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: 4

Answer

$\dfrac{\pi}{3}$

Work Step by Step

Suppose $\theta = \sin^{-1} (\dfrac{\sqrt 3}{2})$ This gives: $\sin \theta= \dfrac{\sqrt 3}{2}$ Now, $\theta = \sin^{-1} (\dfrac{\sqrt 3}{2})=\dfrac{\pi}{3}$ Therefore, $\sin^{-1} (\dfrac{\sqrt 3}{2})=\dfrac{\pi}{3}$
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