Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 627: 66

Answer

$\sqrt {1-4x^2}$

Work Step by Step

Suppose $\theta =\cos^{-1} (2x)$ This gives: $\cos \theta=2x $ Since, $ r=\sqrt {x^2+y^2}$ This implies that $ a=\sqrt {(\dfrac{1}{2})^2-x^2}=\dfrac{\sqrt {1-4x^2}}{2}$ Therefore, we have $\sin [\cos^{-1} (2x)]=\sqrt {1-4x^2}$
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