Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 63


$cos(arcsin~2x)=\sqrt {1-4x^2}$

Work Step by Step

Let $u=arcsin~2x~~$ (Range: $-\frac{\pi}{2}\leq u\leq\frac{\pi}{2}$). Then: $2x=sin~u~~$ $cos^2u+sin^2u=1$ $cos^2u=1-sin^2u$ $cos^2u=1-(2x)^2$ $cos^2u=1-4x^2$ $cos~u=+\sqrt {1-4x^2}$ $cos(arcsin~2x)=\sqrt {1-4x^2}$
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