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 - Page 693: 46

Answer

$2\cos 98.5^{o}\cdot\sin 3.5^{o}$

Work Step by Step

Sum-to-Product: $\displaystyle \sin A-\sin B=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$ ---------- $\displaystyle \frac{A+B}{2}=\frac{102^{o}+95^{o}}{2}=98.5^{o}$ $\displaystyle \frac{A-B}{2}=\frac{102^{o}-95^{o}}{2}=3.5^{o}$ $\sin 102^{o}-\sin95^{o}=2\cos 98.5^{o}\cdot\sin 3.5^{o}$
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