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 - 10倍角の公式(2)

イメージ 1

イメージ 2

三角関数の10倍角の公式をド・モアブルの定理を用いて導出してみました。
展開時の係数は、パスカルの三角形を用います。

準備

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

$\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$

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

$\cos n \theta + i \sin n \theta = ( \cos \theta + i \sin \theta )^{n}$

導出

ド・モアブルの定理で $n = 10$ を代入し展開

$\cos 10 \theta + i \sin 10 \theta = ( \cos \theta + i \sin \theta )^{10} \\ = \cos ^{10} \theta + 10 \cos^{9} \theta ( i \sin \theta ) + 45 \cos^{8} \theta ( i \sin \theta )^{2} + 120 \cos^{7} \theta ( i \sin \theta )^{3} + 210 \cos^{6} \theta ( i \sin \theta )^{4} \\ + 252 \cos^{5} \theta ( i \sin \theta )^{5} + 210 \cos^{4} \theta ( i \sin \theta )^{6} + 120 \cos^{3} \theta ( i \sin \theta )^{7} + 45 \cos^{2} \theta ( i \sin \theta )^{8} \\ + 10 \cos \theta ( i \sin \theta )^{9} + ( i \sin \theta )^{10} \\ = \cos ^{10} \theta + 10 i \cos^{9} \theta \sin \theta - 45 \cos^{8} \theta \sin^{2} \theta - 120 i \cos^{7} \theta \sin^{3} \theta + 210 \cos^{6} \theta \sin^{4} \theta \\ + 252 i \cos^{5} \theta \sin^{5} \theta - 210 \cos^{4} \theta \sin^{6} \theta - 120 i \cos^{3} \theta \sin^{7} \theta + 45 \cos^{2} \theta \sin^{8} \theta + 10 i \cos \theta \sin^{9} \theta \\ - \sin^{10} \theta \\ = \cos ^{10} \theta - 45 \cos^{8} \theta \sin^{2} \theta + 210 \cos^{6} \theta \sin^{4} \theta - 210 \cos^{4} \theta \sin^{6} \theta + 45 \cos^{2} \theta \sin^{8} - \sin^{10} \theta \\ + i \left( 10 \cos^{9} \theta \sin \theta - 120 \cos^{7} \theta \sin^{3} \theta + 252 \cos^{5} \theta \sin^{5} \theta - 120 \cos^{3} \theta \sin^{7} \theta + 10 \cos \theta \sin^{9} \theta \right)$

ここで実部と虚部に分けて、展開していきます。

実部

$\cos ^{10} \theta - 45 \cos^{8} \theta \sin^{2} \theta + 210 \cos^{6} \theta \sin^{4} \theta - 210 \cos^{4} \theta \sin^{6} \theta + 45 \cos^{2} \theta \sin^{8} \theta - \sin^{10} \theta \\ = \cos ^{10} \theta - 45 \cos^{8} \theta \left( 1 - \cos ^{2} \theta \right) + 210 \cos^{6} \theta \left( 1 - \cos ^{2} \theta \right)^{2} - 210 \cos^{4} \theta \left( 1 - \cos ^{2} \theta \right)^{3} \\ + 45 \cos^{2} \theta \left( 1 - \cos ^{2} \theta \right)^{4} - \left( 1 - \cos ^{2} \theta \right)^{5} \\ = \cos ^{10} \theta - 45 \cos^{8} \theta \left( 1 - \cos ^{2} \theta \right) + 210 \cos^{6} \theta \left( 1 - 2 \cos ^{2} \theta + 2 \cos ^{4} \theta \right) \\ - 210 \cos^{4} \theta \left( 1 - 3 \cos ^{2} \theta + 3 \cos ^{4} \theta - \cos ^{6} \theta \right) \\ + 45 \cos^{2} \theta \left( 1 - 4 \cos ^{2} \theta + 6 \cos ^{4} \theta - 4 \cos ^{6} \theta + \cos ^{8} \theta \right) \\ - \left( 1 - 5 \cos ^{2} \theta + 10 \cos ^{4} \theta - 10 \cos ^{6} \theta + 5 \cos ^{8} \theta - \cos ^{10} \theta \right) \\ = ( 1+ 45 + 210 + 210 + 45 + 1 ) \cos ^{10} \theta - ( 45 + 210 \cdot 2 + 210 \cdot 3 + 45 \cdot 4 + 5 ) \cos ^{8} \theta \\ + ( 210 + 210 \cdot 3 + 45 \cdot 6 + 10 ) \cos ^{6} \theta - ( 210 + 45 \cdot 4 + 10 ) \cos ^{4} \theta + ( 45 + 5 ) \cos ^{2} \theta - 1 \\ = 512 \cos ^{10} \theta - 1280 \cos ^{8} \theta + 1120 \cos ^{6} \theta - 400 \cos ^{4} \theta + 50 \cos ^{2} \theta - 1$

∴ $\cos 10 \theta = 512 \cos ^{10} \theta - 1280 \cos ^{8} \theta + 1120 \cos ^{6} \theta - 400 \cos ^{4} \theta + 50 \cos ^{2} \theta - 1$

虚部

$10 \cos^{9} \theta \sin \theta - 120 \cos^{7} \theta \sin^{3} \theta + 252 \cos^{5} \theta \sin^{5} \theta - 120 \cos^{3} \theta \sin^{7} \theta + 10 \cos \theta \sin^{9} \theta \\ = 2 \sin \theta \cos \theta \left( 5 \cos^{8} \theta - 60 \cos^{6} \theta \sin^{2} \theta + 126 \cos^{4} \theta \sin^{4} \theta - 60 \cos^{2} \theta \sin^{6} \theta + 5 \sin^{8} \theta \right)$

ここで、 $A = 5 \cos^{8} \theta - 60 \cos^{6} \theta \sin^{2} \theta + 126 \cos^{4} \theta \sin^{4} \theta - 60 \cos^{2} \theta \sin^{6} \theta + 5 \sin^{8} \theta$ と置き、 $A$ を展開していきます。

$A = 5 \left( 1 - \sin^{2} \theta \right)^{4} - 60 \left( 1 - \sin^{2} \theta \right)^{3} \sin^{2} \theta + 126 \left( 1 - \sin^{2} \theta \right)^{2} \sin^{4} \theta - 60 \left( 1 - \sin^{2} \theta \right) \sin^{6} \theta + 5 \sin^{8} \theta \\ = 5 \left( 1 - 4 \sin^{2} \theta + 6 \sin^{4} \theta - \sin^{6} \theta + \sin^{8} \theta \right) - 60 \left( 1 - 3 \sin^{2} \theta + 3 \sin^{4} \theta - \sin^{6} \theta \right) \sin^{2} \theta \\ + 126 \left( 1 - 2 \sin^{2} \theta + \sin^{4} \theta \right) \sin^{4} \theta - 60 \left( 1 - \sin^{2} \theta \right) \sin^{6} \theta + 5 \sin^{8} \theta \\ = ( 5 + 60 + 126 + 60 + 5 ) \sin^{8} \theta - ( 5 \cdot 4 + 60 \cdot 3 + 126 \cdot 2 + 60 ) \sin^{6} \theta + ( 5 \cdot 6 + 60 \cdot 3 + 126 ) \sin^{4} \theta \\ - ( 5 \cdot 4 + 60 ) \sin^{2} \theta + 5 \\ = 256 \sin^{8} \theta - 512 \sin^{6} \theta + 336 \sin^{4} \theta - 80 \sin^{2} \theta + 5$

∴ $\sin 10 \theta = 2 \sin \theta \cos \theta A \\ = 2 \sin \theta \cos \theta \left( 256 \sin^{8} \theta - 512 \sin^{6} \theta + 336 \sin^{4} \theta - 80 \sin^{2} \theta + 5 \right) \\ = \cos \theta \left( 512 \sin^{9} \theta - 1024 \sin^{7} \theta + 672 \sin^{5} \theta - 160 \sin^{3} \theta + 10 \sin \theta \right)$

$\tan 10 \theta$ の導出

これら $\sin 10 \theta$ ， $\cos 10 \theta$ から、

$\displaystyle \tan 10 \theta = \frac{\sin 10 \theta}{\cos 10 \theta} = \frac{2 \sin \theta \cos \theta \left( 256 \sin^{8} \theta - 512 \sin^{6} \theta + 336 \sin^{4} \theta - 80 \sin^{2} \theta + 5 \right)}{512 \cos ^{10} \theta - 1280 \cos ^{8} \theta + 1120 \cos ^{6} \theta - 400 \cos ^{4} \theta + 50 \cos ^{2} \theta - 1} \\ \displaystyle = \frac{ \dfrac{ 2 \sin \theta }{ \cos \theta } \left( \dfrac{ 256 \sin^{8} \theta }{ \cos^{8} \theta } - \dfrac{ 512 \sin^{6} \theta }{ \cos^{6} \theta } \cdot \dfrac{ 1 }{ \cos^{2} \theta } + \dfrac{ 336 \sin^{4} \theta }{ \cos^{4} \theta } \cdot \dfrac{ 1 }{ \cos^{4} \theta } - \dfrac{ 80 \sin^{2} \theta }{ \cos^{2} \theta } \cdot \dfrac{ 1 }{ \cos^{6} \theta } + \dfrac{ 5 }{ \cos^{8} \theta } \right) }{ 512 - \dfrac{ 1280 }{ \cos^{2} \theta } + \dfrac{ 1120 }{ \cos^{4} \theta } - \dfrac{ 400 }{ \cos^{6} \theta } + \dfrac{ 50 }{ \cos^{8} \theta } - \dfrac{ 1 }{ \cos^{10} \theta } } \\ = \frac{ 2 \tan \theta \left\{ 256 \tan^{8} \theta - 512 \tan^{6} \theta \left( \tan^{2} \theta + 1 \right) + 336 \tan^{4} \theta \left( \tan^{2} \theta + 1 \right)^{2} - 80 \tan^{2} \theta \left( \tan^{2} \theta + 1 \right)^{3} + 5 \left( \tan^{2} \theta + 1 \right)^{4} \right\} }{ 512 - 1280 \left( \tan^{2} \theta + 1 \right) + 1120 \left( \tan^{2} \theta + 1 \right)^{2} - 400 \left( \tan^{2} \theta + 1 \right)^{3} + 50 \left( \tan^{2} \theta + 1 \right)^{4} - \left( \tan^{2} \theta + 1 \right)^{5} } \\ \small = \frac{ 2 \tan \theta \left\{ 256 \tan^{8} \theta -512 \tan^{6} \theta \left( \tan^{2} \theta + 1 \right) + 336 \tan^{4} \theta \left( \tan^{4} \theta + 2 \tan^{2} \theta + 1 \right) - 80 \tan^{2} \theta \left( \tan^{6} \theta + 3 \tan^{4} \theta + 3 \tan^{2} \theta + 1 \right) + 5 \left( \tan^{8} \theta + 4 \tan^{6} \theta + 6 \tan^{4} \theta + 4 \tan^{2} \theta + 1 \right) \right\} }{ 512 - 1280 \left( \tan^{2} \theta + 1 \right) + 1120 \left( \tan^{4} \theta + 2 \tan^{2} \theta + 1 \right) - 400 \left( \tan^{6} \theta + 3 \tan^{4} \theta + 3 \tan^{2} \theta + 1 \right) + 50 \left( \tan^{8} \theta + 4 \tan^{6} \theta + 6 \tan^{4} \theta + 4 \tan^{2} \theta + 1 \right) - \left( \tan^{10} \theta + 5 \tan^{8} \theta + 10 \tan^{6} \theta + 10 \tan^{4} \theta + 5 \tan^{2} \theta + 1 \right) } \\ = \frac{ 2 \tan \theta \left\{ ( 256 - 512 + 336 - 80 + 5 ) \tan^{8} \theta - ( 512 -336 \cdot 2 + 80 \cdot 3 - 5 \cdot 4 ) \tan^{6} \theta + ( 336 - 80 \cdot 3 + 5 \cdot 6 ) \tan^{4} \theta - ( 80 - 5 \cdot 4 ) \tan^{2} \theta + 5 \right\} }{ ( 512 - 1280 + 1120 - 400 + 50 - 1 ) - ( 1280 - 1120 \cdot 2 + 400 \cdot 3 - 50 \cdot 4 + 5 ) \tan^{2} \theta + ( 1120 - 400 \cdot 3 + 50 \cdot 6 - 10 ) \tan^{4} \theta - ( 400 - 50 \cdot 4 + 10 ) \tan^{6} \theta + ( 50 - 5 ) \tan^{8} \theta - \tan^{10} \theta } \\ \displaystyle = \frac{ 2 \tan \theta \left( 5 \tan^{8} \theta - 60 \tan^{6} \theta + 126 \tan^{4} \theta - 60 \tan^{2} \theta + 5 \right) }{ 1 - 45 \tan^{2} \theta + 210 \tan^{4} \theta - 210 \tan^{6} \theta + 45 \tan^{8} \theta - \tan^{10} \theta } \\ \displaystyle = \frac{ 10 \tan \theta - 120 \tan^{3} \theta + 252 \tan^{5} \theta - 120 \tan^{7} \theta + 10 \tan^{9} \theta }{ 1 - 45 \tan^{2} \theta + 210 \tan^{4} \theta - 210 \tan^{6} \theta + 45 \tan^{8} \theta - \tan^{10} \theta }$