回転行列と三角凾数の加法定理

角度\(\alpha\)と角度\(\beta\)の回転行列を掛ける

$$\begin{array}{rcl} \begin{pmatrix} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{pmatrix} \begin{pmatrix} cos(\beta) & -sin(\beta) \\ sin(\beta) & cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) & -cos(\alpha)sin(\beta)-sin(\alpha)cos(\beta) \\ sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta) & -sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) & -(sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)) \\ sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta) & -sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha+\beta) & -sin(\alpha+\beta) \\ sin(\alpha+\beta) & cos(\alpha+\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(\alpha+\beta)の回転行列と等しい．\\ \end{array}$$ 対応する要素同士が加法定理となる． $$\begin{array}{rcl} cos(\alpha+\beta)&=&cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\ sin(\alpha+\beta)&=&sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)\\ \end{array}$$

角度\(\alpha\)と角度\(-\beta\)の回転行列を掛ける

$$\begin{array}{rcl} \begin{pmatrix} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{pmatrix} \begin{pmatrix} cos(-\beta) & -sin(-\beta) \\ sin(-\beta) & cos(-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\alpha)cos(-\beta)-sin(\alpha)sin(-\beta) & -cos(\alpha)sin(-\beta)-sin(\alpha)cos(-\beta) \\ sin(\alpha)cos(-\beta)+cos(\alpha)sin(-\beta) & -sin(\alpha)sin(-\beta)+cos(\alpha)cos(-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) & cos(\alpha)sin(\beta)-sin(\alpha)cos(\beta) \\ sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta) & sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,cos(-\beta)=cos(\beta), sin(-\beta)=-sin(\beta)を適用．\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) & -(sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)) \\ sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta) & cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha-\beta) & -sin(\alpha-\beta) \\ sin(\alpha-\beta) & cos(\alpha-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(\alpha-\beta)の回転行列と等しい．\\ \end{array}$$ $$\begin{array}{rcl} cos(\alpha -\beta)&=&cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)\\ sin(\alpha-\beta)&=&sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)\\ \end{array}$$

角度\(\alpha\)と角度\(\beta\)が一致(\(\alpha=\beta=\theta\))

$$\begin{array}{rcl} \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{pmatrix} \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\theta)cos(\theta)-sin(\theta)sin(\theta) & -cos(\theta)sin(\theta)-sin(\theta)cos(\theta) \\ sin(\theta)cos(\theta)+cos(\theta)sin(\theta) & -sin(\theta)sin(\theta)+cos(\theta)cos(\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\theta)^2-sin(\theta)^2 & -2sin(\theta)cos(\theta) \\ 2sin(\theta)cos(\theta) & cos(\theta)^2-sin(\theta)^2 \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(2\theta) & -sin(2\theta) \\ sin(2\theta) & cos(2\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(2\theta)の回転行列と等しい．\\ \end{array}$$ 対応する要素同士が倍角公式となる． $$\begin{array}{rcl} cos(2\theta)&=&cos(\theta)^2-sin(\theta)^2\\ sin(2\theta)&=&2sin(\theta)cos(\theta)\\ \end{array}$$