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# If $P(x)=\begin{bmatrix}\cos x & \sin x\\-\sin x & \cos x\end{bmatrix}$,then show that$P(x).P(y)=P(x+y)=P(y).P(x).$

Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Given:
$P(x)=\begin{bmatrix}cos x & sin x\\-sin x & cos x\end{bmatrix}$
$P(y)=\begin{bmatrix}cos y & sin y\\-sin y & cos y\end{bmatrix}$
$P(x).P(y)=\begin{bmatrix}cos x & sin x\\-sin x & cos x\end{bmatrix}\begin{bmatrix}cos y & sin y\\-sin y & cos y\end{bmatrix}$
$\Rightarrow \begin{bmatrix}cos x cos y-sin xsin y &cos x sin y+sin x cos y\\-sin x cosy-cos x sin y & -sin x sin y+cos x cos y\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}cos (x+y) & sin (x+y)\\-sin (x +y)& cos (x+y)\end{bmatrix}$
P(x+y)=$\begin{bmatrix}cos x & sin x\\-sin x & cos x\end{bmatrix}+\begin{bmatrix}cos y & sin y\\-sin y & cos y\end{bmatrix}$
$\rightarrow \begin{bmatrix}cos( x+y) & sin (x+y)\\-sin (x+y) & cos( x+y)\end{bmatrix}$
P(y).P(x)=$\begin{bmatrix}cos y & sin y\\-sin y & cos y\end{bmatrix}\begin{bmatrix}cos x & sin x\\-sin x & cos x\end{bmatrix}$
$\Rightarrow \begin{bmatrix}cos y cos x-cos y sin x &cos y sin x+sin y cos x\\-sin y cosx-cos y sin x & -sin x sin y+cos x cos y\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}cos (x+y) & sin (x+y)\\-sin (x +y)& cos (x+y)\end{bmatrix}$
Hence P(x).P(y)=P(x+y)=P(y).P(x).