§7.6–The inverse trigonometric functions Mark R. Woodard Furman U Fall 2010 Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 1 / 11 Outline 1 The inverse sine function 2 The inverse cosine function 3 The inverse tangent function 4 The other inverse trig functions 5 Miscellaneous problems Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 2 / 11 The inverse sine function Definition Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 3 / 11 The inverse sine function Definition The sine function is one-to-one on [−π/2, π/2] and has range [−1, 1] on this domain. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 3 / 11 The inverse sine function Definition The sine function is one-to-one on [−π/2, π/2] and has range [−1, 1] on this domain. We define sin−1 to be the inverse of sine on this domain. It follows that sin−1 has domain [−1, 1] and range [−π/2, π/2]. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 3 / 11 The inverse sine function Definition The sine function is one-to-one on [−π/2, π/2] and has range [−1, 1] on this domain. We define sin−1 to be the inverse of sine on this domain. It follows that sin−1 has domain [−1, 1] and range [−π/2, π/2]. Cancellation equations Because of these restrictions, we must be a little careful with the inverse relationships: sin sin−1 (x) = x − 1 ≤ x ≤ 1 sin−1 sin(x) = x − π/2 ≤ x ≤ π/2 Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 3 / 11 The inverse sine function Problem Evaluate the following: Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 11 The inverse sine function Problem Evaluate the following: sin sin−1 (.3) Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 11 The inverse sine function Problem Evaluate the following: sin sin−1 (.3) sin−1 sin(14π/3) . Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 11 The inverse sine function Problem Evaluate the following: sin sin−1 (.3) sin−1 sin(14π/3) . Problem Find cos 2 sin−1 (1/4) . Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 11 The inverse sine function Problem Evaluate the following: sin sin−1 (.3) sin−1 sin(14π/3) . Problem Find cos 2 sin−1 (1/4) . Problem Evaluate cos sin−1 (x) . Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 11 The inverse sine function Problem Evaluate the following: sin sin−1 (.3) sin−1 sin(14π/3) . Problem Find cos 2 sin−1 (1/4) . Problem Evaluate cos sin−1 (x) . Solution √ Let θ = sin−1 (x). Then cos(θ) = ± 1 − x 2 . Since cosine is nonnegative √ −1 for θ ∈ [−π/2, π/2], it follows that cos(sin (x)) = 1 − x 2 . Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 11 The inverse sine function Theorem Dx sin−1 (x) = √ Mark R. Woodard (Furman U) 1 . 1 − x2 §7.6–The inverse trigonometric functions Fall 2010 5 / 11 The inverse sine function Theorem Dx sin−1 (x) = √ 1 . 1 − x2 Proof. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 5 / 11 The inverse sine function Theorem Dx sin−1 (x) = √ 1 . 1 − x2 Proof. Let y = sin−1 (x). Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 5 / 11 The inverse sine function Theorem Dx sin−1 (x) = √ 1 . 1 − x2 Proof. Let y = sin−1 (x). Then sin(y ) = x. By the chain rule, cos(y )y 0 = 1 hence y 0 = Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions 1 cos(y ) . Fall 2010 5 / 11 The inverse sine function Theorem Dx sin−1 (x) = √ 1 . 1 − x2 Proof. Let y = sin−1 (x). Then sin(y ) = x. By the chain rule, cos(y )y 0 = 1 hence y 0 = √ However, cos(y ) = 1 − x 2 , which gives the desired result. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions 1 cos(y ) . Fall 2010 5 / 11 The inverse sine function Theorem Dx sin−1 (x) = √ 1 . 1 − x2 Proof. Let y = sin−1 (x). Then sin(y ) = x. By the chain rule, cos(y )y 0 = 1 hence y 0 = √ However, cos(y ) = 1 − x 2 , which gives the desired result. 1 cos(y ) . Problem Find the derivative of y = x sin−1 (x 2 ). Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 5 / 11 The inverse cosine function Definition Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 6 / 11 The inverse cosine function Definition The cosine function is one-to-one on the interval [0, π] and has range [−1, 1] on that domain. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 6 / 11 The inverse cosine function Definition The cosine function is one-to-one on the interval [0, π] and has range [−1, 1] on that domain. Let cos−1 denote the inverse of the cosine function restricted to the domain [0, π]. Thus the domain of cos−1 is [−1, 1] and its range is [0, π]. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 6 / 11 The inverse cosine function Definition The cosine function is one-to-one on the interval [0, π] and has range [−1, 1] on that domain. Let cos−1 denote the inverse of the cosine function restricted to the domain [0, π]. Thus the domain of cos−1 is [−1, 1] and its range is [0, π]. The cancellation equations cos cos−1 (x) = x cos−1 cos(x) = x Mark R. Woodard (Furman U) −1≤x ≤1 0≤x ≤π §7.6–The inverse trigonometric functions Fall 2010 6 / 11 The inverse cosine function Problem Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 7 / 11 The inverse cosine function Problem Evaluate cos−1 cos(14π/3) Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 7 / 11 The inverse cosine function Problem Evaluate cos−1 cos(14π/3) √ Show that sin cos−1 (x) = 1 − x 2 Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 7 / 11 The inverse cosine function Problem Evaluate cos−1 cos(14π/3) √ Show that sin cos−1 (x) = 1 − x 2 Theorem Dx cos−1 (x) = − √ Mark R. Woodard (Furman U) 1 1 − x2 §7.6–The inverse trigonometric functions Fall 2010 7 / 11 The inverse cosine function Problem Evaluate cos−1 cos(14π/3) √ Show that sin cos−1 (x) = 1 − x 2 Theorem Dx cos−1 (x) = − √ 1 1 − x2 Proof. Let y = cos−1 (x). Then cos(y ) = x and, by taking derivatives, √ 0 = 1; hence y 0 = −1/ sin(y ). Since sin(y ) = − sin(y )y √ 1 − x 2, 0 2 y = −1/ 1 − x , as was to be shown. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 7 / 11 The inverse tangent function Definition Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 8 / 11 The inverse tangent function Definition The tangent is one-to-one on the interval (−π/2, π/2) and has range (−∞, ∞) on this domain. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 8 / 11 The inverse tangent function Definition The tangent is one-to-one on the interval (−π/2, π/2) and has range (−∞, ∞) on this domain. Let tan−1 be the inverse of the tangent function on this restricted domain. Thus the domain of tan−1 is (−∞, ∞) and its range is (−π/2, π/2). Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 8 / 11 The inverse tangent function Definition The tangent is one-to-one on the interval (−π/2, π/2) and has range (−∞, ∞) on this domain. Let tan−1 be the inverse of the tangent function on this restricted domain. Thus the domain of tan−1 is (−∞, ∞) and its range is (−π/2, π/2). The cancellation equations tan tan−1 (x) = x tan−1 tan(x) = x Mark R. Woodard (Furman U) −∞<x <∞ − π/2 < x < π/2. §7.6–The inverse trigonometric functions Fall 2010 8 / 11 The inverse tangent function Problem Show that sec2 tan−1 (x) = 1 + x 2 Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 9 / 11 The inverse tangent function Problem Show that sec2 tan−1 (x) = 1 + x 2 Theorem Dx tan−1 (x) = Mark R. Woodard (Furman U) 1 , 1 + x2 x ∈ R. §7.6–The inverse trigonometric functions Fall 2010 9 / 11 The inverse tangent function Problem Show that sec2 tan−1 (x) = 1 + x 2 Theorem Dx tan−1 (x) = 1 , 1 + x2 x ∈ R. Proof. Let y = tan−1 (x). Then tan(y ) = x and, by the chain rule, sec2 (y )y 0 = 1. Thus y 0 = 1/ sec2 (y ) = 1/(1 + x 2 ), as was to be shown. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 9 / 11 The other inverse trig functions The other inverse trig functions The remaining inverse trig functions (cot−1 , sec−1 , and csc−1 ) are defined in similar ways. See the text for a summary of these functions, their definitions, and derivatives. Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 10 / 11 Miscellaneous problems Problem Find y 0 in each case: Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11 Miscellaneous problems Problem Find y 0 in each case: y = tan−1 (e x ) Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11 Miscellaneous problems Problem Find y 0 in each case: y = tan−1 (e x ) √ y = 1 − x 2 sin−1 (x) Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11 Miscellaneous problems Problem Find y 0 in each case: y = tan−1 (e x ) √ y = 1 − x 2 sin−1 (x) y = sin−1 (x) + cos−1 (x) Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11
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