Answer
$-\dfrac{\pi}{6}$
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, $\sin^{-1} (-\dfrac{1}{2})=-\dfrac{\pi}{6}$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
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: 5
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