## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Mid-Chapter Check Point - Page 683: 9

#### Answer

See the explanation below.

#### Work Step by Step

The expression on the right side $\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$ can be further simplified by using the sum formula: $\sin \left( C+D \right)=\sin C\cos D+\cos C\sin D$ Thus, after applying the sum formula, the expression is given below: \begin{align} & \frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]=\frac{1}{2}\left[ \sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta \right] \\ & =\frac{1}{2}\left[ 2\sin \alpha \cos \beta \right] \\ & =\sin \alpha \cos \beta \end{align} Hence, the left side is equal to the right side $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$.

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