Solve the following equations

$f(x)=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$

$\therefore f(y)=\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$

Now,

$f(x) f(y)$

$=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$

$=\left[\begin{array}{cc}\cos x \cos y-\sin x \sin y & \sin x \cos y+\cos x \sin y \\ -\sin x \cos y-\cos x \sin y & \cos x \cos y-\sin x \sin y\end{array}\right]$

$=\left[\begin{array}{cc}\cos x \cos y-\sin x \sin y & \sin x \cos y+\cos x \sin y \\ -(\sin x \cos y+\cos x \sin y) & \cos x \cos y-\sin x \sin y\end{array}\right]$

$=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]$

It is given that, $f(x) f(y)=f(z)$

$\therefore\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]=\left[\begin{array}{cc}\cos z & \sin z \\ -\sin z & \cos z\end{array}\right]$

$\Rightarrow z=x+y$

If $f(x)=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$ and $f(x) f(y)=f(z)$, then $z=x+y$