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 $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$.
