三角関数 - 3倍角の公式

カテゴリー[ 昆虫| 田園| 花| 街| 数学・幾何学| 寺院| 城| 祭り| 鉄道| 海| 風力発電]

イメージ 1

三角関数の3倍角の公式です。
加法定理と倍角公式から導出します。

準備

$\displaystyle \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\displaystyle \cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\displaystyle \sin 2 \alpha = 2 \sin \alpha \cos \alpha$

$\displaystyle \cos 2 \alpha = 2 \cos ^ {2} \alpha - 1 = 1 - 2 \sin ^ {2} \alpha$

$\displaystyle \tan 2 \alpha = \frac { 2 \tan \alpha }{ 1 - \tan ^ {2} \alpha }$

導出

$\displaystyle \sin 3\alpha = \sin ( 2 \alpha + \alpha ) = \sin 2 \alpha \cos \alpha + \cos 2 \alpha \sin \alpha = ( 2 \sin \alpha \cos \alpha ) \cos \alpha + \left( 1 - 2 \sin ^ {2} \alpha \right) \sin \alpha \\ \displaystyle = 2 \sin \alpha \cos ^{2} \alpha + \left( 1 - 2 \sin ^ {2} \alpha \right) \sin \alpha = 2 \sin \alpha \left( 1 - \sin ^ {2} \alpha \right) + \left( 1 - 2 \sin ^ {2} \alpha \right) \sin \alpha \\ \displaystyle = \sin \alpha \left\{ 2 \left( 1 - \sin ^ {2} \alpha \right) + 1 - 2 \sin ^ {2} \alpha \right\} = \sin \alpha \left( 2 - 2 \sin ^ {2} \alpha + 1 - 2 \sin ^ {2} \alpha \right) \\ \displaystyle = \sin \alpha \left( 3 - 4 \sin ^ {2} \alpha \right) = 3 \sin \alpha - 4 \sin ^ {3} \alpha$

$\displaystyle \cos 3\alpha = \cos ( 2 \alpha + \alpha ) = \cos 2 \alpha \cos \alpha - \sin 2 \alpha \sin \alpha = \left( 2 \cos ^ {2} \alpha - 1 \right) \cos \alpha - ( 2 \sin \alpha \cos \alpha ) \sin \alpha \\ \displaystyle = \left( 2 \cos ^ {2} \alpha - 1 \right) \cos \alpha - 2 \sin ^ {2} \alpha \cos \alpha = \left( 2 \cos ^ {2} \alpha - 1 \right) \cos \alpha - 2 \left( 1 - \cos ^ {2} \alpha \right) \cos \alpha \\ \displaystyle = \cos \alpha \left\{ 2 \cos ^ {2} \alpha - 1 - 2 \left( 1 - \cos ^ {2} \alpha \right) \right\} = \cos \alpha \left( 4 \cos ^ {2} \alpha - 3 \right) = 4 \cos ^ {3} \alpha - 3 \cos \alpha$

$\displaystyle \tan 3\alpha = \tan ( 2 \alpha + \alpha ) = \frac{ \tan 2 \alpha + \tan \alpha }{ 1 - \tan 2 \alpha \tan \alpha } = \frac{ \frac { 2 \tan \alpha }{ 1 - \tan ^ {2} \alpha } + \tan \alpha }{ 1 - \left( \frac { 2 \tan \alpha }{ 1 - \tan ^ {2} \alpha } \right) \tan \alpha } = \frac{ 2 \tan \alpha + \tan \alpha \left( 1 - \tan ^ {2} \alpha \right) }{ 1 - \tan ^ {2} \alpha - 2 \tan ^ {2} \alpha } \\ \displaystyle = \frac{ \tan \alpha \left( 2 + 1 - \tan ^ {2} \alpha \right) }{ 1 - 3 \tan ^ {2} \alpha } = \frac{ \tan \alpha \left( 3 - \tan ^ {2} \alpha \right) }{ 1 - 3 \tan ^ {2} \alpha } = \frac{ 3 \tan \alpha - \tan ^ {3} \alpha }{ 1 - 3 \tan ^ {2} \alpha }$