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: 58

Answer

$$\frac{{\sqrt {9 - {x^2}} }}{3}$$

Work Step by Step

$$\eqalign{ & \cos \left( {{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right) \cr & {\text{From the triangle shown below}} \cr & \sin \theta = \frac{x}{3} \cr & \theta = {\sin ^{ - 1}}\left( {\frac{x}{3}} \right),{\text{ then}} \cr & \cos \left( {{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right) = \cos \theta \cr & {\text{From the triangle we obtain }}\cos \theta \cr & \cos \left( {{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right) = \frac{{\sqrt {9 - {x^2}} }}{3} \cr} $$
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