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三角関数 - 8倍角の公式(2)

三角関数の8倍角の公式をド・モアブルの定理を用いて導出します。

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 $n = 8$ を代入し展開

$\cos 8 \theta + i \sin 8 \theta = ( \cos \theta + i \sin \theta )^{8} \\ = \cos^{8} \theta + 8 \cos^{7} \theta ( i \sin \theta ) + 28 \cos^{6} \theta ( i \sin \theta )^{2} + 56 \cos^{5} \theta ( i \sin \theta )^{3} + 70 \cos^{4} \theta ( i \sin \theta )^{4} \\ + 56 \cos^{3} \theta ( i \sin \theta )^{5} + 28 \cos^{2} \theta ( i \sin \theta )^{6} + 8 \cos \theta ( i \sin \theta )^{7} + ( i \sin \theta )^{8} \\ = \cos^{8} \theta + 8 i \cos^{7} \theta \sin \theta - 28 \cos^{6} \theta \sin^{2} \theta - 56 i \cos^{5} \theta \sin^{3} \theta + 70 \cos^{4} \theta \sin^{4} \theta + 56 i \cos^{3} \theta \sin^{5} \theta \\ - 28 \cos^{2} \theta \sin^{6} \theta - 8 i \cos \theta \sin^{7} \theta+ \sin^{8} \theta \\ = \cos^{8} \theta - 28 \cos^{6} \theta \sin^{2} \theta + 70 \cos^{4} \theta \sin^{4} \theta - 28 \cos^{2} \theta \sin^{6} \theta + \sin^{8} \theta \\ + i \left( 8 \cos^{7} \theta \sin \theta - 56 \cos^{5} \theta \sin^{3} \theta + 56 \cos^{3} \theta \sin^{5} \theta - 8 \cos \theta \sin^{7} \theta \right)$

ここで実部と虚部に分けて、展開していきます。

実部

$\cos^{8} \theta - 28 \cos^{6} \theta \sin^{2} \theta + 70 \cos^{4} \theta \sin^{4} \theta - 28 \cos^{2} \theta \sin^{6} \theta + \sin^{8} \theta \\ = \cos^{8} \theta - 28 \cos^{6} \theta \left( 1 - \cos^{2} \theta \right) + 70 \cos^{4} \theta \left( 1 - \cos^{2} \theta \right)^{2} - 28 \cos^{2} \theta \left( 1 - \cos^{2} \theta \right)^{3} + \left( 1 - \cos^{2} \theta \right)^{4} \\ = \cos^{8} \theta - 28 \cos^{6} \theta \left( 1 - \cos^{2} \theta \right) + 70 \cos^{4} \theta \left( 1 - 2 \cos^{2} \theta + \cos^{4} \theta \right) \\ - 28 \cos^{2} \theta \left( 1 - 3 \cos^{2} \theta + 3 \cos^{4} \theta + \cos^{6} \theta \right) + \left( 1 - 4 \cos^{2} \theta + 6 \cos^{4} \theta - 4 \cos^{6} \theta + \cos^{8} \theta \right) \\ = ( 1 + 28 + 70 + 28 + 1 ) \cos^{8} \theta - ( 28 + 70 \cdot 2 + 28 \cdot 3 + 4 ) \cos^{6} \theta + ( 70 + 28 \cdot 3 + 6 ) \cos^{4} \theta - ( 28 + 4 ) \cos^{2} \theta + 1 \\ = 128 \cos^{8} \theta - 256 \cos^{6} \theta + 160 \cos^{4} \theta - 32 \cos^{2} \theta + 1$

∴ $\cos 8 \theta = 128 \cos^{8} \theta - 256 \cos^{6} \theta + 160 \cos^{4} \theta - 32 \cos^{2} \theta + 1$

虚部

$8 \cos^{7} \theta \sin \theta - 56 \cos^{5} \theta \sin^{3} \theta + 56 \cos^{3} \theta \sin^{5} \theta - 8 \cos \theta \sin^{7} \theta \\ = 8 \cos \theta \sin \theta \left( \cos^{6} \theta - 7 \cos^{4} \theta \sin^{2} \theta + 7 \cos^{2} \theta \sin^{4} \theta - \sin^{6} \theta \right) \\ = 8 \cos \theta \sin \theta \left\{ \left( 1 - \sin^{2} \theta \right)^{3} - 7 \left( 1 - \sin^{2} \theta \right)^{2} \sin^{2} \theta + 7 \left( 1 - \sin^{2} \theta \right) \sin^{4} \theta - \sin^{6} \theta \right\} \\ = 8 \cos \theta \sin \theta \left\{ \left( 1 - 3 \sin^{2} \theta + 3 \sin^{4} \theta - \sin^{6} \theta \right) - 7 \left( 1 - 2 \sin^{2} \theta + \sin^{4} \theta \right) \sin^{2} \theta + 7 \left( 1 - \sin^{2} \theta \right) \sin^{4} \theta - \sin^{6} \theta \right\} \\ = 8 \cos \theta \sin \theta \left\{ 1 - ( 3 + 7 ) \sin^{2} \theta + ( 3 + 7 \cdot 2 + 7 ) \sin^{4} \theta - ( 1 + 7 + 7 + 1 ) \sin^{6} \theta \right\} \\ = 8 \cos \theta \sin \theta \left( 1 - 10 \sin^{2} \theta + 24 \sin^{4} \theta - 16 \sin^{6} \theta \right) \\ = \cos \theta \left( 8 \sin \theta - 80 \sin^{3} \theta + 192 \sin^{5} \theta - 128 \sin^{7} \theta \right)$

∴ $\sin 8 \theta = \cos \theta \left( 8 \sin \theta - 80 \sin^{3} \theta + 192 \sin^{5} \theta - 128 \sin^{7} \theta \right)$

$\tan 8 \theta$ の導出

これら $\sin 8 \theta$ ， $\cos 8 \theta$ から、

$\displaystyle \tan 8 \theta = \frac{\sin 8 \theta}{\cos 8 \theta} = \frac{ 8 \cos \theta \sin \theta \left( 1 - 10 \sin^{2} \theta + 24 \sin^{4} \theta - 16 \sin^{6} \theta \right) }{ 128 \cos^{8} \theta - 256 \cos^{6} \theta + 160 \cos^{4} \theta - 32 \cos^{2} \theta + 1 } \\ \displaystyle = \frac{ \dfrac{ 8 \sin \theta }{ \cos \theta } \left( \dfrac{ 1 }{ \cos^{6} \theta } - \dfrac{ 10 \sin^{2} \theta }{ \cos^{2} \theta } \cdot \dfrac{ 1 }{ \cos^{4} \theta } + \dfrac{ 24 \sin^{4} \theta }{ \cos^{4} \theta } \cdot \dfrac{ 1 }{ \cos^{2} \theta } - \dfrac{ 16 \sin^{6} \theta }{ \cos^{6} \theta } \right) }{ 128 - \dfrac{ 256 }{ \cos^{2} \theta } + \dfrac{ 160 }{ \cos^{4} \theta } - \dfrac{ 32 }{ \cos^{6} \theta } + \dfrac{ 1 }{ \cos^{8} \theta } } \\ \displaystyle = \frac{ 8 \tan \theta \left\{ \left( 1 + \tan^{2} \theta \right)^{3} - 10 \tan^{2} \theta \left( 1 + \tan^{2} \theta \right)^{2} + 24 \tan^{4} \left( 1 + \tan^{2} \theta \right) - 16 \tan^{6} \theta \right\} }{ 128 - 256 \left( 1 + \tan^{2} \theta \right) + 160 \left( 1 + \tan^{2} \theta \right)^{2} - 32 \left( 1 + \tan^{2} \theta \right)^{3} + \left( 1 + \tan^{2} \theta \right)^{4} } \\ = \frac{ 8 \tan \theta \left\{ \left( 1 + 3 \tan^{2} \theta + 3 \tan^{4} \theta + \tan^{6} \theta \right) - 10 \tan^{2} \theta \left( 1 + 2 \tan^{2} \theta + \tan^{4} \theta \right) + 24 \tan^{4} \left( 1 + \tan^{2} \theta \right) - 16 \tan^{6} \theta \right\} }{ 128 - 256 \left( 1 + \tan^{2} \theta \right) + 160 \left( 1 + 2 \tan^{2} \theta + \tan^{4} \theta \right) - 32 \left( 1 + 3 \tan^{2} \theta + 3 \tan^{4} \theta + \tan^{6} \theta \right) + \left( 1 + 4 \tan^{2} \theta + 6 \tan^{4} \theta + 4 \tan^{6} \theta + \tan^{8} \theta \right) } \\ = \frac{ 8 \tan \theta \left\{ 1 - ( - 1 + 10 ) \tan^{2} \theta + ( 3 - 10 \cdot 2 + 24 ) \tan^{4} \theta - ( - 1 + 10 - 24 + 16 ) \tan^{6} \theta \right\} }{ ( 128 - 256 + 160 - 32 + 1 ) - ( 256 - 160 \cdot 2 + 32 \cdot 3 - 4 ) \tan^{2} \theta + ( 160 - 32 \cdot 3 + 6 ) \tan^{4} \theta - ( 32 - 4 ) \tan^{6} \theta + \tan^{8} \theta } \\ \displaystyle = \frac{ 8 \tan \theta \left( 1 - 7 \tan^{2} \theta + 7 \tan^{4} \theta - \tan^{6} \theta \right) }{ 1 - 28 \tan^{2} \theta + 70 \tan^{4} \theta -28 \tan^{6} \theta + \tan^{8} \theta } \\ \displaystyle = \frac{ 8 \tan \theta - 56 \tan^{3} \theta + 56 \tan^{5} \theta - 8 \tan^{7} \theta }{ 1 - 28 \tan^{2} \theta + 70 \tan^{4} \theta -28 \tan^{6} \theta + \tan^{8} \theta }$