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# Lesson 16: Inverse Trigonometric Functions

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We go over the trigonometric function, their inverses, and the derivatives of the inverse functions. The surprising fact is that these derivatives are simpler functions than the functions themselves.

We go over the trigonometric function, their inverses, and the derivatives of the inverse functions. The surprising fact is that these derivatives are simpler functions than the functions themselves.

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• 1. Section 3.5 Inverse Trigonometric Functions V63.0121.034, Calculus I October 28, 2009 Announcements . . . . . .
• 2. What functions are invertible? In order for f−1 to be a function, there must be only one a in D corresponding to each b in E. Such a function is called one-to-one The graph of such a function passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is continuous, then f−1 is continuous. . . . . . .
• 3. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications . . . . . .
• 4. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . x . π π s . in − . − . 2 2 . . . . . .
• 5. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . . x . π π s . in − . . − . 2 2 . . . . . .
• 6. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . y . =x . . . . x . π π s . in − . . − . 2 2 . . . . . .
• 7. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . rcsin a . . . . x . π π s . in − . . − . 2 2 . The domain of arcsin is [−1, 1] [ π π] The range of arcsin is − , 2 2 . . . . . .
• 8. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . c . os . . x . 0 . . π . . . . . .
• 9. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . . c . os . . x . 0 . . π . . . . . . .
• 10. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . y . =x . c . os . . x . 0 . . π . . . . . . .
• 11. arccos Arccos is the inverse of the cosine function after restriction to [0, π] . . rccos a y . . c . os . . . x . 0 . . π . The domain of arccos is [−1, 1] The range of arccos is [0, π] . . . . . .
• 12. arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 t .an . . . . . .
• 13. arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 t .an . . . . . .
• 14. arctan Arctan is the inverse of the tangent function after restriction to y . =x [−π/2, π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 t .an . . . . . .
• 15. arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. y . π . a . rctan 2 . x . π − . 2 The domain of arctan is (−∞, ∞) ( π π) The range of arctan is − , 2 2 π π lim arctan x = , lim arctan x = − x→∞ 2 x→−∞ 2 . . . . . .
• 16. arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 s . ec . . . . . .
• 17. arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. y . . . x . 3π π π 3π − . − . . . . 2 2 2 2 s . ec . . . . . .
• 18. arcsec Arcsecant is the inverse of secant after restriction to y . =x [0, π/2) ∪ (π, 3π/2]. y . . . x . 3π π π 3π − . − . . . . 2 2 2 2 s . ec . . . . . .
• 19. arcsec 3π . Arcsecant is the inverse of secant after restriction to 2 [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 . . . . . .
• 20. Values of Trigonometric Functions π π π π x 0 6 √4 √3 2 1 2 3 sin x 0 1 √2 √2 2 3 2 1 cos x 1 0 2 2 √2 1 tan x 0 √ 1 3 undef 3 √ 1 cot x undef 3 1 √ 0 3 2 2 sec x 1 √ √ 2 undef 3 2 2 2 csc x undef 2 √ √ 1 2 3 . . . . . .
• 21. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 . . . . . .
• 22. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 Solution π 6 . . . . . .
• 23. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 Solution π 6 π − 4 . . . . . .
• 24. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 Solution π 6 π − 4 3π 4 . . . . . .
• 25. Caution: Notational ambiguity . in2 x =.(sin x)2 s . in−1 x = (sin x)−1 s 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. 1 I use csc x for and arcsin x for the inverse of sin x. sin x . . . . . .
• 26. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications . . . . . .
• 27. Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is deﬁned in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) . . . . . .
• 28. Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is deﬁned in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) “Proof”. If y = f−1 (x), then f (y ) = x , So by implicit differentiation dy dy 1 1 f′ (y) = 1 =⇒ = ′ = ′ −1 dx dx f (y) f (f (x)) . . . . . .
• 29. The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) . . . . . .
• 30. The derivative of arcsin 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: . . . . . . .
• 31. The derivative of arcsin 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 . . . . . . . .
• 32. The derivative of arcsin 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 . . . . . . .
• 33. The derivative of arcsin 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 . . . . . .
• 34. The derivative of arcsin 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 . . . . . .
• 35. The derivative of arcsin 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 . So d 1 y . = arcsin x arcsin(x) = √ dx 1 − x2 . √ . 1 − x2 . . . . . .
• 36. Graphing arcsin and its derivative 1 .√ 1 − x2 . . rcsin a . | . . | − . 1 1 . . . . . . .
• 37. 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) . . . . . .
• 38. 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 . √ . 1 − x2 So d 1 y . = arccos x arccos(x) = − √ . dx 1 − x2 x . . . . . . .
• 39. Graphing arcsin and arccos a . rccos a . rcsin . | . . | − . 1 1 . . . . . . .
• 40. Graphing arcsin and arccos a . rccos Note (π ) cos θ = sin −θ a . rcsin 2 π =⇒ arccos x = − arcsin x 2 . | . . | So it’s not a surprise that their − . 1 1 . derivatives are opposites. . . . . . .
• 41. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y . . . . . .
• 42. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: . . . . . . .
• 43. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: x . . 1 . . . . . . .
• 44. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: x . y . = arctan x . 1 . . . . . . .
• 45. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: √ . 1 + x2 x . y . = arctan x . 1 . . . . . . .
• 46. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: 1 cos(arctan x) = √ 1 + x2 √ . 1 + x2 x . y . = arctan x . 1 . . . . . . .
• 47. The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y To simplify, look at a right triangle: 1 cos(arctan x) = √ 1 + x2 √ . 1 + x2 x . So d 1 y . = arctan x arctan(x) = . dx 1 + x2 1 . . . . . . .
• 48. Graphing arctan and its derivative y . . /2 π a . rctan . 1 x . 1 + x2 − . π/2 . . . . . .
• 49. Example √ Let f(x) = arctan x. Find f′ (x). . . . . . .
• 50. Example √ Let f(x) = arctan x. Find f′ (x). Solution d √ 1 d√ 1 1 arctan x = (√ )2 x= · √ dx 1+ x dx 1+x 2 x 1 = √ √ 2 x + 2x x . . . . . .
• 51. The derivative of arcsec 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)) . . . . . .
• 52. The derivative of arcsec 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: . . . . . . .
• 53. The derivative of arcsec 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: . . . . . . .
• 54. The derivative of arcsec 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 . . 1 . . . . . . .
• 55. The derivative of arcsec 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 . . . . . . .
• 56. The derivative of arcsec 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 . . . . . . .
• 57. The derivative of arcsec 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 So d 1 y . = arcsec x arcsec(x) = √ . dx x x2 − 1 1 . . . . . . .
• 58. Another Example Example Let f(x) = earcsec x . Find f′ (x). . . . . . .
• 59. Another Example Example Let f(x) = earcsec x . Find f′ (x). Solution 1 f′ (x) = earcsec x · √ x x2 − 1 . . . . . .
• 60. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications . . . . . .
• 61. Application Example One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter 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 batter’s angle of gaze need to change to follow the ball as it crosses home plate? . . . . . .
• 62. Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. y . 1 . 30 ft/sec . θ . . 2 . ft . . . . . .
• 63. Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · 2 2 dt dt 1 + ( y /2 ) y . 1 . 30 ft/sec . θ . . 2 . ft . . . . . .
• 64. Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · 2 2 dt dt 1 + ( y /2 ) When y = 0 and y′ = −130, y . then dθ 1 1 = · (−130) = −65 rad/sec 1 . 30 ft/sec dt y =0 1+0 2 . θ . . 2 . ft . . . . . .
• 65. Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · 2 2 dt dt 1 + ( y /2 ) When y = 0 and y′ = −130, y . then dθ 1 1 = · (−130) = −65 rad/sec 1 . 30 ft/sec dt y =0 1+0 2 . θ The human eye can only . track at 3 rad/sec! . 2 . ft . . . . . .
• 66. Recap y y′ 1 arcsin x √ 1 − x2 1 Remarkable that the arccos x − √ 1 − x2 derivatives of these 1 arctan x transcendental functions 1 + x2 are algebraic (or even 1 arccot x − rational!) 1 + x2 1 arcsec x √ x x2 − 1 1 arccsc x − √ x x2 − 1 . . . . . .