = A2 + B 2 sin(α + φ). 67. Consider the triangle having a, b, and d as sides. The angle formed by sides a and b is π − θ so √ from the law of cosines, d2 = a2 + b2 − 2ab cos(π − θ) = a2 + b2 + 2ab cos θ, d = a2 + b2 + 2ab cos θ.