Answer
See the explanation below.
Work Step by Step
Let us consider the right-hand side of the provided expression:
$2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$
One of the product-to-sum formulas is $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$. Therefore, here $\alpha $ is $\frac{\alpha -\beta }{2}$ and $\beta $ is $\frac{\alpha +\beta }{2}$.
Now, the expression can be simplified as shown below:
$\begin{align}
& 2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}=2\cdot \frac{1}{2}\left[ \sin \left( \frac{\alpha -\beta }{2}+\frac{\alpha +\beta }{2} \right)+\sin \left( \frac{\alpha -\beta }{2}-\frac{\alpha +\beta }{2} \right) \right] \\
& =2\cdot \frac{1}{2}\left[ \sin \left( \frac{\alpha -\beta +\alpha +\beta }{2} \right)+\sin \left( \frac{\alpha -\beta -\alpha -\beta }{2} \right) \right] \\
& =\sin \left( \frac{2\alpha }{2} \right)+\sin \left( \frac{-2\beta }{2} \right) \\
& =\sin \alpha +\sin \left( -\beta \right)
\end{align}$
One of the even-odd identities of trigonometry is $\sin \left( -x \right)=-\sin x$. Therefore, applying this identity:
$\sin \alpha +\sin \left( -\beta \right)=\sin \alpha -\sin \beta $
Thus, the right-hand side of the expression is equal to the left-hand side:
$\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$.
