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: 62



Work Step by Step

$$A=8\sin7x\sin9x$$ Again, we switch the position of $\sin 7x$ and $\sin 9x$ to avoid later complexity. $$A=8\sin9x\sin7x$$ 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 $$A=8\times\frac{1}{2}[\cos(9x-7x)-\cos(9x+7x)]$$ $$A=4(\cos2x-\cos16x)$$ $$A=4\cos2x-4\cos16x$$
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