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: 1



Work Step by Step

Suppose $\theta = \sin^{-1} (\dfrac{1}{2})$ This gives: $\sin \theta= \dfrac{1}{2}$ Now, $\theta = \sin^{-1} (\dfrac{1}{2})=\dfrac{\pi}{6}$ Therefore, $\sin^{-1} (\dfrac{1}{2})=\dfrac{\pi}{6}$
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