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- 1. Section 3.5 Inverse Trigonometric Functions V63.0121.021, Calculus I New York University November 2, 2010 Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 2 / 32 Objectives Know the deﬁnitions, domains, ranges, and other properties of the inverse trignometric functions: arcsin, arccos, arctan, arcsec, arccsc, arccot. Know the derivatives of the inverse trignometric functions. V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 32 Notes Notes Notes 1 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 2. 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 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 32 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. V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 5 / 32 Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 32 Notes Notes Notes 2 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 3. arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. x y sin − π 2 π 2 y = x arcsin The domain of arcsin is [−1, 1] The range of arcsin is − π 2 , π 2 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 32 arccos Arccos is the inverse of the cosine function after restriction to [0, π] x y cos 0 π y = x arccos The domain of arccos is [−1, 1] The range of arccos is [0, π] V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 8 / 32 arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. x y tan − 3π 2 − π 2 π 2 3π 2 y = x arctan − π 2 π 2 The domain of arctan is (−∞, ∞) The range of arctan is − π 2 , π 2 lim x→∞ arctan x = π 2 , lim x→−∞ arctan x = − π 2 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 9 / 32 Notes Notes Notes 3 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 4. arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. x y sec − 3π 2 − π 2 π 2 3π 2 y = x π 2 3π 2 The domain of arcsec is (−∞, −1] ∪ [1, ∞) The range of arcsec is 0, π 2 ∪ π 2 , π lim x→∞ arcsec x = π 2 , lim x→−∞ arcsec x = 3π 2 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 10 / 32 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 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 11 / 32 Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos − √ 2 2 Solution π 6 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 12 / 32 Notes Notes Notes 4 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 5. 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 . I use csc x for 1 sin x and arcsin x for the inverse of sin x. V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 32 Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 32 The Inverse Function Theorem 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 (f −1 ) (b) = 1 f (f −1(b)) In Leibniz notation we have dx dy = 1 dy/dx Upshot: Many times the derivative of f −1 (x) can be found by implicit diﬀerentiation and the derivative of f : V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 32 Notes Notes Notes 5 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 6. Illustrating the Inverse Function Theorem Example Use the inverse function theorem to ﬁnd the derivative of the square root function. Solution (Newtonian notation) Let f (x) = x2 so that f −1 (y) = √ y. Then f (u) = 2u so for any b > 0 we have (f −1 ) (b) = 1 2 √ b Solution (Leibniz notation) If the original function is y = x2 , then the inverse function is deﬁned by x = y2 . Diﬀerentiate implicitly: 1 = 2y dy dx =⇒ dy dx = 1 2 √ x V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 32 Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = 1 − x2 So d dx arcsin(x) = 1 √ 1 − x2 1 x y = arcsin x 1 − x2 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 19 / 32 Graphing arcsin and its derivative The domain of f is [−1, 1], but the domain of f is (−1, 1) lim x→1− f (x) = +∞ lim x→−1+ f (x) = +∞ | −1 | 1 arcsin 1 √ 1 − x2 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 20 / 32 Notes Notes Notes 6 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 7. Composing with arcsin Example Let f (x) = arcsin(x3 + 1). Find f (x). Solution V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 32 Derivation: The derivative of arccos Let y = arccos x, so x = cos y. Then − sin y dy dx = 1 =⇒ dy dx = 1 − sin y = 1 − sin(arccos x) To simplify, look at a right triangle: sin(arccos x) = 1 − x2 So d dx arccos(x) = − 1 √ 1 − x2 1 1 − x2 x y = arccos x V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 22 / 32 Graphing arcsin and arccos | −1 | 1 arcsin arccos Note cos θ = sin π 2 − θ =⇒ arccos x = π 2 − arcsin x So it’s not a surprise that their derivatives are opposites. V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 23 / 32 Notes Notes Notes 7 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 8. Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: cos(arctan x) = 1 √ 1 + x2 So d dx arctan(x) = 1 1 + x2 x 1 y = arctan x 1 + x2 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 24 / 32 Graphing arctan and its derivative x y arctan 1 1 + x2 π/2 −π/2 The domain of f and f are both (−∞, ∞) Because of the horizontal asymptotes, lim x→±∞ f (x) = 0 V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 25 / 32 Composing with arctan Example Let f (x) = arctan √ x. Find f (x). Solution V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 32 Notes Notes Notes 8 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 9. Derivation: The derivative of arcsec Try this ﬁrst. V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 32 Another Example Example Let f (x) = earcsec 3x . Find f (x). Solution V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 28 / 32 Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 29 / 32 Notes Notes Notes 9 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010
- 10. 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? V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 30 / 32 Solution V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 31 / 32 Summary y y arcsin x 1 √ 1 − x2 arccos x − 1 √ 1 − x2 arctan x 1 1 + x2 arccot x − 1 1 + x2 arcsec x 1 x √ x2 − 1 arccsc x − 1 x √ x2 − 1 Remarkable that the derivatives of these transcendental functions are algebraic (or even rational!) V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 32 Notes Notes Notes 10 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.021, Calculus I November 2, 2010

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