加法定理の証明

$3$\displaystyle{\sin $ {\alpha}+{\beta} $ }$

$3$\displaystyle{\begin{array}{lll}\sin $ {\alpha}+{\beta} $ & \vspace{6}\\ & \vspace{6}\\ \text{\hspace{5}}=\cos \{ 90- $ {\alpha}+{\beta} $ \} & \vspace{6}\\ \end{array}}$
　 (三角関数計算の基礎を参照)

$3$\displaystyle{\text{\hspace{5}}=\cos \{ $ 90- {\alpha} $ - {\beta} \} }$

$3$\displaystyle{\begin{array}{lll}\text{\hspace{5}}=\cos $ 90- {\alpha} $ \cos {\beta}+& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\sin $ 90- {\alpha} $ \sin {\beta}& \vspace{6}\\ \end{array}}$

$3$\displaystyle{\text{\hspace{5}}=\sin {\alpha}\cos {\beta}+\cos {\alpha}\sin {\beta}}$

となる。

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