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三角関数 - 6倍角の公式(2)

三角関数の6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 $n = 6$ を代入し展開

$\cos 6 \theta + i \sin 6 \theta = ( \cos \theta + i \sin \theta )^{6} \\ \small = \cos^{6} \theta + 6 \cos^{5} \theta ( i \sin \theta ) + 15 \cos^{4} \theta ( i \sin \theta)^{2} + 20 \cos^{3} \theta ( i \sin \theta)^{3} + 15 \cos^{2} \theta ( i \sin \theta)^{4} + 6 \cos \theta ( i \sin \theta)^{5} + ( i \sin \theta)^{6} \\ = \cos^{6} \theta + 6 i \cos^{5} \theta \sin \theta - 15 \cos^{4} \theta \sin^{2} \theta - 20 i \cos^{3} \theta \sin^{3} \theta + 15 \cos^{2} \theta \sin^{4} \theta + 6 i \cos \theta \sin^{5} \theta - \sin^{6} \theta \\ = \cos^{6} \theta - 15 \cos^{4} \theta \sin^{2} \theta + 15 \cos^{2} \theta \sin^{4} \theta - \sin^{6} \theta + i \left( 6 \cos^{5} \theta \sin \theta - 20 \cos^{3} \theta \sin^{3} \theta + 6 \cos \theta \sin^{5} \theta \right)$

ここで実部と虚部に分けて、展開していきます。

実部

$\cos^{6} \theta - 15 \cos^{4} \theta \sin^{2} \theta + 15 \cos^{2} \theta \sin^{4} \theta - \sin^{6} \theta \\ = \cos^{6} \theta - 15 \cos^{4} \theta \left( 1 - \cos^{2} \theta \right) + 15 \cos^{2} \theta \left( 1 - \cos^{2} \theta \right)^{2} - \left( 1 - \cos^{2} \theta \right)^{3} \\ = \cos^{6} \theta - 15 \cos^{4} \theta \left( 1 - \cos^{2} \theta \right) + 15 \cos^{2} \theta \left( 1 - 2 \cos^{2} \theta + \cos^{4} \theta \right) - \left( 1 - 3 \cos^{2} \theta +3 \cos^{4} \theta - \cos^{6} \theta \right) \\ = \left( 1 + 15 + 15 + 1 \right) \cos^{6} \theta - \left( 15 + 15 \cdot 2 + 3 \right) \cos^{4} \theta + \left( 15 + 3 \right) \cos^{2} \theta - 1 \\ = 32 \cos^{6} \theta - 48 \cos^{4} \theta + 18 \cos^{2} \theta - 1$

∴ $\cos 6 \theta = 32 \cos^{6} \theta - 48 \cos^{4} \theta + 18 \cos^{2} \theta - 1$

虚部

$6 \cos^{5} \theta \sin \theta - 20 \cos^{3} \theta \sin^{3} \theta + 6 \cos \theta \sin^{5} \theta = 2 \cos \theta \sin \theta \left( 3 \cos^{4} \theta - 10 \cos^{2} \theta \sin^{2} \theta + 3 \sin^{4} \theta \right) \\ = 2 \cos \theta \sin \theta \left\{ 3 \left( 1 - \sin^{2} \theta \right)^{2} - 10 \left( 1 - \sin^{2} \theta \right) \sin^{2} \theta + 3 \sin^{4} \theta \right\} \\ = 2 \cos \theta \sin \theta \left\{ 3 \left( 1 - 2 \sin^{2} \theta + \sin^{4} \theta \right) - 10 \sin^{2} \theta \left( 1 - \sin^{2} \theta \right) + 3 \sin^{4} \theta \right\} \\ = 2 \cos \theta \sin \theta \left\{ ( 3 + 10 + 3 ) \sin^{4} \theta - ( 3 \cdot 2 + 10 ) \sin^{2} \theta + 3 \right\} \\ = 2 \cos \theta \sin \theta \left( 16 \sin^{4} \theta - 16 \sin^{2} \theta + 3 \right) = \cos \theta \left( 32 \sin^{5} \theta - 32 \sin^{3} \theta + 6 \sin \theta \right)$
∴ $\sin 6 \theta = \cos \theta \left( 32 \sin^{5} \theta - 32 \sin^{3} \theta + 6 \sin \theta \right)$

$\tan 6 \theta$ の導出

$\displaystyle \tan 6 \theta = \frac{ \sin 6 \theta }{ \cos 6 \theta } = \frac{ 2 \cos \theta \sin \theta \left( 16 \sin^{4} \theta - 16 \sin^{2} \theta + 3 \right) }{ 32 \cos^{6} \theta - 48 \cos^{4} \theta + 18 \cos^{2} \theta - 1 } \\ \displaystyle = \dfrac{ \dfrac{ 2 \sin \theta }{ \cos \theta } \left( \dfrac{ 16 \sin^{4} \theta }{ \cos^{4} \theta } - \dfrac {16 \sin^{2} \theta }{ \cos^{2} \theta } \cdot \dfrac{ 1 }{ \cos^{2} \theta } + \dfrac{ 3 }{ \cos^{4} \theta } \right) }{ 32 - \dfrac{ 48 }{ \cos^{2} \theta } + \dfrac{ 18 }{ \cos^{4} \theta } - \dfrac{ 1 }{ \cos^{6} \theta } } \\ \displaystyle = \frac{ 2 \tan \theta \left\{ 16 \tan^{4} \theta - 16 \tan^{2} \theta \left( 1 + \tan^{2} \theta \right) + 3 \left( 1 + \tan^{2} \theta \right)^{2} \right\} }{ 32 - 48 \left( 1 + \tan^{2} \theta \right) + 18 \left( 1 + \tan^{2} \theta \right)^{2} - \left( 1 + \tan^{2} \theta \right)^{3} } \\ \displaystyle = \frac{ 2 \tan \theta \left\{ 16 \tan^{4} \theta - 16 \tan^{2} \theta \left( 1 + \tan^{2} \theta \right) + 3 \left( 1 + 2 \tan^{2} \theta + \tan^{4} \theta \right) \right\} }{ 32 - 48 \left( 1 + \tan^{2} \theta \right) + 18 \left( 1 + 2 \tan^{2} \theta + \tan^{4} \theta \right) - \left( 1 + 3 \tan^{2} \theta + 3 \tan^{4} \theta + \tan^{6} \theta \right) } \\ \displaystyle = \frac{ 2 \tan \theta \left\{ 3 - ( 16 - 3 \cdot 2 ) \tan^{2} \theta + ( 16 - 16 + 3 ) \tan^{4} \theta \right\} }{ ( 32 - 48 + 18 - 1 ) - ( 48 - 18 \cdot 2 + 3 ) \tan^{2} \theta + ( 18 - 3 ) \tan^{4} \theta - \tan^{6} \theta } \\ \displaystyle = \frac{ 2 \tan \theta \left( 3 - 10 \tan^{2} \theta + 3 \tan^{4} \theta \right) }{ 1 - 15 \tan^{2} \theta + 15 \tan^{4} \theta - \tan^{6} \theta } = \frac{ 6 \tan \theta - 20 \tan^{3} \theta + 6 \tan^{5} \theta }{ 1 - 15 \tan^{2} \theta + 15 \tan^{4} \theta - \tan^{6} \theta }$