$\cos (\alpha +\beta )=\cos\alpha\cos\beta – \sin\alpha\sin\beta$

$\cos (\alpha -\beta )=\cos\alpha\cos\beta + \sin\alpha\sin\beta$

$\sin(\alpha +\beta )=\sin\alpha\cos\beta + \cos\alpha\sin\beta$

$\sin(\alpha -\beta )=\sin\alpha\cos\beta – \cos\alpha\sin\beta$

两式相加，相减

$\cos (\alpha +\beta ) + \cos (\alpha – \beta )=\cos\alpha\cos\beta – \sin\alpha\sin\beta + \cos\alpha\cos\beta + \sin\alpha\sin\beta = 2\cos\alpha\cos\beta$

$\cos (\alpha +\beta ) – \cos (\alpha – \beta )=\cos\alpha\cos\beta – \sin\alpha\sin\beta – \cos\alpha\cos\beta – \sin\alpha\sin\beta = -2\sin\alpha\sin\beta$

$\sin(\alpha +\beta ) +\sin(\alpha -\beta )=\sin\alpha\cos\beta + \cos\alpha\sin\beta + \sin\alpha\cos\beta – \cos\alpha\sin\beta = 2\sin\alpha\cos\beta$

$\sin(\alpha +\beta ) -\sin(\alpha -\beta )=\sin\alpha\cos\beta + \cos\alpha\sin\beta -\sin\alpha\cos\beta + \cos\alpha\sin\beta = 2\cos\alpha\sin\beta$

整理一下可得到

$\cos\alpha\cos\beta = \frac{1}{2}\left [ \cos (\alpha +\beta ) +\cos (\alpha – \beta ) \right ]$

$\sin\alpha\sin\beta = -\frac{1}{2}\left [\cos (\alpha +\beta )-\cos (\alpha – \beta )\right ]$

$\sin\alpha\cos\beta = \frac{1}{2}\left [ \sin(\alpha +\beta ) +\sin(\alpha -\beta ) \right ]$

$\cos\alpha\sin\beta = \frac{1}{2}\left [\sin(\alpha +\beta ) -\sin(\alpha -\beta )\right ]$

令 ${\alpha}’=\alpha +\beta$ ${\beta}’=\alpha -\beta$

则 $\alpha = \frac{{\alpha}’+{\beta}’}{2}$ $\beta= \frac{{\alpha}’-{\beta}’}{2}$

则

$\cos ({\alpha}’) +\cos ({\beta}’) =2\cos\frac{{\alpha}’+{\beta}’}{2}\cos\frac{{\alpha}’-{\beta}’}{2}$

$\cos({\alpha}’) -\cos({\beta}’) =-2\sin\frac{{\alpha}’+{\beta}’}{2}\sin\frac{{\alpha}’-{\beta}’}{2}$

$\sin({\alpha}’) +\sin({\beta}’) =2\sin\frac{{\alpha}’+{\beta}’}{2}\cos\frac{{\alpha}’-{\beta}’}{2}$

$\sin({\alpha}’) -\sin({\beta}’) =2\cos\frac{{\alpha}’+{\beta}’}{2}\sin\frac{{\alpha}’-{\beta}’}{2}$