Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises - Page 679: 52

Answer

$-\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $

Work Step by Step

$\sin \left( -\dfrac {5\pi }{12}\right) =-\sin \left( \dfrac {5\pi }{12}\right) =-\sin \left( \dfrac {\pi }{4}+\dfrac {\pi }{6}\right) =-\left( \sin \dfrac {\pi }{4}\cos \dfrac {\pi }{6}+\cos \dfrac {\pi }{4}\sin \dfrac {\pi }{6}\right) =$ $=-(\dfrac {\sqrt {2}}{2}\dfrac {\sqrt {3}}{2}+\dfrac {\sqrt {2}}{2}\dfrac {1}{2})=-\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $
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