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

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

3倍角の公式(Triple-angle formulae)

$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^3 &=& \displaystyle \cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\\ &=& \displaystyle \cos^3\theta+i3\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\\ &=& \displaystyle \left(\cos^3\theta-3\cos\theta\sin^2\theta\right)+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)\\ &=& \displaystyle \cos(3\theta)+i\sin(3\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(3\theta)&=& \cos^3\theta-3\cos\theta\sin^2\theta\\ &=& \cos^3\theta-3\cos\theta(1-\cos^2\theta)\\ &=& \cos^3\theta-3\cos\theta+3\cos^3\theta\\ &=& 4\cos^3\theta-3\cos\theta\\ \displaystyle \sin(3\theta)&=& 3\cos^2\theta\sin\theta-\sin^3\theta\\ &=& 3\sin\theta(1-\sin^2\theta)-\sin^3\theta\\ &=& 3\sin\theta-3\sin^3\theta-\sin^3\theta\\ &=& 3\sin\theta-4\sin^3\theta\\ \end{array}$$

4倍角の場合

$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^4 &=& \displaystyle \cos^4\theta+4\cos^3\theta(i\sin\theta)+6\cos^2\theta(i\sin\theta)^2+4\cos\theta(i\sin\theta)^3+(i\sin\theta)^4\\ &=& \displaystyle \cos^4\theta+i4\cos^3\theta\sin\theta-6\cos^2\theta\sin^2\theta-i4\cos\theta\sin^3\theta+\sin^4\theta\\ &=& \displaystyle \left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)+i\left(4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\right)\\ &=& \displaystyle \cos(4\theta)+i\sin(4\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(4\theta)&=&\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ &=&\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-\cos^2\theta)^2\\ &=&\cos^4\theta-6\cos^2\theta+6\cos^4\theta+1-2\cos^2\theta+\cos^4\theta\\ &=&\cos^4\theta+6\cos^4\theta+\cos^4\theta-6\cos^2\theta-2\cos^2\theta+1\\ &=&8\cos^4\theta-8\cos^2\theta+1\\ \displaystyle \sin(4\theta)&=&4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\\ &=&\cos\theta(4\cos^2\theta\sin\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta(1-\sin^2\theta)-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-4\sin^3\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-8\sin^2\theta)\\ \end{array}$$