Invoke: 和差角公式: https://www.cnblogs.com/Preparing/p/18176139
由和差角公式推衍而得
\[\begin{align}
序0: \enspace
\sin (\alpha+\beta)+\sin (\alpha-\beta)
\\
\Rightarrow \sin \alpha \cos \beta+\sin \beta \cos \alpha+\sin \alpha \cos \beta-\sin \beta \cos \alpha
\\
= 2 \sin \alpha \cos \beta
\\ \\
所得公式1:
\quad \sin \alpha \cos \beta=\frac{\sin (\alpha+\beta) + \sin (\alpha-\beta)}{2}
\\ \\ \\
序1: \enspace \sin (\alpha+\beta)-\sin (\alpha-\beta)
\\
\Rightarrow \sin \alpha \cos \beta+\sin \beta \cos \alpha-(\sin \alpha \cos \beta-\sin \beta \cos \alpha)
\\
=2 \sin \beta \cos \alpha
\\ \\
所得公式2: \enspace
\cos \alpha\sin \beta=\frac{\sin (\alpha+\beta)-\sin (\alpha-\beta)}{2}
\\ \\ \\
序2: \enspace
\cos (\alpha+\beta)+\cos (\alpha-\beta)
\\
\Rightarrow \cos \beta \cos \alpha-\sin \alpha \sin \beta+\cos \beta
\cos \alpha+\sin \alpha \sin \beta
\\
=2 \cos \alpha \cos \beta
\\ \\
所得公式3: \enspace
\cos \alpha\cos \beta =\frac{\cos (\alpha+\beta)+\cos (\alpha-\beta)}{2}
\\ \\ \\
序3: \enspace
\cos(\alpha+\beta)-\cos(\alpha-\beta)
\\
\Rightarrow\cos\beta\cos\alpha-\sin\alpha\sin\beta-(\cos\beta\cos\alpha+\sin\alpha
\sin\beta)
\\
=-2\sin\alpha\sin\beta
\\ \\
所得公式4: \enspace
\sin\alpha\sin\beta=\frac{\cos(\alpha+\beta)-\cos(\alpha-\beta)}{-2}
\end{align}
\]
summarize
\[\begin{align}
所得公式1:
\quad \sin \alpha \cos \beta=\frac{\sin (\alpha+\beta) + \sin (\alpha-\beta)}{2}
\\ \\
所得公式2: \enspace
\cos \alpha\sin \beta=\frac{\sin (\alpha+\beta)-\sin (\alpha-\beta)}{2}
\\ \\
所得公式3: \enspace
\cos \alpha\cos \beta =\frac{\cos (\alpha+\beta)+\cos (\alpha-\beta)}{2}
\\ \\
所得公式4: \enspace
\sin\alpha\sin\beta=\frac{\cos(\alpha+\beta)-\cos(\alpha-\beta)}{-2}
\end{align}
\]