\cos \alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ]
or
Derivation 1
Use the sum and difference identities for cosine functions
\cos(\alpha +\beta ) = \cos \alpha\cos \beta -\sin \alpha \sin \beta
\cos(\alpha -\beta ) = \cos \alpha\cos \beta +\sin \alpha \sin \beta
Let x = \cos \alpha\cos \beta, y = \sin \alpha \sin \beta
Solve the system of equations
\begin{alignedat}{2} &x-y = \cos(\alpha +\beta ) \\ &x+y = \cos(\alpha -\beta ) \end{alignedat}
x= \dfrac{1}{2} [ \cos(\alpha +\beta )+ \cos(\alpha -\beta )]
y = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]
Therefore,
\cos\alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ]
\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]