If P (x) = then show that

Given,

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

So, $\quad P(y)=\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$

Now,

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

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

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

$=P(x+y)$

Also, ...(i)

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

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

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

Therefore, form (i) and (ii), we get

$P(x) \cdot P(y)=P(x+y)=P(y) \cdot P(x)$