# Simplify

Question:

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

Solution:

$\cos \theta\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\sin \theta\left[\begin{array}{lr}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right]$

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

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

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \quad\left(\because \cos ^{2} \theta+\sin ^{2} \theta=1\right)$