Calculus: Early Transcendentals (2nd Edition)

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

Chapter 1 - Functions - Review Exercises - Page 53: 51

Answer

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

Work Step by Step

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