Answer
The required solution is $\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]=\sin \alpha \sin \beta $.
Work Step by Step
We know that the sum and difference formulas are $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ and $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $.
Add the left and right sides of these two identities:
$\begin{align}
& \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\
& =2\sin \alpha \cos \beta
\end{align}$
Now, multiply both sides by $\frac{1}{2}$.
$\begin{align}
& \frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]=\frac{1}{2}\cdot 2\sin \alpha \cos \beta \\
& =\sin \alpha \cos \beta
\end{align}$
Therefore, the product-to-sum formula is derived, which is $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$.
Hence, by adding the left and right sides of the identities, the product-to-sum formula is derived for $\sin \alpha \cos \beta $.
