Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Trigonometry and Periodic Functions - 7.1 The Inverse Sine, Cosine, and Tangent Functions - 7.1 Assess Your Understanding - Page 451: 74

Answer

$ -\frac{\sqrt 3}{2}$

Work Step by Step

$sin^{-1}(x)=-\frac{\pi}{3} \longrightarrow x=sin(-\frac{\pi}{3})=-\frac{\sqrt 3}{2}$
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