Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.2 Algebra Techniques for Finding Limits - 13.2 Assess Your Understanding - Page 903: 59

Answer

$\frac{\pi}{3}$

Work Step by Step

As $sin(\frac{\pi}{3})=\frac{\sqrt 3}{2}$, we have $sin^{-1}(\frac{\sqrt 3}{2})=\frac{\pi}{3}$
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