Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Review Exercises - Page 708: 43



Work Step by Step

Using the Product-to-Sum Formula $sin(u)sin(v)=\frac{1}{2}[cos(u-v)-cos(u+v)]$, we have $sin(6x)sin(4x)=\frac{1}{2}[cos(6x-4x)-cos(6x+4x)]=\frac{1}{2}[cos(2x)-cos(10x)]$
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