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: 38

Answer

$\sin 225^{o}+\sin 55^{o}$

Work Step by Step

Product-to-Sum: $\displaystyle \cos A\sin B=\frac{1}{2}[\sin(A+B)-\sin(A-B)]$ ---------------- $2\cos 85^{o}\sin 140^{o}=$ $=2(\displaystyle \frac{1}{2}[\sin(85^{o}+140^{o})-\sin(85^{o}-140^{o})])$ $=\sin 225^{o}-\sin(-55^{o})$ $=\sin 225^{o}+\sin 55^{o}$
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