Derivation of Product to Sum Identities
$$ \cos\alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ]$$
$$\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]$$

Question

Derivation of Product to Sum Identities
$$ \cos\alpha\cos \beta = \dfrac{1}{2}  [\cos(\alpha+\beta  ) +\cos(\alpha - \beta) ]$$
$$\sin \alpha\sin \beta = \dfrac{1}{2}  [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]$$

Uniteasy Admin · Accepted Answer

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
```katex
\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,
$$ \cos\alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ]$$
$$\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]$$

Question

Answer