Answer
See the explanation below.
Work Step by Step
Let us consider the right-hand side of the given expression:
$2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$
One of the product-to-sum formulas is $\cos \alpha \cos \beta =\frac{1}{2}\left[ \cos \left( \alpha -\beta \right)+\cos \left( \alpha +\beta \right) \right]$. Therefore, here $\alpha $ is $\frac{\alpha +\beta }{2}$ and $\beta $ is $\frac{\alpha -\beta }{2}$.
Now, the expression can be calculated as shown below:
$\begin{align}
& 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=2\cdot \frac{1}{2}\left[ \cos \left( \frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2} \right)+\cos \left( \frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2} \right) \right] \\
& =2\cdot \frac{1}{2}\left[ \cos \left( \frac{\alpha +\beta -\alpha +\beta }{2} \right)+\cos \left( \frac{\alpha +\beta +\alpha -\beta }{2} \right) \right] \\
& =\cos \left( \frac{2\beta }{2} \right)+\cos \left( \frac{2\alpha }{2} \right) \\
& =\cos \beta +\cos \alpha
\end{align}$
Then, rearrange the terms:
$\cos \beta +\cos \alpha =\cos \alpha +\cos \beta $
Thus, the right-hand side of the expression is equal to the left-hand side:
$\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$.
