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: 49

Answer

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

Work Step by Step

$\sin \dfrac {7\pi }{12}=\sin \left( \dfrac {\pi }{3}+\dfrac {\pi }{4}\right) =\sin \dfrac {\pi }{3}\cos \dfrac {\pi }{4}+\cos \dfrac {\pi }{3}\sin \dfrac {\pi }{4}=\dfrac {\sqrt {3}}{2}\dfrac {\sqrt {2}}{2}+\dfrac {1}{2}\dfrac {\sqrt {2}}{2}=\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $
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