Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 61

Answer

$$2x\sqrt {1 - {x^2}} $$

Work Step by Step

$$\eqalign{ & \sin \left( {2{{\cos }^{ - 1}}x} \right) \cr & {\text{Use the Hint }}\sin 2\theta = 2\sin \theta \cos \theta \cr & \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \left( {{{\cos }^{ - 1}}x} \right)\cos \left( {{{\cos }^{ - 1}}x} \right) \cr & \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \left( {{{\cos }^{ - 1}}x} \right)x \cr & {\text{From the triangle shown below}} \cr & \cos \theta = \frac{x}{1} \cr & \cos \theta = x \cr & \theta = {\cos ^{ - 1}}x,{\text{ then}} \cr & \sin \left( {{{\cos }^{ - 1}}x} \right) = \sin \theta \cr & {\text{From the triangle we obtain }}\cos \theta \cr & \sin \left( {{{\cos }^{ - 1}}x} \right) = \frac{{\sqrt {1 - {x^2}} }}{1} \cr & \sin \left( {{{\cos }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \cr & {\text{Therefore,}} \cr & \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \left( {{{\cos }^{ - 1}}x} \right)x \cr & \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2x\sqrt {1 - {x^2}} \cr} $$
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