# Prove the following

Question:

If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then show that $\mathrm{A}^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

Solution:

Given, $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

Now, $A^{2}=A \cdot A$

$=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$=\left[\begin{array}{cc}\cos ^{2} \theta-\sin ^{2} \theta & \cos \theta \sin \theta+\sin \theta \cos \theta \\ -\sin \theta \cos \theta-\cos \theta \sin \theta & -\sin ^{2} \theta+\cos ^{2} \theta\end{array}\right]$

$=\left[\begin{array}{cc}\cos 2 \theta & 2 \sin \theta \cos \theta \\ 2 \sin \theta \cos \theta & \cos 2 \theta\end{array}\right]=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

Hence proved.