Answer
$ \frac{\pi}{9}$
Work Step by Step
$sin^{-1}[sin(-\frac{10\pi}{9})]=sin^{-1}[sin(2\pi-\frac{10\pi}{9})]=sin^{-1}[sin(\frac{8\pi}{9})]=sin^{-1}[sin(\frac{\pi}{9})]=\frac{\pi}{9}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 10 - Systems of Equations and Inequalities - Section 10.1 Systems of Linear Equations: Substitution and Elimination - 10.1 Assess Your Understanding - Page 737: 89
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