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 691: 59


See the explanation below.

Work Step by Step

Let us consider the right-hand side of the given expression: $2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ One of the product-to-sum formulas is $\cos \alpha \cos \beta =\frac{1}{2}\left[ \cos \left( \alpha -\beta \right)+\cos \left( \alpha +\beta \right) \right]$. Therefore, here $\alpha $ is $\frac{\alpha +\beta }{2}$ and $\beta $ is $\frac{\alpha -\beta }{2}$. Now, the expression can be calculated as shown below: $\begin{align} & 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=2\cdot \frac{1}{2}\left[ \cos \left( \frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2} \right)+\cos \left( \frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2} \right) \right] \\ & =2\cdot \frac{1}{2}\left[ \cos \left( \frac{\alpha +\beta -\alpha +\beta }{2} \right)+\cos \left( \frac{\alpha +\beta +\alpha -\beta }{2} \right) \right] \\ & =\cos \left( \frac{2\beta }{2} \right)+\cos \left( \frac{2\alpha }{2} \right) \\ & =\cos \beta +\cos \alpha \end{align}$ Then, rearrange the terms: $\cos \beta +\cos \alpha =\cos \alpha +\cos \beta $ Thus, the right-hand side of the expression is equal to the left-hand side: $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$.
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