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: 39

Answer

$\displaystyle \sin\frac{\pi}{2}-\sin\frac{\pi}{6}$

Work Step by Step

Product-to-Sum: $\sin A \displaystyle \cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$ ---------------- $2\displaystyle \sin\frac{\pi}{6}\cos\frac{\pi}{3}=$ $=2(\displaystyle \frac{1}{2}[\sin(\frac{\pi}{6}+\frac{\pi}{3})+\sin(\frac{\pi}{6}-\frac{\pi}{3})])$ $=\displaystyle \sin\frac{3\pi}{6}+\sin(-\frac{\pi}{6})$ $=\displaystyle \sin\frac{\pi}{2}-\sin\frac{\pi}{6}$
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