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 626: 35

Answer

$\dfrac{\pi}{6}$

Work Step by Step

Suppose $\theta =\sin (\dfrac{5\pi}{6})$ This gives: $\sin^{-1} \theta=\sin^{-1}[\sin(\dfrac{5\pi}{6})]$ The inverse property states that $\sin^{-1}(\sin a)=a $ satisfies for every $ a $ in the range $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ . Therefore, we have $\sin^{-1}[\sin(\dfrac{5\pi}{6})]=\dfrac{\pi}{6}$
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