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# If $A=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix},\;then\;show\;that\;A^2=\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}.$

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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• $cos2\theta=cos^2\theta-sin^2\theta$
• $sin2\theta=cos\theta sin\theta+sin\theta cos\theta$
Step1:
Given
$A=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}$

$A^2=A.A=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}$

$\Rightarrow \begin{bmatrix}\cos\theta.cos\theta+sin\theta(-sin\theta) &cos\theta( \sin\theta)+sin\theta(cos\theta)\\-\sin\theta.cos\theta+ cos\theta(-sin\theta)&sin\theta(sin\theta)+ \cos\theta(cos\theta)\end{bmatrix}$

$\Rightarrow\begin{bmatrix} cos^2\theta-sin^2\theta&\cos\theta sin\theta+sin\theta cos\theta\\-sin\theta cos\theta-cos\theta sin\theta&-sin^2\theta+cos^2\theta\end{bmatrix}$

$\Rightarrow\begin{bmatrix} cos^2\theta-sin^2\theta&\cos\theta sin\theta+sin\theta cos\theta\\-(sin\theta cos\theta+cos\theta sin\theta)&cos^2-sin^2\theta\end{bmatrix}$
Step2:
We know that

$cos2\theta=cos^2\theta-sin^2\theta$

$sin2\theta=cos\theta sin\theta+sin\theta cos\theta$

Hence we have

$A^2=\begin{bmatrix}cos2\theta & sin2\theta\\-sin2\theta & cos2\theta\end{bmatrix}$