Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises: 37

Answer

$\sin 160^{\mathrm{o}}-\sin 44^{\mathrm{o}}$

Work Step by Step

Product-to-Sum: $\sin A \displaystyle \cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$ ---------------- $2\sin 58^{\mathrm{o}}\cos 102^{\mathrm{o}}=$ $=2(\displaystyle \frac{1}{2}[\sin(58^{\mathrm{o}}+102^{\mathrm{o}})+\sin(58^{\mathrm{o}}-102^{\mathrm{o}})])$ $=\sin 160^{\mathrm{o}}+\sin(-44^{\mathrm{o}})$ $=\sin 160^{\mathrm{o}}-\sin 44^{\mathrm{o}}$
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