Question
If \[ P(x)= \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \] show that \[ P(x)P(y)=P(x+y)=P(y)P(x). \]
Solution
Step 1: Compute \(P(x)P(y)\)
\[ P(x)P(y)= \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix} \] \[ = \begin{bmatrix} \cos x\cos y – \sin x\sin y & \cos x\sin y + \sin x\cos y \\ -\sin x\cos y – \cos x\sin y & -\sin x\sin y + \cos x\cos y \end{bmatrix} \]Step 2: Use Identities
\[ \cos(x+y)=\cos x\cos y – \sin x\sin y \] \[ \sin(x+y)=\sin x\cos y + \cos x\sin y \]Step 3: Substitute
\[ P(x)P(y)= \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix} = P(x+y) \]Step 4: Show Commutativity
Similarly, \[ P(y)P(x)=P(y+x)=P(x+y) \]Final Result
\[
P(x)P(y)=P(x+y)=P(y)P(x)
\]
Hence proved.