Question

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

Answer 1

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

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

Now,

$f\left(x\right)f\left(y\right)$

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

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

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

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

It is given that, f(x)f(y) = f(​z)

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

⇒ z = x + y

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