Question
If $A \alpha=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$, show that $A \alpha \cdot A \beta=A \alpha+\beta$
✓
Answer
$A_\alpha=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$\begin{aligned} & \therefore \quad A_\beta=\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta\end{array}\right] \\ & \therefore \quad A_\alpha \cdot A_\beta=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta\end{array}\right] \\ & =\left[\begin{array}{cc}\cos \alpha \cos \beta-\sin \alpha \sin \beta & \cos \alpha \sin \beta+\sin \alpha \cos \beta \\ -\sin \alpha \cos \beta-\cos \alpha \sin \beta & -\sin \alpha \sin \beta+\cos \alpha \cos \beta\end{array}\right] \\ & =\left[\begin{array}{cc}\cos \alpha \cos \beta-\sin \alpha \sin \beta & \cos \alpha \sin \beta+\sin \alpha \cos \beta \\ -[\sin \alpha \cos \beta+\cos \alpha \sin \beta] & \cos \alpha \sin \beta-\sin \alpha \cos \beta\end{array}\right] \\ & =\left[\begin{array}{cc}\cos (\alpha+\beta) & \sin (\alpha+\beta) \\ -\sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right]\end{aligned}$
$\therefore \quad A_\alpha A_\beta=A_{(\alpha+\beta)}$
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