Wehave, $\mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos \mathrm{x} & \sin \mathrm{x} \\ -\sin \mathrm{x} & \cos \mathrm{x}\end{array}\right]$
$$
\therefore \mathrm{P}(\mathrm{y})=\left[\begin{array}{cc}
\cos \mathrm{y} & \sin \mathrm{y} \\
-\sin \mathrm{y} & \cos \mathrm{y}
\end{array}\right]
$$
Now, $\mathrm{P}(\mathrm{x}) \cdot \mathrm{P}(\mathrm{y})=\left[\begin{array}{cc}\cos \mathrm{x} & \sin \mathrm{x} \\ -\sin \mathrm{x} & \cos \mathrm{x}\end{array}\right]\left[\begin{array}{cc}\cos \mathrm{y} & \sin \mathrm{y} \\ -\sin \mathrm{y} & \cos \mathrm{y}\end{array}\right]$
$$
\begin{aligned}
&=\left[\begin{array}{cc}
\cos x \cdot \cos y-\sin x \cdot \sin y & \cos x \cdot \sin y+\sin x \cdot \cos y \\
-\sin x \cdot \cos y-\cos x \cdot \sin y & -\sin x \cdot \sin y+\cos x \cdot \cos y
\end{array}\right] \\
&=\left[\begin{array}{cc}
\cos (x+y) & \sin (x+y) \\
-\sin (x+y) & \cos (x+y)
\end{array}\right]
\end{aligned}
$$
$$
\left[\begin{array}{l}
\because \cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y \\
\text { and } \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y
\end{array}\right]
$$


and $\quad P(x+y)=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]$
Also, $\quad \mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})$
$$
\begin{aligned}
&=\left[\begin{array}{cc}
\cos y & \sin y \\
-\sin y & \cos y
\end{array}\right]\left[\begin{array}{cc}
\cos x & \sin x \\
-\sin x & \cos x
\end{array}\right] \\
=& {\left[\begin{array}{cc}
\cos y \cdot \cos x-\sin y \cdot \sin x & \cos y \cdot \sin x+\sin y \cdot \cos x \\
-\sin y \cdot \cos x-\sin x \cdot \cos y & -\sin y \cdot \sin x+\cos y \cdot \cos x
\end{array}\right] } \\
&=\left[\begin{array}{cc}
\cos (x+y) & \sin (x+y) \\
-\sin (x+y) & \cos (x+y)
\end{array}\right]
\end{aligned}
$$
Thus, we see from the Eqs. (i), (ii) and (iii) that, $\mathrm{P}(\mathrm{x}) \cdot \mathrm{P}(\mathrm{y})=\mathrm{P}(\mathrm{x}+\mathrm{y})=\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x}) \quad$ Hence proved.