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# Lesson 16: Inverse Trigonometric Functions (Section 041 handout)

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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 (Section 041 handout)

1. 1. Section 3.5 Inverse Trigonometric Functions V63.0121.041, Calculus I New York University November 1, 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 31 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 3 / 31 Notes Notes Notes 1 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
2. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 31 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 5 / 31 Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 6 / 31 Notes Notes Notes 2 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
3. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 31 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] V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 31 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 31 Notes Notes Notes 3 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
4. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 11 / 31 Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos − √ 2 2 Solution π 6 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 31 Notes Notes Notes 4 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
5. 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 15 / 31 Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 16 / 31 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)) Upshot: Many times the derivative of f −1 (x) can be found by implicit diﬀerentiation and the derivative of f : V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 31 Notes Notes Notes 5 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
6. 6. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 31 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 31 Composing with arcsin Example Let f (x) = arcsin(x3 + 1). Find f (x). Solution V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 20 / 31 Notes Notes Notes 6 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
7. 7. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 31 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. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 31 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 31 Notes Notes Notes 7 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
8. 8. 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 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 31 Composing with arctan Example Let f (x) = arctan √ x. Find f (x). Solution V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 25 / 31 Derivation: The derivative of arcsec Try this ﬁrst. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: tan(arcsec x) = √ x2 − 1 1 So d dx arcsec(x) = 1 x √ x2 − 1 x 1 y = arcsec x x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 31 Notes Notes Notes 8 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
9. 9. Another Example Example Let f (x) = earcsec 3x . Find f (x). Solution f (x) = earcsec 3x · 1 3x (3x)2 − 1 · 3 = 3earcsec 3x 3x √ 9x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 31 Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 31 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? V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 29 / 31 Notes Notes Notes 9 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010
10. 10. Solution V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 31 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!) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 31 Notes Notes Notes 10 Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010