Answer
$\cos \beta+\cos \alpha$
Work Step by Step
We need to prove the Sum to Product Identity:
$ \cos \alpha +\cos \beta =2 \cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha- \beta}{2}$
In order to prove that, we will simplify the right hand side by using the product to sum formula as follows:
$ \cos \alpha \cos \beta =\dfrac{1}{2} [\cos (\alpha- \beta ) +\cos (\alpha+\beta )]$
$2 \cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha- \beta}{2}=(2) (\dfrac{1}{2}) [(\cos \dfrac{(\alpha- \beta )}{2} +\cos \dfrac{(\alpha+\beta )}{2})+ (\cos \dfrac{(\alpha- \beta )}{2} +\cos \dfrac{(\alpha+\beta )}{2})]\\=\cos (\dfrac{2 \beta}{2})+\cos (\dfrac{2 \alpha}{2})\\=\cos \beta+\cos \alpha$
