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]

三角関数 - 5倍角の公式

イメージ 1

三角関数の5倍角の公式です。
加法定理、倍角の公式、3倍角の公式から導出できます。

準備

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

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

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

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

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

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

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

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

導出

$\displaystyle \sin 5 \alpha = \sin \left( 3 \alpha + 2 \alpha \right) = \sin 3 \alpha \cos 2 \alpha + \cos 3 \alpha \sin 2 \alpha \\ \displaystyle = \sin \alpha \left( 3 - 4 \sin ^ {2} \alpha \right) \left( 1 - 2 \sin ^ {2} \alpha \right) + \cos \alpha \left( 1 - 4 \sin ^ 2 \alpha \right) \cdot 2 \sin \alpha \cos \alpha \\ \displaystyle = \sin \alpha \left( 3 - 10 \sin ^ {2} \alpha +8 \sin ^ {4} \alpha \right) + 2 \sin \alpha \left( 1 - \sin ^ {2} \alpha \right) \left( 1 - 4 \sin ^ {2} \alpha \right) \\ \displaystyle = \sin \alpha \left\{ 3 - 10 \sin ^ {2} \alpha + 8 \sin ^ {4} \alpha + 2 \left( 1 - 5 \sin ^ {2} \alpha + 4 \sin ^ {4} \alpha \right) \right\} \\ \displaystyle = 16 \sin ^ {5} \alpha - 20 \sin ^ {3} \alpha + 5 \sin \alpha$

$\displaystyle \cos 5 \alpha = \cos \left( 3 \alpha + 2 \alpha \right) = \cos 3 \alpha \cos 2 \alpha - \sin 3 \alpha \sin 2 \alpha \\ \displaystyle = \cos \alpha \left( 4 \cos ^ {2} \alpha - 3 \right) \left( 2 \cos ^ {2} \alpha - 1 \right) - \sin \alpha \left( 4 \cos ^ {2} \alpha - 1 \right) \cdot 2 \sin \alpha \cos \alpha \\ \displaystyle = \cos \alpha \left\{ 8 \cos ^ {4} \alpha - 10 \cos ^ {2} \alpha + 3 + 2 \left( - \sin ^ {2} \alpha \right) \left( 4 \cos ^ {2} \alpha - 1 \right) \right\} \\ \displaystyle = \cos \alpha \left\{ 8 \cos ^ {4} \alpha - 10 \cos ^ {2} \alpha + 3 + 2 \left( \cos ^ {2} \alpha - 1 \right) \left( 4 \cos ^ {2} \alpha - 1 \right) \right\} \\ \displaystyle = \cos \alpha \left\{ 8 \cos ^ {4} \alpha - 10 \cos ^ {2} \alpha + 3 + 2 \left( 4 \cos ^ {4} \alpha - 5 \cos ^ {2} \alpha + 1 \right) \right\} \\ \displaystyle = 16 \cos ^ {5} \alpha - 20 \cos ^ {3} \alpha + 5 \cos \alpha$

$\displaystyle \tan 5 \alpha = \tan \left( 3 \alpha + 2 \alpha \right) = \frac {\tan 3 \alpha + \tan 2 \alpha}{ 1 - {\tan 3 \alpha} {\tan 2 \alpha} } = \frac {{ \frac{ \tan \alpha \left( 3 - \tan ^ {2} \alpha \right) }{ 1 - 3 \tan ^ {2} \alpha } } + { \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } }}{ 1 - { \frac{ \tan \alpha \left( 3 - \tan ^ {2} \alpha \right) }{ 1 - 3 \tan ^ {2} \alpha } } \cdot { \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } } } \\ \displaystyle = \frac{ { \tan \alpha \left( 3 - \tan ^ {2} \alpha \right) } \left( 1 - \tan ^ {2} \alpha \right) + { 2 \tan \alpha } \left( 1 - 3 \tan ^ {2} \alpha \right) } { \left( 1 - 3 \tan ^ {2} \alpha \right) \left( 1 - \tan ^ {2} \alpha \right) - { \tan \alpha \left( 3 - \tan ^ {2} \alpha \right) } \cdot { 2 \tan \alpha } } \\ \displaystyle = \frac{ \tan \alpha \left\{ \left( 3 - \tan ^ {2} \alpha \right) \left( 1 - \tan ^ {2} \alpha \right) + 2 \left( 1 - 3 \tan ^ {2} \alpha \right) \right\} } { \left( 1 - 3 \tan ^ {2} \alpha \right) \left( 1 - \tan ^ {2} \alpha \right) - 2 \tan ^ {2} \alpha \left( 3 - \tan ^ {2} \alpha \right) } \\ \displaystyle = \frac{ \tan \alpha \left( 3 - 4 \tan ^ {2} \alpha + \tan ^ {4} \alpha + 2 - 6 \tan ^ {2} \alpha \right) }{ 1 - 4 \tan ^ {2} \alpha + 3 \tan ^ {4} \alpha - 6 \tan ^ {2} \alpha + 2 \tan ^ {4} \alpha } = \frac{ \tan \alpha \left( 5 - 10 \tan ^ {2} \alpha + \tan ^ {4} \alpha \right) }{ 1 - 10 \tan ^ {2} \alpha + 5 \tan ^ {4} \alpha } \\ \displaystyle = \frac{ 5 \tan \alpha - 10 \tan ^ {3} \alpha + \tan ^ {5} \alpha }{ 1 - 10 \tan ^ {2} \alpha + 5 \tan ^ {4} \alpha }$

LibreOffice 数式(Math) のソース：

sin 5 %alpha = sin ( 3 %alpha + 2 %alpha ) = sin 3 %alpha cos 2 %alpha + cos 3 %alpha sin 2 %alpha

sin %alpha ( 3 - 4 sin ^2 %alpha ) ( 1 - 2 sin ^2 %alpha ) + cos %alpha ( 1 - 4 sin ^2 %alpha ) cdot 2 sin %alpha cos %alpha

sin %alpha ( 3 - 10 sin ^2 %alpha + 8 sin ^4 %alpha ) + 2 sin %alpha ( 1 - sin ^2 %alpha ) ( 1 - 4 sin ^2 %alpha )

sin %alpha lbrace 3 - 10 sin ^2 %alpha + 8 sin ^4 %alpha + 2 ( 1 - 5 sin ^2 %alpha + 4 sin ^4 %alpha ) rbrace

16 sin ^5 %alpha - 20 sin ^3 %alpha + 5 sin %alpha

cos 5 %alpha = cos ( 3 %alpha + 2 %alpha ) = cos 3 %alpha cos 2 %alpha - sin 3 %alpha sin 2 %alpha

cos %alpha ( 4 cos ^2 %alpha - 3 ) ( 2 cos ^2 - 1 ) - sin %alpha ( 4 cos ^2 %alpha - 1 ) cdot 2 sin %alpha cos %alpha

cos %alpha lbrace 8 cos ^4 %alpha - 10 cos ^2 %alpha + 3 + 2 ( - sin ^2 %alpha ) ( 4 cos ^2 %alpha - 1 ) rbrace

cos %alpha lbrace 8 cos ^4 %alpha - 10 cos ^2 %alpha + 3 + 2 ( cos ^2 %alpha - 1 ) ( 4 cos ^2 %alpha - 1 ) rbrace

cos %alpha lbrace 8 cos ^4 %alpha - 10 cos ^2 %alpha + 3 + 2 ( 4 cos ^4 %alpha - 5 cos ^2 %alpha + 1 ) rbrace

16 cos ^5 %alpha - 20 cos ^3 %alpha + 5 cos %alpha

tan 5 %alpha = tan ( 3 %alpha + 2 %alpha ) = { tan 3 %alpha + tan 2 %alpha } over { 1 - tan 3%alpha tan 2 %alpha } = { { 3 tan %alpha - tan ^3 %alpha } over { 1 - 3 tan ^2 %alpha } + { 2 tan %alpha } over { 1 - tan ^2 %alpha } } over { 1 - { 3 tan %alpha - tan ^3 %alpha } over { 1 - {3 tan ^2 %alpha } } cdot { { 2 tan %alpha } over { 1 - tan ^2 %alpha } } }

{ ( 3 tan %alpha - tan ^3 %alpha ) ( 1 - tan ^2 %alpha ) + 2 tan %alpha ( 1 - 3 tan ^2 %alpha ) } over { ( 1 - 3 tan ^2 %alpha ) ( 1 - tan ^2 %alpha ) - ( 3 tan %alpha - tan ^3 %alpha )cdot 2 tan %alpha }

{ tan %alpha lbrace ( 3 - tan ^2 %alpha ) ( 1 - tan ^2 %alpha ) + 2 ( 1 - 3 tan ^2 %alpha ) rbrace } over { 1 - 4 tan ^2 %alpha + 3 tan ^4 %alpha -6 tan ^2 %alpha + 2 tan ^4 %alpha } = { tan %alpha ( 3 - 4 tan ^2 %alpha + tan ^4 %alpha + 2 - 6 tan ^2 %alpha) } over { 1 - 10 tan ^2 %alpha + 5 tan ^4 %alpha }

{ tan ^5 %alpha - 10 tan ^3 %alpha + 5 tan %alpha } over { 5 tan ^4 %alpha - 10 tan ^2 %alpha + 1 }