Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 40


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Work Step by Step

Let us consider the left side of the given expression: $\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$ By using the identities of trigonometry, $\cos \,\left( \alpha +\beta \right)=\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta $ $\cos \,\left( \alpha -\beta \right)=\cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta $ , the above expression can be further simplified as: $\begin{align} & \cos \,\left( \alpha +\beta \right)+\cos \,\left( \alpha -\beta \right)=\left( \cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta \right)+\left( \cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta \right) \\ & =\cos \,\alpha \,\cos \beta +\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta +\sin \,\alpha \,\sin \,\beta \\ & =2\cos \,\alpha \,\cos \beta \end{align}$ Hence, the left side of the given expression is equal to the right side, which is $\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)=2\cos \,\alpha \cos \,\beta $.
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