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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).`
Solution
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)`
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