## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 51

#### Answer

See the full explanation below.

#### Work Step by Step

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

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