Answer
The required solution is $\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right) \right]=\cos \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 $.
Subtract the left and right sides of these two identities:
$\begin{align}
& \sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta -\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right) \\
& =\sin \alpha \cos \beta +\cos \alpha \sin \beta -\sin \alpha \cos \beta +\cos \alpha \sin \beta \\
& =2\cos \alpha \sin \beta
\end{align}$
Now, multiply both sides by $\frac{1}{2}$.
$\begin{align}
& \frac{1}{2}\left[ \sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right) \right]=\frac{1}{2}\cdot 2\cos \alpha \sin \beta \\
& =\cos \alpha \sin \beta
\end{align}$
Thus, the product-to-sum formula is derived, which is $\cos \alpha \sin \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right) \right]$.
