Question

If $A_{\alpha }=\left[\begin{array}{cc}cosn\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{array}\right]$ , then the prove following .
$A_{\alpha }{A}_{\beta }=A_{\alpha +\beta }=A_{\beta }{A}_{\alpha }$

Answer 1

$A_{\alpha }{A}_{\beta }=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]\left[\begin{array}{cc}cos\beta & sin\beta \\ -sin\beta & cos\beta \end{array}\right]$
$=\left[\begin{array}{cc}cos(\alpha +\beta )& sin(\alpha +\beta )\\ -sin(\alpha +\beta )& cos(\alpha +\beta )\end{array}\right]=A_{\alpha +\beta }$
Also $A_{\beta }\cdot A_{\alpha }=A_{\alpha +\beta }$ which can be shown as above . Hence proved .