Question
If $A (\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ then prove that $A ^2(\alpha)= A (2 \alpha)$

Answer

$\begin{array}{l} A ^2(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \end{array}$
$ = \\ {\left[\begin{array}{cc}\cos ^2 \alpha-\sin ^2 \alpha & \cos \alpha \sin \alpha+\sin \alpha \cos \alpha \\ -\cos \alpha \sin \alpha-\sin \alpha \cos \alpha & -\sin ^2 \alpha \cos ^2 \alpha\end{array}\right]} $
$ =\left[\begin{array}{cc}\cos (\alpha+\alpha) & \sin (\alpha+\alpha) \\ -\sin (\alpha+\alpha) & \cos (\alpha+\alpha)\end{array}\right] \\ {\left[\begin{array}{cc}\because \sin ( A + B )=\sin A \cos B +\cos A \sin B \\ \cos ( A + B )=\cos A \cos B -\sin A \sin B \end{array}\right]} $
$ =\left[\begin{array}{cc}\cos (2 \alpha) & \sin (2 \alpha) \\ -\sin (2 \alpha) & \cos (2 \alpha)\end{array}\right] $
$ = A (2 \alpha)$

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