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: 39


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Work Step by Step

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