Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Quiz: 8

Answer

$\sin \left( \dfrac {\pi }{3}+\theta \right) -\sin \left( \dfrac {\pi }{3}-\theta \right) =\left( \sin \dfrac {\pi }{3}\cos \theta +\cos \dfrac {\pi }{3}\sin \theta \right) -\left( \sin \dfrac {\pi }{3}\cos \theta -\cos \dfrac {\pi }{3}\sin \theta \right)=2\cos \dfrac {\pi }{3}\sin \theta =2\times \left( \dfrac {1}{2}\right) \times \sin \theta =\sin \theta $

Work Step by Step

$\sin \left( \dfrac {\pi }{3}+\theta \right) -\sin \left( \dfrac {\pi }{3}-\theta \right) =\left( \sin \dfrac {\pi }{3}\cos \theta +\cos \dfrac {\pi }{3}\sin \theta \right) -\left( \sin \dfrac {\pi }{3}\cos \theta -\cos \dfrac {\pi }{3}\sin \theta \right)=2\cos \dfrac {\pi }{3}\sin \theta =2\times \left( \dfrac {1}{2}\right) \times \sin \theta =\sin \theta $
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