Question

 If `P(x)=[[cos x,sin x],[-sin x,cos x]],` then show that `P(x),P(y)=P(x+y)=P(y)P(x).`

Answer 1

 Given :  \[P(x) = \begin{bmatrix}\ cosx & \ sinx \\ - \ sinx & \ cosx\end{bmatrix}\]

Then ,

\[P(y) = \begin{bmatrix}\ cosy & \ siny \\ - \ siny & \ cosy\end{bmatrix}\] 

Now,

\[P(x) P(y) = \begin{bmatrix}\ cosx & \ sinx \\ - \ sinx & \ cosx\end{bmatrix}\begin{bmatrix}\ cosy &  \ siny \\ - \ siny & \ cosy\end{bmatrix}\]
\[ = \begin{bmatrix}\ cosx\ cosy - \ sinx\ siny & \ cosx\ siny + \ sinx\ cosy \\ -  \ sinx\ cosy - \ cosx\ siny & - \ sinx\ siny + \ cosx\ cosy\end{bmatrix}\]
\[ = \begin{bmatrix}\cos\left( x + y \right) & \sin\left( x + y \right) \\ - \sin\left( x + y \right) & \cos\left( x + y \right)\end{bmatrix} . . . (1)\]

Also, 

\[P(x + y) = \begin{bmatrix}\cos\left( x + y \right) & \sin\left( x + y \right) \\ - \sin\left( x + y \right) & \cos\left( x + y \right)\end{bmatrix} . . . (2)\]

Now,

\[P(y) P(x) = \begin{bmatrix}\ cosy & \ siny \\ - \ siny & \ cosy\end{bmatrix}\begin{bmatrix}\ cosx & \ sinx \\ - \ sinx & \ cosx\end{bmatrix}\]
\[ = \begin{bmatrix}\ cosy\ cosx - \ siny\ sinx & \ cosy\ sinx + \ siny\ cosx \\ - \ siny \ cosx - \ cosy\ sinx & -  \ siny\ sinx + \ cosy\ cosx\end{bmatrix}\]
\[ = \begin{bmatrix}\cos\left( x + y \right) & \sin\left( x + y \right) \\ - \sin\left( x + y \right) & \cos\left( x + y \right)\end{bmatrix} . . . (3)\]

from (1),(2) and (3) ,we get

`p(x)   p(y)=p(x+y)=p(y)  p(x)`