三角関数 - 6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6倍角の公式です。
倍角の公式、3倍角の公式から導出できます。

準備

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

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

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

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

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

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

導出

$\displaystyle \sin 6 \alpha = \sin 3 ( 2 \alpha ) = \sin 2 \alpha \left( 3 - 4 \sin ^ {2} 2 \alpha \right) = 2 \sin \alpha \cos \alpha \left\{ 3 - 4 \left( 2 \sin \alpha \cos \alpha \right) ^ {2} \right\} \\ \displaystyle = 2 \sin \alpha \cos \alpha \left( 3 - 16 \sin ^ {2} \alpha \cos ^ {2} \alpha \right) = 2 \sin \alpha \cos \alpha \left\{ 3 - 16 \sin ^ {2} \alpha \left( 1 - \sin ^ {2} \alpha \right) \right\} \\ \displaystyle = 2 \sin \alpha \cos \alpha \left( 3 - 16 \sin ^ {2} \alpha + 16 \sin ^ {4} \alpha \right) = \cos \alpha \left( 32 \sin ^ {5} \alpha - 32 \sin ^ {3} \alpha + 6 \sin \alpha \right)$

$\displaystyle \cos 6 \alpha = \cos 3 ( 2 \alpha ) = \cos 2 \alpha \left( 4 \cos ^ {2} 2 \alpha - 3 \right) = \left( 2 \cos ^ {2} \alpha - 1 \right) \left\{ 4 \left( 2 \cos ^ {2} \alpha - 1 \right) ^{2} - 3 \right\} \\ \displaystyle = \left( 2 \cos ^ {2} \alpha - 1 \right) \left( 16 \cos ^ {4} \alpha - 16 \cos ^ {2} \alpha + 1 \right) = 32 \cos ^ {6} \alpha - 32 \cos ^ {4} \alpha + 2 \cos ^ {2} \alpha - 16 \cos ^ {4} \alpha + 16 \cos ^ {2} \alpha - 1 \\ \displaystyle = 32 \cos ^ {6} \alpha - 48 \cos ^ {4} \alpha + 18 \cos ^ {2} \alpha - 1$

$\displaystyle \tan 6 \alpha = \tan 3 ( 2 \alpha ) = \frac{ \tan 2 \alpha \left( 3 - \tan ^ {2} 2 \alpha \right) }{ 1 - 3 \tan ^ {2} 2 \alpha } = \frac{ \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } \cdot \left\{ 3 - \left( \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } \right) ^{2} \right\} }{ 1 - 3 \left( \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } \right) ^{2} } \\ \displaystyle = { \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } } \cdot \frac{ 3 - \left( \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } \right) ^{2} }{ 1 - 3 \left( \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } \right) ^{2} } = { \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } } \cdot \frac { 3 \left( 1 - \tan ^ {2} \alpha \right) ^ {2} - \left( 2 \tan \alpha \right) ^ {2} }{ \left( 1 - \tan ^ {2} \alpha \right) ^ {2} - 3 \left( 2 \tan \alpha \right) ^ {2} } \\ \displaystyle = { \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } } \cdot \frac{ 3 - 6 \tan ^ {2} \alpha + 3 \tan ^ {4} \alpha - 4 \tan ^ {2} \alpha }{ 1 - 2 \tan ^ {2} \alpha + \tan ^ {4} \alpha - 12 \tan ^ {2} \alpha } = { \frac{ 2 \tan \alpha } { 1 - \tan ^ {2} \alpha } } \cdot \frac{ 3 - 10 \tan ^ {2} \alpha + 3 \tan ^ {4} \alpha }{ 1 - 14 \tan ^ {2} \alpha + \tan ^ {4} \alpha } \\ \displaystyle = \frac{ 2 \tan \alpha \left( 3 - 10 \tan ^ {2} \alpha + 3 \tan ^ {4} \alpha \right) }{ 1 - 14 \tan ^ {2} \alpha + \tan ^ {4} \alpha - \tan ^ {2} \alpha + 14 \tan ^ {4} \alpha - \tan ^ {6} \alpha } \\ \displaystyle = \frac{ 6 \tan \alpha - 20 \tan ^ {3} \alpha + 6 \tan ^ {5} \alpha }{ 1 - 15 \tan ^ {2} \alpha + 15 \tan ^ {4} \alpha - \tan ^ {6} \alpha }$

LibreOffice 数式(Math) のソース：

alignl sin 6 %alpha = sin 3 ( 2 %alpha ) = sin 2 %alpha ( 3 - 4 sin ^2 2 %alpha ) = 2 sin %alpha cos %alpha lbrace 3 - 4 ( 2 sin %alpha cos %alpha ) ^2 rbrace
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alignl phantom { y } = 2 sin %alpha cos %alpha ( 3 - 16 sin ^2 %alpha cos ^2 %alpha ) = 2 sin %alpha cos %alpha lbrace 3 - 16 sin ^2 %alpha ( 1 - sin ^2 %alpha ) rbrace
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alignl phantom { y } = 2 sin %alpha cos %alpha ( 3 - 16 sin ^2 %alpha + 16 sin ^4 %alpha ) = cos %alpha ( 32 sin ^5 %alpha - 32 sin ^3 %alpha + 6 sin %alpha )

alignl cos 6 %alpha = cos 3 ( 2 %alpha ) = cos 2 %alpha ( 4 cos ^2 2 %alpha - 3 ) = ( 2 cos ^2%alpha - 1 ) lbrace 4 ( 2 cos ^2 %alpha - 1 )^2 -3 rbrace
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alignl phantom { y } = ( 2 cos ^2 %alpha - 1 ) ( 16 cos ^4 %alpha - 16 %alpha ^2 %alpha + 1 ) = 32 cos %alpha ^6 - 32 cos ^4 %alpha + 2 cos ^2 %alpha - 16 cos ^4 %alpha + 16 cos ^2 %alpha - 1
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alignl phantom { y } = 32 cos %alpha ^6 - 48 cos ^4 %alpha + 18 cos ^2 %alpha - 1

alignl tan 6 %alpha = tan 3 ( 2 %alpha ) = { alignc { tan 2 %alpha ( 3 - tan ^2 2 %alpha ) } over { 1 - 3 tan ^2 2 %alpha } } = { alignc { { { 2 tan %alpha } over { 1 - tan ^2 %alpha } } cdot left lbrace 3 - left ( { 2 tan %alpha } over { 1 - tan ^2 %alpha } right ) ^2 right rbrace } over { 1 - 3 left ( { 2 tan %alpha } over { 1 - tan ^2 %alpha } right )^2 } }
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alignl { phantom { y } = { alignc { { { 2 tan %alpha } over { 1 - tan ^2 %alpha } } cdot left lbrace { 3 ( 1 - tan ^2 %alpha ) ^2 - ( 2 tan %alpha ) ^2 } over { ( 1 - tan ^2 %alpha ) ^2 } right rbrace } over { { ( 1 - tan ^2 %alpha ) ^2 - 3 ( 2 tan %alpha ) ^2 } over { ( 1 - tan ^2 %alpha ) ^2 } } } } = { alignc { 2 tan %alpha ( 3 - 6 tan ^2 %alpha + 3 tan ^4 %alpha - 4 tan ^2 %alpha ) } over { ( 1 - tan ^2 %alpha ) ( 1 - 2 tan ^2 %alpha + tan ^4 %alpha - 12 tan ^2 %alpha ) } }
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alignl phantom { y } = { alignc { 2 tan %alpha ( 3 - 10 tan ^2 %alpha + 3 tan ^4 %alpha ) } over { ( 1 - tan ^2 %alpha ) ( 1 - 14 tan ^2 %alpha + tan ^4 %alpha ) } } = { alignc { 2 tan %alpha ( 3 - 10 tan ^2 %alpha + 3 tan ^4 %alpha ) } over { 1 - 14 tan ^2 %alpha + tan ^4 %alpha - tan ^2 %alpha + 14 tan ^4 %alpha - tan ^6 %alpha } }
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alignl phantom { y } = { alignc { 6 tan %alpha - 20 tan ^3 %alpha + 6 tan ^5 %alpha } over { 1 - 15 tan ^2 %alpha + 15 tan ^4 %alpha - tan ^6 %alpha } }