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 628: 121

Answer

$ x=\sin (\dfrac{\pi}{8})$

Work Step by Step

We have $2 \sin^{-1} (x)=\dfrac{\pi}{4}$ This gives: $ \sin^{-1} (x)=\dfrac{\pi}{8}$ Therefore, we have $ x=\sin (\dfrac{\pi}{8})$
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