# cos関数の合成 (composition of cosine functions)

２つのcos関数 ${r}_{1}\mathrm{cos}{\theta }_{1}$${r}_{2}\mathrm{cos}{\theta }_{2}$ の合成

${r}_{1}\mathrm{cos}{\theta }_{1}+{r}_{2}\mathrm{cos}{\theta }_{2}=r\mathrm{cos}\left(\theta +\phi \right)$

ここで，右辺の $r$ , $\theta$ , $\phi$ $\left(-\pi <\phi \le \pi \right)$ は次式で与えられる．

$r=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}+2{r}_{1}{r}_{2}\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)}$

$\theta =\frac{{\theta }_{1}+{\theta }_{2}}{2}$

$\mathrm{tan}\phi =\frac{{r}_{1}-{r}_{2}}{{r}_{1}+{r}_{2}}\mathrm{tan}\frac{{\theta }_{1}-{\theta }_{2}}{2}$

ただし， ${\theta }_{1}-{\theta }_{2}=±\pi$ のとき， $r={r}_{1}-{r}_{2}$$\phi =±\pi /2$ とする．

または，次式のようにとっても構わない．

$r=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}+2{r}_{1}{r}_{2}\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)}$

$\theta =\frac{{\theta }_{1}-{\theta }_{2}}{2}$

$\mathrm{tan}\phi =\frac{{r}_{1}-{r}_{2}}{{r}_{1}+{r}_{2}}\mathrm{tan}\frac{{\theta }_{1}+{\theta }_{2}}{2}$

ただし， ${\theta }_{1}+{\theta }_{2}=±\pi$ のとき， $r={r}_{1}-{r}_{2}$$\phi =±\pi /2$ とする．

■ 導出

${r}_{1}\mathrm{cos}{\theta }_{1}+{r}_{2}\mathrm{cos}{\theta }_{2}$

$=\frac{{r}_{1}+{r}_{2}}{2}\left(\mathrm{cos}{\theta }_{1}+\mathrm{cos}{\theta }_{2}\right)$ $+\frac{{r}_{1}-{r}_{2}}{2}\left(\mathrm{cos}{\theta }_{1}-\mathrm{cos}{\theta }_{2}\right)$

$=\left({r}_{1}+{r}_{2}\right)\mathrm{cos}\frac{{\theta }_{1}-{\theta }_{2}}{2}\mathrm{cos}\frac{{\theta }_{1}+{\theta }_{2}}{2}$ $-\left({r}_{1}-{r}_{2}\right)\mathrm{sin}\frac{{\theta }_{1}-{\theta }_{2}}{2}\mathrm{sin}\frac{{\theta }_{1}+{\theta }_{2}}{2}$   （ $\because$ 和積の公式

$a=\left({r}_{1}+{r}_{2}\right)\mathrm{cos}\frac{{\theta }_{1}-{\theta }_{2}}{2}$ ,   $b=\left({r}_{1}-{r}_{2}\right)\mathrm{sin}\frac{{\theta }_{1}-{\theta }_{2}}{2}$ ,   $\theta =\frac{{\theta }_{1}+{\theta }_{2}}{2}$

とおくと

${r}_{1}\mathrm{cos}{\theta }_{1}+{r}_{2}\mathrm{cos}{\theta }_{2}=a\mathrm{cos}\theta -b\mathrm{sin}\theta$

$=\sqrt{{a}^{2}+{b}^{2}}\left(\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}\mathrm{cos}\theta -\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}\mathrm{sin}\theta \right)$

$=r\left(\mathrm{cos}\theta \mathrm{cos}\phi -\mathrm{sin}\theta \mathrm{sin}\phi \right)$ $=r\mathrm{cos}\left(\theta +\phi \right)$   （ $\because$ 加法定理

が得られる．ここで， $r=\sqrt{{a}^{2}+{b}^{2}}$ ,   $\mathrm{cos}\phi =\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}$ ,   $\mathrm{sin}\phi =\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}$  とおいた（三角関数の合成）．

また，

${a}^{2}+{b}^{2}$$={\left({r}_{1}+{r}_{2}\right)}^{2}{\mathrm{cos}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}$$+{\left({r}_{1}-{r}_{2}\right)}^{2}{\mathrm{sin}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}$

$=\left({r}_{1}^{2}+2{r}_{1}{r}_{2}+{r}_{2}^{2}\right){\mathrm{cos}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}$$+\left({r}_{1}^{2}-2{r}_{1}{r}_{2}+{r}_{2}^{2}\right){\mathrm{sin}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}$

$=\left({r}_{1}^{2}+{r}_{2}^{2}\right)\left({\mathrm{cos}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}+{\mathrm{sin}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)$$+2{r}_{1}{r}_{2}\left({\mathrm{cos}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}-{\mathrm{sin}}^{2}\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)$

$={r}_{1}^{2}+{r}_{2}^{2}+2{r}_{1}{r}_{2}\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)$   （ $\because$ 2倍角の公式

より

$r=\sqrt{{a}^{2}+{b}^{2}}$ $=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}+2{r}_{1}{r}_{2}\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)}$

であり， $\phi$

$\mathrm{tan}\phi =\frac{\mathrm{sin}\phi }{\mathrm{cos}\phi }=\frac{b}{a}$ $=\frac{\left({r}_{1}-{r}_{2}\right)\mathrm{sin}\frac{{\theta }_{1}-{\theta }_{2}}{2}}{\left({r}_{1}+{r}_{2}\right)\mathrm{cos}\frac{{\theta }_{1}-{\theta }_{2}}{2}}$ $=\frac{{r}_{1}-{r}_{2}}{{r}_{1}+{r}_{2}}\mathrm{tan}\frac{{\theta }_{1}-{\theta }_{2}}{2}$

を満たす角である．（導出完了）

（※）導出の最初の展開式において

$a=\left({r}_{1}+{r}_{2}\right)\mathrm{cos}\frac{{\theta }_{1}+{\theta }_{2}}{2}$ ,   $b=\left({r}_{1}-{r}_{2}\right)\mathrm{sin}\frac{{\theta }_{1}+{\theta }_{2}}{2}$ ,   $\theta =\frac{{\theta }_{1}-{\theta }_{2}}{2}$

とおいて同様に進めると，

$r=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}+2{r}_{1}{r}_{2}\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)}$ ,   $\mathrm{tan}\phi =\frac{{r}_{1}-{r}_{2}}{{r}_{1}+{r}_{2}}\mathrm{tan}\frac{{\theta }_{1}+{\theta }_{2}}{2}$

が得られる．

また，sin関数の形で合成

${r}_{1}\mathrm{cos}{\theta }_{1}+{r}_{2}\mathrm{cos}{\theta }_{2}=r\mathrm{sin}\left(\theta +\phi \right)$

したり，sinの合成 ${r}_{1}\mathrm{sin}{\theta }_{1}+{r}_{2}\mathrm{sin}{\theta }_{2}$ や，sinとcosの合成 ${r}_{1}\mathrm{sin}{\theta }_{1}+{r}_{2}\mathrm{cos}{\theta }_{2}$ もできる．

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