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: 11


The required solution is $2\sin 2x\cos 5x$.

Work Step by Step

One of the sum-to-product formula is $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is $7x$ and the value of $\beta $ is $3x$. Thus, the expression can be evaluated as provided below: $\begin{align} & \sin 7x-\sin 3x=2\sin \frac{7x-3x}{2}\cos \frac{7x+3x}{2} \\ & =2\sin \frac{4x}{2}\cos \frac{10x}{2} \\ & =2\sin 2x\cos 5x \end{align}$ Hence, the given expression can be written as $2\sin 2x\cos 5x$. So, it is not possible to find the exact value.
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