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Lesson 15: Inverse Trigonometric Functions
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Lesson 15: Inverse Trigonometric Functions

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  • 1. Section 3.5 Inverse Trigonometric Functions V63.0121.006/016, Calculus I March 11, 2010 Announcements Exams returned in recitation There is WebAssign due Tuesday March 23 and written HW due Thursday March 25 . . . . . .
  • 2. Announcements Exams returned in recitation There is WebAssign due Tuesday March 23 and written HW due Thursday March 25 next quiz is Friday April 2 . . . . . .
  • 3. What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. . . . . . .
  • 4. What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function 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 . . . . . .
  • 5. 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. . . . . . .
  • 6. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications . . . . . .
  • 7. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . x . π π s . in − . . 2 2 . . . . . .
  • 8. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . . x . π π s . in − . . . 2 2 . . . . . .
  • 9. 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 . . . . . .
  • 10. 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 . . . . . .
  • 11. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . c . os . . x . 0 . . π . . . . . .
  • 12. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . . c . os . . x . 0 . . π . . . . . . .
  • 13. arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . y . =x . c . os . . x . 0 . . π . . . . . . .
  • 14. 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] . . . . . .
  • 15. 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 . . . . . .
  • 16. 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 . . . . . .
  • 17. 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 . . . . . .
  • 18. 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 . . . . . .
  • 19. 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 . . . . . .
  • 20. 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 . . . . . .
  • 21. 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 . . . . . .
  • 22. 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 . . . . . .
  • 23. 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 . . . . . .
  • 24. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 . . . . . .
  • 25. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 Solution π 6 . . . . . .
  • 26. What is arctan(−1)? . 3 . π/4 . . . . − . π/4 . . . . . .
  • 27. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 √ 2 s . in(3π/4) = 2 . √ . 2 . os(3π/4) = − c 2 . − . π/4 . . . . . .
  • 28. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 √ But, the range of arctan ( π π) 2 s . in(3π/4) = is − , 2 2 2 . √ . 2 . os(3π/4) = − c 2 . − . π/4 . . . . . .
  • 29. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 But, the range of arctan ( π π) √ is − , 2 2 2 c . os(π/4) = . 2 Another angle whose . π tangent is −1 is − , and √ 4 2 this is in the right range. . in(π/4) = − s 2 . − . π/4 . . . . . .
  • 30. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 But, the range of arctan ( π π) √ is − , 2 2 2 c . os(π/4) = . 2 Another angle whose . π tangent is −1 is − , and √ 4 2 this is in the right range. . in(π/4) = − s π 2 So arctan(−1) = − 4 . − . π/4 . . . . . .
  • 31. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 Solution π 6 π − 4 . . . . . .
  • 32. Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 Solution π 6 π − 4 3π 4 . . . . . .
  • 33. 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, and never for (sin x)−1 . 1 I use csc x for and arcsin x for the inverse of sin x. sin x . . . . . .
  • 34. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications . . . . . .
  • 35. 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)) . . . . . .
  • 36. 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)) . . . . . .
  • 37. 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) . . . . . .
  • 38. 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: . . . . . . .
  • 39. 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 . . . . . . . .
  • 40. 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 . . . . . . .
  • 41. 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 . . . . . .
  • 42. 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 . . . . . .
  • 43. 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 . . . . . .
  • 44. Graphing arcsin and its derivative 1 .√ 1 − x2 The domain of f is [−1, 1], but the domain . . rcsin a of f′ is (−1, 1) lim f′ (x) = +∞ x →1 − lim f′ (x) = +∞ . | . . | x→−1+ − . 1 1 . . . . . . . .
  • 45. 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) . . . . . .
  • 46. 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 . . . . . . .
  • 47. Graphing arcsin and arccos . . rccos a . . rcsin a . | . |. . − . 1 1 . . . . . . . .
  • 48. Graphing arcsin and arccos . . rccos a Note (π ) cos θ = sin −θ . . rcsin a 2 π =⇒ arccos x = − arcsin x 2 . | . |. . So it’s not a surprise that their − . 1 1 . derivatives are opposites. . . . . . . .
  • 49. 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 . . . . . .
  • 50. 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: . . . . . . .
  • 51. 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 . . . . . . .
  • 52. 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 . . . . . . .
  • 53. 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 . . . . . . .
  • 54. 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 . . . . . . .
  • 55. 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 . . . . . . .
  • 56. Graphing arctan and its derivative y . . /2 π a . rctan 1 . 1 + x2 . x . − . π/2 The domain of f and f′ are both (−∞, ∞) Because of the horizontal asymptotes, lim f′ (x) = 0 x→±∞ . . . . . .
  • 57. Example √ Let f(x) = arctan x. Find f′ (x). . . . . . .
  • 58. 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 . . . . . .
  • 59. The derivative of arcsec Try this first. . . . . . .
  • 60. 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)) . . . . . .
  • 61. 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: . . . . . . .
  • 62. 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: . . . . . . .
  • 63. 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 . . 1 . . . . . . .
  • 64. 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 . . . . . . .
  • 65. 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 . . . . . . .
  • 66. 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 So d 1 y . = arcsec x arcsec(x) = √ . dx x x2 − 1 1 . . . . . . .
  • 67. Another Example Example Let f(x) = earcsec x . Find f′ (x). . . . . . .
  • 68. Another Example Example Let f(x) = earcsec x . Find f′ (x). Solution 1 f′ (x) = earcsec x · √ x x2 − 1 . . . . . .
  • 69. Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications . . . . . .
  • 70. 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? . . . . . .
  • 71. 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 . . . . . .
  • 72. 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 . . . . . .
  • 73. 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 . . . . . .
  • 74. 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 . . . . . .
  • 75. Recap y y′ 1 arcsin x √ 1 − x2 1 arccos x − √ Remarkable that the 1 − x2 derivatives of these 1 transcendental functions arctan x 1 + x2 are algebraic (or even 1 rational!) arccot x − 1 + x2 1 arcsec x √ x x2 − 1 1 arccsc x − √ x x2 − 1 . . . . . .