证明：
$\sin \alpha \cos \beta =\dfrac{1}{2}\left( \sin \left(\alpha +\beta \right)+\sin \left( \alpha-\beta \right) \right)$
$\cos \alpha \sin \beta =\dfrac{1}{2}\left( \sin \left(\alpha +\beta \right)-\sin \left( \alpha-\beta \right) \right)$
$\cos \alpha \cos \beta =\dfrac{1}{2}\left( \cos \left(\alpha +\beta \right)+\cos \left( \alpha-\beta \right) \right)$
$\sin \alpha \sin \beta =-\dfrac{1}{2}\left( \cos \left(\alpha +\beta \right)-\cos \left( \alpha-\beta \right) \right)$
$\sin \alpha +\sin \beta =2\sin \dfrac{\alpha +\beta}{2}\cos \dfrac{\alpha -\beta }{2}$
$\sin \alpha -\sin \beta =2\cos \dfrac{\alpha +\beta}{2}\sin \dfrac{\alpha -\beta }{2}$
$\cos \alpha +\cos \beta =2\cos \dfrac{\alpha +\beta}{2}\cos \dfrac{\alpha -\beta }{2}$
$\cos \alpha -\cos \beta =-2\sin \dfrac{\alpha +\beta}{2}\sin \dfrac{\alpha -\beta }{2}$
$\sin \alpha \cos \beta =\dfrac{1}{2}\left( \sin \left(\alpha +\beta \right)+\sin \left( \alpha-\beta \right) \right)$
$\cos \alpha \sin \beta =\dfrac{1}{2}\left( \sin \left(\alpha +\beta \right)-\sin \left( \alpha-\beta \right) \right)$
$\cos \alpha \cos \beta =\dfrac{1}{2}\left( \cos \left(\alpha +\beta \right)+\cos \left( \alpha-\beta \right) \right)$
$\sin \alpha \sin \beta =-\dfrac{1}{2}\left( \cos \left(\alpha +\beta \right)-\cos \left( \alpha-\beta \right) \right)$
$\sin \alpha +\sin \beta =2\sin \dfrac{\alpha +\beta}{2}\cos \dfrac{\alpha -\beta }{2}$
$\sin \alpha -\sin \beta =2\cos \dfrac{\alpha +\beta}{2}\sin \dfrac{\alpha -\beta }{2}$
$\cos \alpha +\cos \beta =2\cos \dfrac{\alpha +\beta}{2}\cos \dfrac{\alpha -\beta }{2}$
$\cos \alpha -\cos \beta =-2\sin \dfrac{\alpha +\beta}{2}\sin \dfrac{\alpha -\beta }{2}$
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