1-1

次の式を示せ。但し $\delta\equiv \delta_2-\delta_1$

$$\begin{pmatrix}\sin\delta_2 & \cos\delta_2 \\ -\sin\delta_1 & -\... ...\end{pmatrix} = \begin{pmatrix}1 & -\cos\delta \\ -\cos\delta &1 \end{pmatrix}$$ (1)

1-1解答

$$\begin{pmatrix}\sin\delta_2 & \cos\delta_2 \\ -\sin\delta_1 & -\c... ...trix}\sin\delta_2 & -\sin\delta_2 \\ \cos\delta_2 & -\cos\delta_1 \end{pmatrix} = \begin{pmatrix}\sin^2\delta_2+\cos^2\delta_2 & -\left( \sin\del... ...+ \cos\delta_1\cos\delta_2\right) & \sin^2\delta_1+\cos^2\delta_1 \end{pmatrix}$$

$$= \begin{pmatrix}1 & -\cos\left(\delta_2-\delta_1\right) \\ -\cos\left(\delta_2-\delta_1\right) &1 \end{pmatrix}$$

$$= \begin{pmatrix}1 & -\cos\delta \\ -\cos\delta &1 \end{pmatrix} \begin{pmatrix}1 & -\cos\delta \\ -\cos\delta &1 \end{pmatrix}$$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp