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: 16


The required solution is $-2\sin \frac{x}{2}\cos \frac{3x}{2}$.

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 $x$ and the value of $\beta $ is $2x$. Now, the expression can be evaluated as provided below: $\begin{align} & \sin x-\sin 2x=2\sin \frac{x-2x}{2}\cos \frac{x+2x}{2} \\ & =2\sin \frac{-x}{2}\cos \frac{3x}{2} \end{align}$ Then, applying the even-odd identity, which is $sin(-x)=-\sin x$, the expression can be further evaluated as given below: $2\sin \frac{-x}{2}\cos \frac{3x}{2}==-2\sin \frac{x}{2}\cos \frac{3x}{2}$ Hence, the provided expression can be written as $-2\sin \frac{x}{2}\cos \frac{3x}{2}$. So, it is not possible to find the exact value.
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