Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 61


$$6\sin4x\sin5x=3\cos x-3\cos9x$$

Work Step by Step

$$A=6\sin4x\sin5x$$ One thing you can do here is to switch the position of $\sin 4x$ and $\sin 5x$ so that later we might not encounter negative values inside trigonometric functions. What I mean is this $$A=6\sin5x\sin4x$$ The product-to-sum identity that will be applied here is $$\sin X\sin Y=\frac{1}{2}[\cos(X-Y)-\cos(X+Y)]$$ Therefore, A would be $$A=6\times\frac{1}{2}[\cos(5x-4x)-\cos(5x+4x)]$$ $$A=3(\cos x-\cos 9x)$$ $$A=3\cos x-3\cos9x$$
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