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# Lesson 16: Inverse Trigonometric Functions (slides)

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We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.

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### Lesson 16: Inverse Trigonometric Functions (slides)

1. 1. Sec on 3.6 Inverse Trigonometric Func ons V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 28, 2011.
2. 2. Announcements Midterm has been returned. Please see FAQ on Blackboard (under ”Exams and Quizzes”) Quiz 3 this week in recita on on Sec on 2.6, 2.8, 3.1, 3.2 Quiz 4 April 14–15 on Sec ons 3.3, 3.4, 3.5, and 3.7 Quiz 5 April 28–29 on Sec ons 4.1, 4.2, 4.3, and 4.4
3. 3. Objectives Know the deﬁni ons, domains, ranges, and other proper es of the inverse trignometric func ons: arcsin, arccos, arctan, arcsec, arccsc, arccot. Know the deriva ves of the inverse trignometric func ons.
4. 4. Outline Inverse Trigonometric Func ons Deriva ves of Inverse Trigonometric Func ons Arcsine Arccosine Arctangent Arcsecant Applica ons
5. 5. What is an inverse function? Deﬁni on Let f be a func on with domain D and range E. The inverse of f is the func on f−1 deﬁned by: f−1 (b) = a, where a is chosen so that f(a) = b.
6. 6. What is an inverse function? Deﬁni on Let f be a func on with domain D and range E. The inverse of f is the func on f−1 deﬁned by: f−1 (b) = a, where a is chosen so that f(a) = b. So f−1 (f(x)) = x, f(f−1 (x)) = x
7. 7. What functions are invertible? In order for f−1 to be a func on, there must be only one a in D corresponding to each b in E. Such a func on is called one-to-one The graph of such a func on passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is con nuous, then f−1 is con nuous.
8. 8. Graphing the inverse function y −1 If b = f(a), then f (b) = a. . x
9. 9. Graphing the inverse function y −1 If b = f(a), then f (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . (b, a) (a, b) . x
10. 10. Graphing the inverse function y y=x −1 If b = f(a), then f (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . On the xy-plane, the point (b, a) (b, a) is the reﬂec on of (a, b) in the line y = x. (a, b) . x
11. 11. Graphing the inverse function y y=x −1 If b = f(a), then f (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . On the xy-plane, the point (b, a) (b, a) is the reﬂec on of (a, b) in the line y = x. (a, b) . x
12. 12. Graphing the inverse function y y=x −1 If b = f(a), then f (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . On the xy-plane, the point (b, a) (b, a) is the reﬂec on of (a, b) in the line y = x. (a, b) . x
13. 13. Graphing the inverse function y y=x −1 If b = f(a), then f (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . On the xy-plane, the point (b, a) (b, a) is the reﬂec on of (a, b) in the line y = x. (a, b) Therefore: . x Fact The graph of f−1 is the reﬂec on of the graph of f in the line y = x.
14. 14. Graphing the inverse function y y=x −1 If b = f(a), then f (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . On the xy-plane, the point (b, a) (b, a) is the reﬂec on of (a, b) in the line y = x. (a, b) Therefore: . x Fact The graph of f−1 is the reﬂec on of the graph of f in the line y = x.
15. 15. arcsin Arcsin is the inverse of the sine func on a er restric on to [−π/2, π/2]. y . x π π sin − 2 2
16. 16. arcsin Arcsin is the inverse of the sine func on a er restric on to [−π/2, π/2]. y . x π π sin − 2 2
17. 17. arcsin Arcsin is the inverse of the sine func on a er restric on to [−π/2, π/2]. y y=x . x π π sin − 2 2
18. 18. arcsin Arcsin is the inverse of the sine func on a er restric on to [−π/2, π/2]. y arcsin . x π π sin − 2 2 The domain of arcsin is [−1, 1] [ π π] The range of arcsin is − , 2 2
19. 19. arccos Arccos is the inverse of the cosine func on a er restric on to [0, π] y cos . x 0 π
20. 20. arccos Arccos is the inverse of the cosine func on a er restric on to [0, π] y cos . x 0 π
21. 21. arccos Arccos is the inverse of the cosine func on a er restric on to [0, π] y y=x cos . x 0 π
22. 22. arccos Arccos is the inverse of the cosine func on a er restric on to [0, π] arccos y cos . x 0 π The domain of arccos is [−1, 1] The range of arccos is [0, π]
23. 23. arctan Arctan is the inverse of the tangent func on a er restric on to y (−π/2, π/2). . x 3π π π 3π − − 2 2 2 2 tan
24. 24. arctan Arctan is the inverse of the tangent func on a er restric on to y (−π/2, π/2). . x 3π π π 3π − − 2 2 2 2 tan
25. 25. arctan y=x Arctan is the inverse of the tangent func on a er restric on to y (−π/2, π/2). . x 3π π π 3π − − 2 2 2 2 tan
26. 26. arctan Arctan is the inverse of the tangent func on a er restric on to y (−π/2, π/2). π 2 arctan . x π − 2 ( π π ∞) The domain of arctan is (−∞, ) The range of arctan is − , π 2 2 π lim arctan x = , lim arctan x = − x→∞ 2 x→−∞ 2
27. 27. arcsec Arcsecant is the inverse of secant a er restric on to [0, π/2) ∪ [π, 3π/2). y . x 3π π π 3π − − 2 2 2 2 sec
28. 28. arcsec Arcsecant is the inverse of secant a er restric on to [0, π/2) ∪ [π, 3π/2). y . x 3π π π 3π − − 2 2 2 2 sec
29. 29. arcsec Arcsecant is the inverse of secant a er restric on to x y= [0, π/2) ∪ [π, 3π/2). y . x 3π π π 3π − − 2 2 2 2 sec
30. 30. arcsec 3π 2 Arcsecant is the inverse of secant a er restric on to [0, π/2) ∪ [π, 3π/2). y π 2 . x The domain of arcsec is (−∞, −1] ∪ [1, ∞) [ π ) (π ] The range of arcsec is 0, ∪ ,π 2 2 π 3π lim arcsec x = , lim arcsec x = x→∞ 2 x→−∞ 2
31. 31. Values of Trigonometric Functions x 0 π/6 π/4 π/3 π/2 √ √ sin x 0 1/2 2/2 3/2 1 √ √ cos x 1 3/2 2/2 1/2 0 √ √ tan x 0 1/ 3 1 3 undef √ √ cot x undef 3 1 1/ 3 0 √ √ sec x 1 2/ 3 2/ 2 2 undef √ √ csc x undef 2 2/ 2 2/ 3 1
32. 32. Check: Values of inverse trigonometric functions Example Solu on Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2
33. 33. Check: Values of inverse trigonometric functions Example Solu on Find π arcsin(1/2) 6 arctan(−1) ( √ ) 2 arccos − 2
34. 34. What is arctan(−1)? 3π/4 . −π/4
35. 35. What is arctan(−1)? ( ) 3π 3π/4 Yes, tan = −1 4 √ 2 sin(3π/4) = 2 .√ 2 cos(3π/4) = − 2 −π/4
36. 36. What is arctan(−1)? ( ) 3π 3π/4 Yes, tan = −1 4 √ But, the ) 2 ( π π range of arctan is sin(3π/4) = − , 2 2 2 .√ 2 cos(3π/4) = − 2 −π/4
37. 37. What is arctan(−1)? ( ) 3π 3π/4 Yes, tan = −1 4 √ But, the ) ( π π range of arctan is 2 − , cos(π/4) = 2 2 . 2 Another angle whose √ π 2 tangent is −1 is − , and sin(π/4) = − 4 2 this is in the right range. −π/4
38. 38. What is arctan(−1)? ( ) 3π 3π/4 Yes, tan = −1 4 √ But, the ) ( π π range of arctan is 2 − , cos(π/4) = 2 2 . 2 Another angle whose √ π 2 tangent is −1 is − , and sin(π/4) = − 4 2 this is in the right range. π So arctan(−1) = − −π/4 4
39. 39. Check: Values of inverse trigonometric functions Example Solu on Find π arcsin(1/2) 6 π arctan(−1) − ( √ ) 4 2 arccos − 2
40. 40. Check: Values of inverse trigonometric functions Example Solu on Find π arcsin(1/2) 6 π arctan(−1) − ( √ ) 4 2 3π arccos − 2 4
41. 41. Caution: Notational ambiguity sin2 x =.(sin x)2 sin−1 x = (sin x)−1 sinn x means the nth power of sin x, except when n = −1! The book uses sin−1 x for the inverse of sin x, and never for (sin x)−1 . 1 I use csc x for and arcsin x for the inverse of sin x. sin x
42. 42. Outline Inverse Trigonometric Func ons Deriva ves of Inverse Trigonometric Func ons Arcsine Arccosine Arctangent Arcsecant Applica ons
43. 43. The Inverse Function Theorem Theorem (The Inverse Func on Theorem) Let f be diﬀeren able at a, and f′ (a) ̸= 0. Then f−1 is deﬁned in an open interval containing b = f(a), and 1 (f−1 )′ (b) = f′ (f−1 (b)) In Leibniz nota on we have dx 1 = dy dy/dx
44. 44. Illustrating the IFT Example Use the inverse func on theorem to ﬁnd the deriva ve of the square root func on.
45. 45. Illustrating the IFT Example Use the inverse func on theorem to ﬁnd the deriva ve of the square root func on. Solu on (Newtonian nota on) √ Let f(x) = x2 so that f−1 (y) = y. Then f′ (u) = 2u so for any b > 0 we have 1 (f−1 )′ (b) = √ 2 b
46. 46. Illustrating the IFT Example Use the inverse func on theorem to ﬁnd the deriva ve of the square root func on. Solu on (Leibniz nota on) If the original func on is y = x2 , then the inverse func on is deﬁned by x = y2 . Diﬀeren ate implicitly: dy dy 1 1 = 2y =⇒ = √ dx dx 2 x
47. 47. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)
48. 48. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a right triangle: .
49. 49. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a right triangle: y = arcsin x .
50. 50. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a right triangle: 1 x y = arcsin x .
51. 51. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a right triangle: 1 x y = arcsin x .√ 1 − x2
52. 52. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a right triangle: √ cos(arcsin x) = 1 − x2 1 x y = arcsin x .√ 1 − x2
53. 53. The derivative of arcsine Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a right triangle: √ cos(arcsin x) = 1 − x2 1 So xFact d 1 y = arcsin x arcsin(x) = √ .√ dx 1 − x2 1 − x2
54. 54. Graphing arcsin and its derivative 1 √ The domain of f is [−1, 1], 1 − x2 but the domain of f′ is (−1, 1) arcsin lim− f′ (x) = +∞ x→1 | . | lim + f′ (x) = +∞ −1 1 x→−1
55. 55. Composing with arcsin Example Let f(x) = arcsin(x3 + 1). Find f′ (x).
56. 56. Composing with arcsin Example Let f(x) = arcsin(x3 + 1). Find f′ (x). Solu on We have d 1 d 3 arcsin(x3 + 1) = √ (x + 1) dx 1 − (x3 + 1)2 dx 3x2 = √ −x6 − 2x3
57. 57. The derivative of arccos Let y = arccos x, so x = cos y. Then dy dy 1 1 − sin y = 1 =⇒ = = dx dx − sin y − sin(arccos x)
58. 58. The derivative of arccos Let y = arccos x, so x = cos y. Then dy dy 1 1 − sin y = 1 =⇒ = = dx dx − sin y − sin(arccos x)To simplify, look at a right triangle: √ sin(arccos x) = 1 − x2 1 √So 1 − x2Fact d 1 y = arccos x arccos(x) = − √ . dx 1 − x2 x
59. 59. Graphing arcsin and arccos arccos arcsin | . | −1 1
60. 60. Graphing arcsin and arccos arccos Note (π ) cos θ = sin −θ arcsin 2 π =⇒ arccos x = − arcsin x 2 | . | −1 1 So it’s not a surprise that their deriva ves are opposites.
61. 61. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx sec
62. 62. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx secTo simplify, look at a right triangle: .
63. 63. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx secTo simplify, look at a right triangle: y = arctan x .
64. 64. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx secTo simplify, look at a right triangle: x y = arctan x . 1
65. 65. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx secTo simplify, look at a right triangle: √ 1 + x2 x y = arctan x . 1
66. 66. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx secTo simplify, look at a right triangle: 1 cos(arctan x) = √ 1 + x2 √ 1 + x2 x y = arctan x . 1
67. 67. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = 2y = cos2 (arctan x) dx dx secTo simplify, look at a right triangle: 1 cos(arctan x) = √ 1 + x2 √ So 1 + x2 xFact d 1 y = arctan x . arctan(x) = 1 dx 1 + x2
68. 68. Graphing arctan and its derivative y π/2 arctan 1 1 + x2 . x −π/2 The domain of f and f′ are both (−∞, ∞) Because of the horizontal asymptotes, lim f′ (x) = 0 x→±∞
69. 69. Composing with arctan Example √ Let f(x) = arctan x. Find f′ (x).
70. 70. Composing with arctan Example √ Let f(x) = arctan x. Find f′ (x). Solu on d √ 1 d√ 1 1 arctan x = (√ )2 x= · √ dx 1+ x dx 1+x 2 x 1 = √ √ 2 x + 2x x
71. 71. The derivative of arcsec Try this ﬁrst.
72. 72. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))
73. 73. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a right triangle: .
74. 74. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a right triangle: .
75. 75. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a right triangle: y = arcsec x .
76. 76. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a right triangle: x y = arcsec x . 1
77. 77. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) √To simplify, look at a right triangle: x2 − 1 tan(arcsec x) = 1 √ x x2 − 1 y = arcsec x . 1
78. 78. The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) √To simplify, look at a right triangle: x2 − 1 tan(arcsec x) = 1 So √ x x2 − 1Fact d 1 y = arcsec x arcsec(x) = √ . dx x x2 − 1 1
79. 79. Another Example Example Let f(x) = earcsec 3x . Find f′ (x).
80. 80. Another Example Example Let f(x) = earcsec 3x . Find f′ (x). Solu on 1 f′ (x) = earcsec 3x · √ ·3 3x (3x)2 − 1 3earcsec 3x = √ 3x 9x2 − 1
81. 81. Outline Inverse Trigonometric Func ons Deriva ves of Inverse Trigonometric Func ons Arcsine Arccosine Arctangent Arcsecant Applica ons
82. 82. Application Example One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a ba er stands 2 ft away from home plate as a pitch is thrown with a velocity of 130 ft/sec (about 90 mph). At what rate does the ba er’s angle of gaze need to change to follow the ball as it crosses home plate?
83. 83. Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0. y 130 ft/sec θ . 2
84. 84. Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · dt 1 + (y/2)2 2 dt y 130 ft/sec θ . 2
85. 85. Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · dt 1 + (y/2)2 2 dt When y = 0 and y′ = −130, then y dθ 1 1 130 ft/sec = · (−130) = −65 rad/sec dt y=0 1+0 2 θ . 2
86. 86. Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · dt 1 + (y/2)2 2 dt When y = 0 and y′ = −130, then y dθ 1 1 130 ft/sec = · (−130) = −65 rad/sec dt y=0 1+0 2 θ .The human eye can only track at 3 rad/sec! 2
87. 87. Summary y y′ y y′ 1 1 arcsin x √ arccos x − √ 1 − x2 1 − x2 1 1 arctan x arccot x − 1 + x2 1 + x2 1 1 arcsec x √ arccsc x − √ x x2 − 1 x x2 − 1 Remarkable that the deriva ves of these transcendental func ons are algebraic (or even ra onal!)