コンテンツにスキップ

利用者:Ninomy/倍角公式

加法定理

${\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta }$
${\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$

これを用いて各種倍角公式を導く。

2倍角の公式

${\displaystyle \sin 2\theta =2\sin \theta \cos \theta }$
{\displaystyle {\begin{aligned}\cos 2\theta &=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \end{aligned}}}

{\displaystyle {\begin{aligned}\sin 2\theta &=\sin \theta \cos \theta +\cos \theta \sin \theta \\&=2\sin \theta \cos \theta \end{aligned}}}
{\displaystyle {\begin{aligned}\cos 2\theta &=\cos \theta \cos \theta -\sin \theta \sin \theta \\&=\cos ^{2}\theta -\sin ^{2}\theta \\&=\cos ^{2}\theta -(1-\cos ^{2}\theta )\\&=2\cos ^{2}\theta -1\\&=2(1-\sin ^{2}\theta )-1\\&=1-2\sin ^{2}\theta \qquad \Box \end{aligned}}}

半角の公式

${\displaystyle \sin ^{2}{\frac {\theta }{2}}={\frac {1-cos\theta }{2}}}$
${\displaystyle \cos ^{2}{\frac {\theta }{2}}={\frac {1+cos\theta }{2}}}$

cosの2倍角の公式で ${\displaystyle \theta }$${\displaystyle {\frac {\theta }{2}}}$ と置き換えて変形する。

{\displaystyle {\begin{aligned}\cos 2\theta &=1-2\sin ^{2}\theta \\2\sin ^{2}\theta &=1-\cos 2\theta \\\sin ^{2}\theta &={\frac {1-\cos 2\theta }{2}}\\\therefore \sin ^{2}{\frac {\theta }{2}}&={\frac {1-\cos \theta }{2}}\end{aligned}}}

{\displaystyle {\begin{aligned}\cos 2\theta &=2\cos ^{2}\theta -1\\2\cos ^{2}\theta &=1+\cos 2\theta \\\cos ^{2}\theta &={\frac {1+\cos 2\theta }{2}}\\\therefore \cos ^{2}{\frac {\theta }{2}}&={\frac {1+\cos \theta }{2}}\qquad \Box \end{aligned}}}

3倍角の公式

${\displaystyle \sin 3\theta =3\sin \theta -4\sin ^{3}\theta }$
${\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta }$

{\displaystyle {\begin{aligned}\sin 3\theta &=\sin(2\theta +\theta )\\&=\sin 2\theta \cos \theta +\cos 2\theta \sin \theta \\&=2\sin \theta \cos ^{2}\theta +(1-2\sin ^{2}\theta )\sin \theta \\&=2\sin \theta (1-\sin ^{2}\theta )+(1-2\sin ^{2}\theta )\sin \theta \\&=2\sin \theta -2\sin ^{3}\theta +\sin \theta -2\sin ^{3}\theta \\&=3\sin \theta -4\sin ^{4}\theta \end{aligned}}}
{\displaystyle {\begin{aligned}\cos 3\theta &=\cos(2\theta +\theta )\\&=\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\&=(2\cos ^{2}\theta -1)\cos \theta -2\sin ^{2}\theta \cos \theta \\&=(2\cos ^{2}\theta -1)\cos \theta -2(1-\cos ^{2}\theta )\cos \theta \\&=2\cos ^{3}\theta -\cos \theta -2\cos \theta +2\cos ^{3}\theta \\&=4\cos ^{3}\theta -3\cos \theta \qquad \Box \end{aligned}}}

n倍角の公式

これまでの公式を一般化し、任意倍の倍角公式を導く。

オイラーの公式による導出

オイラーの公式

${\displaystyle e^{i\theta }=\cos \theta +i\sin \theta }$

を用いて導く。

オイラーの公式

${\displaystyle e^{i\theta }=\cos \theta +i\sin \theta }$

より、

{\displaystyle {\begin{aligned}e^{-i\theta }&=\cos(-\theta )+i\sin(-\theta )\\&=\cos \theta -i\sin \theta \end{aligned}}}

であるから、

${\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}}$
${\displaystyle \sin \theta ={\frac {e^{i\theta }-e^{-i\theta }}{2i}}}$

である。したがって、

{\displaystyle {\begin{aligned}\cos n\theta &={\frac {e^{in\theta }+e^{-in\theta }}{2}}\\&={\frac {(e^{i\theta })^{n}+(e^{-i\theta })^{n}}{2}}\\&={\frac {(\cos \theta +i\sin \theta )^{n}+(\cos \theta -i\sin \theta )^{n}}{2}}\end{aligned}}}
{\displaystyle {\begin{aligned}\sin n\theta &={\frac {e^{in\theta }-e^{-in\theta }}{2i}}\\&={\frac {(e^{i\theta })^{n}-(e^{-i\theta })^{n}}{2i}}\\&={\frac {(\cos \theta +i\sin \theta )^{n}-(\cos \theta -i\sin \theta )^{n}}{2i}}\qquad \Box \end{aligned}}}