一、求证：\(\sin\alpha\cos\beta=\dfrac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\)

证明：因为\[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\]\[\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta\]将以上两式的左右两边分别相加，得\[\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta\]即\[\sin\alpha\cos\beta=\dfrac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\]同理得到\[\cos\alpha\sin\beta=\dfrac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)]\]\[\cos\alpha\cos\beta=\dfrac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)]\]\[\sin\alpha\sin\beta=\dfrac{1}{2}[\cos(\alpha+\beta)-\cos(\alpha-\beta)]\]由于公式的左边为积的形式，右边为和或差的形式，故把上述四个公式称为 积化和差 公式.

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二、求证：\(\sin\theta+\sin\varphi=2\sin\dfrac{\theta+\varphi}{2}\cos\dfrac{\theta-\varphi}{2}\)

证明：由上一题的证明有\[\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta\]设 \(\alpha+\beta=\theta,\alpha-\beta=\varphi\) .那么\[\alpha=\dfrac{\theta+\varphi}{2},\beta=\dfrac{\theta-\varphi}{2}\]把 \(\alpha,\beta\) 的值代入上式，即得\[\sin\theta+\sin\varphi=2\sin\dfrac{\theta+\varphi}{2}\cos\dfrac{\theta-\varphi}{2}\]同理得\[\sin\theta-\sin\varphi=2\cos\dfrac{\theta+\varphi}{2}\sin\dfrac{\theta-\varphi}{2}\]\[\cos\theta+\cos\varphi=2\cos\dfrac{\theta+\varphi}{2}\cos\dfrac{\theta-\varphi}{2}\]\[\cos\theta-\cos\varphi=-2\sin\dfrac{\theta+\varphi}{2}\sin\dfrac{\theta-\varphi}{2}\]我们把上述四个公式称为和差化积公式.