Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 689: 14

Answer

The required solution is $-2\sin 8x\sin x$.

Work Step by Step

One of the sum-to-product formula is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is $9x$ and the value of $\beta $ is $7x$. Thus, the expression can be evaluated as provided below: $\begin{align} & \cos 9x-\cos 7x=-2\sin \frac{9x+7x}{2}\sin \frac{9x-7x}{2} \\ & =-2\sin \frac{16x}{2}\sin \frac{2x}{2} \\ & =-2\sin 8x\sin x \end{align}$ Hence, the given expression can be written as $-2\sin 8x\sin x$. So, it is not possible to find the exact value.
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