cos関数の合成(複素数を用いた導出)

２つのcos関数 $3$\displaystyle{\hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}}$ , $3$\displaystyle{\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2}}$ の合成式を，複素数を用いて表すと

$3$\displaystyle{ \hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2} }$ $3$\displaystyle{ =Re \[ \hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ {i}\hspace{1}{{\theta}}_{2} } \] }$ ---- (1)

となる（ $3$\displaystyle{{}^\bullet\limits{ }_\bullet {}^\bullet }$ オイラーの公式）．

$3$\displaystyle{ \hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ {i}\hspace{1}{{\theta}}_{2} } }$

$3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}$ {e}^{ {i}\hspace{1}{{\theta}}_{1} }+{e}^{ {i}\hspace{1}{{\theta}}_{2} } $+\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}$ {e}^{ {i}\hspace{1}{{\theta}}_{1} }- {e}^{ {i}\hspace{1}{{\theta}}_{2} } $ }$

$3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}\exp \[ {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] $ \exp \[ {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] +\exp \[ - {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] $ }$

$3$\displaystyle{ +\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}\exp \[ {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] $ \exp \[ {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] - \exp \[ - {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] $ }$

$3$\displaystyle{ =\exp \[ {i}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \] \{ $\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}+{i}$\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \} }$ ---- (2)

上式において

$3$\displaystyle{ a=$\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ b=$\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ {\theta}=\frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ ---- (3)

とおくと

$3$\displaystyle{ \hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ {i}\hspace{1}{{\theta}}_{2} }={e}^{ {i}{\theta} }$a+b{i}$ }$ $3$\displaystyle{ ={e}^{ {i}{\theta} }\cdot r{e}^{ {i}{\varphi} } }$ $3$\displaystyle{ =r{e}^{ {i}${\theta}+{\varphi}$ } }$ ---- (4)

となる．ここで，

$3$\displaystyle{ r=\sqrt{ {a}^{2}+{b}^{2} } }$ $3$\displaystyle{ =\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2}$ } }$ ---- (5)

$3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ ---- (6)

である．したがって，

を得る．（導出完了）

（※）式(1)の複素数を用いた表現において

として同様に進めると，

$3$\displaystyle{ \hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ - {i}\hspace{1}{{\theta}}_{2} }=r{e}^{ {i}${\theta}+{\varphi}$ } }$ ---- (8)

$3$\displaystyle{ r=\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2}$ } }$ , $3$\displaystyle{ {\theta}=\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ ---- (9)

が得られる．

■ 複素平面での幾何学的な意味

図に示すように，複素数の和

$3$\displaystyle{ \hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ {i}\hspace{1}{{\theta}}_{2} }=r{e}^{ {i}${\theta}+{\varphi}$ } }$

は，複素平面上での2つのベクトルの和に対応する．右側の図中の角 $3$\displaystyle{\alpha }$ は，

$3$\displaystyle{ \alpha =\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} }$ ---- (10)

であり，余弦定理により

$3$\displaystyle{ r=\sqrt{ r_{1}^{2}+r_{2}^{2}- 2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos ${\pi}- \alpha $ } }$
$3$\displaystyle{ =\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos \alpha } }$ ---- (11)

である．また，角 $3$\displaystyle{{\phi}}$ は次式を満たす．

$3$\displaystyle{ \tan {\phi}=\frac{\hspace{2} \hspace{1}{r}_{1}\sin \alpha \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}\cos \alpha +\hspace{1}{r}_{2} \hspace{2}} }$ ---- (12)

図から分かるように， $3$\displaystyle{ {\theta}+{\varphi}=\hspace{1}{{\theta}}_{2}+{\phi} }$ であり， $3$\displaystyle{ {\theta}= $\hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2}$ /2 }$ なので

$3$\displaystyle{ {\varphi}=\hspace{1}{{\theta}}_{2}+{\phi}- \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ $3$\displaystyle{ ={\phi}- \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ $3$\displaystyle{ ={\phi}- \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} }$ ---- (13)

となる．したがって，

$3$\displaystyle{ \tan {\varphi}=\tan ${\phi}- \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}$ }$ $3$\displaystyle{ =\frac{\hspace{2} \tan {\phi}- \tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} 1+\tan {\phi}\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$   （ $3$\displaystyle{{}^\bullet\limits{ }_\bullet {}^\bullet }$ 加法定理）
       $3$\displaystyle{ =\frac{\hspace{2} \frac{\hspace{2} \hspace{1}{r}_{1}\sin \alpha \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}\cos \alpha +\hspace{1}{r}_{2} \hspace{2}}- \tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} 1+\frac{\hspace{2} \hspace{1}{r}_{1}\sin \alpha \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}\cos \alpha +\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}\sin \alpha - $\hspace{1}{r}_{1}\cos \alpha +\hspace{1}{r}_{2}$\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}\cos \alpha +\hspace{1}{r}_{2}+\hspace{1}{r}_{1}\sin \alpha \tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$
       $3$\displaystyle{ =\frac{\hspace{2} 2\hspace{1}{r}_{1}\sin \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}\cos \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}- \hspace{1}{r}_{1}$ 2{ \cos }^{2}\frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}- 1 $\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}- \hspace{1}{r}_{2}\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}$ 1- 2{ \sin }^{2}\frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} $+\hspace{1}{r}_{2}+2\hspace{1}{r}_{1}\sin \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}\cos \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$   （ $3$\displaystyle{{}^\bullet\limits{ }_\bullet {}^\bullet }$ 2倍角の公式）
       $3$\displaystyle{ =\frac{\hspace{2} $\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$     ---- (14)

が得られる．

また，式(2)の表現

$3$\displaystyle{ \hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ {i}\hspace{1}{{\theta}}_{2} } }$ $3$\displaystyle{ ={e}^{ {i}{\theta} }\text{\hspace{1}} \{ \text{\hspace{1}}$\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}+{i}$\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}\text{\hspace{1}} \} }$ ---- (15)

（ $3$\displaystyle{ {\theta}= $\hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2}$ /2 }$ ， $3$\displaystyle{ \alpha =\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} }$ ）において，右辺の $3$\displaystyle{ \{ \text{\hspace{1}} \} }$ 内は，長径を $3$\displaystyle{\hspace{1}{r}_{1}+\hspace{1}{r}_{2}}$ ，短径を $3$\displaystyle{\hspace{1}{r}_{1}- \hspace{1}{r}_{2}}$ とした複素平面上の楕円軌道を表している．その楕円軌道上の，角 $3$\displaystyle{\alpha /2}$ のときの位置ベクトルは $3$\displaystyle{xy}$ 平面上で

$3$\displaystyle{ \hspace{1}{ {r}\limits^{\rightarrow } }_{{\varphi}}= $ \(\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}\text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}$\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \text{\hspace{1}}\) }$

と表され，その大きさが

$3$\displaystyle{ r= \| \hspace{1}{ {r}\limits^{\rightarrow } }_{{\varphi}} \| }$
$3$\displaystyle{ =\sqrt{ { $\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$ }^{2}{ \cos }^{2}\frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}}+{ $\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$ }^{2}{ \sin }^{2}\frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} } }$
$3$\displaystyle{ =\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos \alpha } }$

であり， $3$\displaystyle{x}$ 軸とのなす角 $3$\displaystyle{{\varphi}}$ が

$3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2} $\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} $\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2}\alpha \hspace{2}}{\hspace{2}2\hspace{2}} }$

を満たす（ただし， $3$\displaystyle{\alpha =\pm {\pi}}$ のとき $3$\displaystyle{{\varphi}=\pm {\pi}/2 }$ とする）．したがって，

$3$\displaystyle{ \hspace{1}{ {r}\limits^{\rightarrow } }_{{\varphi}}= $ r\cos {\varphi}\text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}r\sin {\varphi} \text{\hspace{1}}$ }$

と表される．さらに，式(15)の右辺の指数関数 $3$\displaystyle{{e}^{ {i}{\theta} }}$ は，複素平面上のベクトルを原点の周りに角 $3$\displaystyle{{\theta}}$ だけ回転させることに対応するので， $3$\displaystyle{\hspace{1}{ {r}\limits^{\rightarrow } }_{{\varphi}}}$ を $3$\displaystyle{{\theta}}$ 回転させると

$3$\displaystyle{ \hspace{1}{ {r}\limits^{\rightarrow } }_{{\varphi}}\rightarrow \hspace{1}{ {r}\limits^{\rightarrow } }_{ {\theta}+{\varphi} } }$ $3$\displaystyle{ = $ r\cos \( {\theta}+{\varphi} $\text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}r\sin $ {\theta}+{\varphi} $ \text{\hspace{1}}\) }$

となる．したがって，

$3$\displaystyle{ \hspace{1}{ {r}\limits^{\rightarrow } }_{ {\theta}+{\varphi} }\text{\hspace{1}}\text{\hspace{1}}\Leftrightarrow \text{\hspace{1}}\text{\hspace{1}}r{e}^{ {i}${\theta}+{\varphi}$ }=\hspace{1}{r}_{1}{e}^{ {i}\hspace{1}{{\theta}}_{1} }+\hspace{1}{r}_{2}{e}^{ {i}\hspace{1}{{\theta}}_{2} } }$

という対応がある．

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最終更新日：2025年4月23日