Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Mid-Chapter Check Point - Page 577: 21


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

Work Step by Step

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