Question

If $A_x=\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]$ and $A_y=\left[\begin{array}{cc}\mathrm{cosy}& \sin y\\ -\sin y& \cos y\end{array}\right]$

Show that $A_x{A}_y=A_{x+y}$.

Answer 1

$\begin{array}{c}{A}_x{A}_y=\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]\left[\begin{array}{cc}\mathrm{cosy}& \sin y\\ -\sin y& \cos y\end{array}\right]\\ =\left[\begin{array}{cc}\cos x\mathrm{cosy}-\sin x\mathrm{siny}& \cos x\sin y+\sin x\cos y\\ -\sin x\mathrm{cosy}-\cos x\mathrm{cosy}& -\sin x\sin y+\cos x\cos y\end{array}\right]\\ =\left[\begin{array}{cc}\cos \left(x+y\right)& \sin \left(x+y\right)\\ -\sin \left(x+y\right)& \cos \left(x+y\right)\end{array}\right]=A_{x+y}\end{array}$