3分でわかる！交流の消費電力について

三角関数

補1

$\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha sin\beta$
$\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

$\alpha =\beta$

$$\begin{array}{rcl}\cos (\alpha +\alpha )& =& \cos 2\alpha \\ & =& \cos \alpha \cos \alpha –\sin \alpha \sin \alpha \\ & =& {\cos }^2\alpha -{\sin }^2\alpha \\ \\ \\ & =& (1-{\sin }^2\alpha )-{\sin }^2\alpha \\ & =& 1-2{\sin }^2\alpha \end{array}$$

$${\sin }^2\alpha ={\displaystyle \frac{1-\cos 2\alpha }{2}}$$

補2

$\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha sin\beta$
$\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

$\beta ={\displaystyle \frac{\pi }{2}}$

$$\begin{array}{rcl}\sin (\alpha \pm {\displaystyle \frac{\pi }{2}})& =& \sin \alpha \cos {\displaystyle \frac{\pi }{2}}\pm \cos \alpha \sin {\displaystyle \frac{\pi }{2}}\\ & =& \pm \cos \alpha \end{array}$$

補3

$\alpha =\beta$

$$\begin{array}{rcl}\sin (\alpha +\alpha )& =& \sin 2\alpha \\ & =& \sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ & =& 2\sin \alpha \cos \alpha \end{array}$$

$$\sin \alpha \cos \alpha ={\displaystyle \frac{1}{2}}\sin 2\alpha$$