ド・モアブルの定理(de Moivre's theorem)と三角凾数(trigonometric function)の3倍角の公式

ド・モアブルの定理

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

3倍角の公式(Triple-angle formulae)

$$\begin{array}{rcl}{\displaystyle (\cos \theta +i\sin \theta {)}^3}& =& {\displaystyle {\cos }^3\theta +3{\cos }^2\theta (i\sin \theta )+3\cos \theta (i\sin \theta {)}^2+(i\sin \theta {)}^3}\\ & =& {\displaystyle {\cos }^3\theta +i3{\cos }^2\theta \sin \theta -3\cos \theta {\sin }^2\theta -i{\sin }^3\theta }\\ & =& {\displaystyle ({\cos }^3\theta -3\cos \theta {\sin }^2\theta )+i(3{\cos }^2\theta \sin \theta -{\sin }^3\theta )}\\ & =& {\displaystyle \cos \left(3\theta \right)+i\sin \left(3\theta \right)\dots \mathrm{こ}\mathrm{れ}\mathrm{が}\mathrm{ド}\mathrm{・}\mathrm{モ}\mathrm{ア}\mathrm{ブ}\mathrm{ル}\mathrm{の}\mathrm{定}\mathrm{理}\mathrm{の}\mathrm{結}\mathrm{果}\mathrm{と}\mathrm{等}\mathrm{し}\mathrm{い}．}\end{array}$$ $$\begin{array}{rcl}{\displaystyle \cos \left(3\theta \right)}& =& {\cos }^3\theta -3\cos \theta {\sin }^2\theta \\ & =& {\cos }^3\theta -3\cos \theta (1-{\cos }^2\theta )\\ & =& {\cos }^3\theta -3\cos \theta +3{\cos }^3\theta \\ & =& 4{\cos }^3\theta -3\cos \theta \\ {\displaystyle \sin \left(3\theta \right)}& =& 3{\cos }^2\theta \sin \theta -{\sin }^3\theta \\ & =& 3\sin \theta (1-{\sin }^2\theta )-{\sin }^3\theta \\ & =& 3\sin \theta -3{\sin }^3\theta -{\sin }^3\theta \\ & =& 3\sin \theta -4{\sin }^3\theta \end{array}$$

4倍角の場合

$$\begin{array}{rcl}{\displaystyle (\cos \theta +i\sin \theta {)}^4}& =& {\displaystyle {\cos }^4\theta +4{\cos }^3\theta (i\sin \theta )+6{\cos }^2\theta (i\sin \theta {)}^2+4\cos \theta (i\sin \theta {)}^3+(i\sin \theta {)}^4}\\ & =& {\displaystyle {\cos }^4\theta +i4{\cos }^3\theta \sin \theta -6{\cos }^2\theta {\sin }^2\theta -i4\cos \theta {\sin }^3\theta +{\sin }^4\theta }\\ & =& {\displaystyle ({\cos }^4\theta -6{\cos }^2\theta {\sin }^2\theta +{\sin }^4\theta )+i(4{\cos }^3\theta \sin \theta -4\cos \theta {\sin }^3\theta )}\\ & =& {\displaystyle \cos \left(4\theta \right)+i\sin \left(4\theta \right)\dots \mathrm{こ}\mathrm{れ}\mathrm{が}\mathrm{ド}\mathrm{・}\mathrm{モ}\mathrm{ア}\mathrm{ブ}\mathrm{ル}\mathrm{の}\mathrm{定}\mathrm{理}\mathrm{の}\mathrm{結}\mathrm{果}\mathrm{と}\mathrm{等}\mathrm{し}\mathrm{い}．}\end{array}$$ $$\begin{array}{rcl}{\displaystyle \cos \left(4\theta \right)}& =& {\cos }^4\theta -6{\cos }^2\theta {\sin }^2\theta +{\sin }^4\theta \\ & =& {\cos }^4\theta -6{\cos }^2\theta (1-{\cos }^2\theta )+(1-{\cos }^2\theta {)}^2\\ & =& {\cos }^4\theta -6{\cos }^2\theta +6{\cos }^4\theta +1-2{\cos }^2\theta +{\cos }^4\theta \\ & =& {\cos }^4\theta +6{\cos }^4\theta +{\cos }^4\theta -6{\cos }^2\theta -2{\cos }^2\theta +1\\ & =& 8{\cos }^4\theta -8{\cos }^2\theta +1\\ {\displaystyle \sin \left(4\theta \right)}& =& 4{\cos }^3\theta \sin \theta -4\cos \theta {\sin }^3\theta \\ & =& \cos \theta (4{\cos }^2\theta \sin \theta -4{\sin }^3\theta )\\ & =& \cos \theta (4\sin \theta (1-{\sin }^2\theta )-4{\sin }^3\theta )\\ & =& \cos \theta (4\sin \theta -4{\sin }^3\theta -4{\sin }^3\theta )\\ & =& \cos \theta (4\sin \theta -8{\sin }^2\theta )\end{array}$$