Compute $R^{2016}$ of a given counterclockwise rotation. 
Write out the matrix $R$ of counterclockwise rotation by 30$^{\circ}$ in $\mathbb{R}^2$. Compute ${R}^{2016}$. Now this is an easy question to answer overall; 30 goes into 360 12 times and one twelfth of 2016 is precisely 168 - hence the transformation is equivalent to the identity; however the question asks write out the matrix, given as follows
  $$A_\theta = \left[ \begin{matrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{matrix} \right]$$
  where $\theta=$30$^{\circ}$
  $$A_\theta = 2016\left[ \begin{matrix} \cos(n30^{\circ}) & -\sin(n30^{\circ})\\ \sin(n30^{\circ}) & \cos(n30^{\circ})\end{matrix} 
\right]$$
  where $n=12$ and 
  $$A_\theta = 2016\left[ \begin{matrix} \cos(2\pi ) & -\sin(2 \pi)\\ \sin(2\pi) & \cos(2 \pi)\end{matrix} 
\right]$$
  $$A_\theta = 2016\left[ \begin{matrix} 1 & 0\\ 0 & 1\end{matrix} 
\right]$$
  but where to from here?

 A: This is how your words translate to matrix operations:
$$ \left[ \begin{matrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{matrix} \right]^{2016} = \left(\left[ \begin{matrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{matrix} \right]^{12}\right)^{168}$$
$$ = \left[ \begin{matrix} \cos(12\theta) & -\sin(12\theta)\\ \sin(12\theta) & \cos(12\theta)\end{matrix} \right]^{168} =_{_\theta \ _= \ _{30^o}} \left[ \begin{matrix} 1 & 0\\ 0 & 1\end{matrix} \right]^{168}= \left[ \begin{matrix} 1 & 0\\ 0 & 1\end{matrix} \right] $$
A: We define the matrix-valued function
$$\mathrm R (\theta) := \begin{bmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{bmatrix}$$
Note that this is the matrix representation of $\exp (i \theta) = \cos (\theta) + i \sin (\theta)$. Raising this complex number to the $n$-th power, we obtain
$$\exp (i n \theta) = \cos (n \theta) + i \sin (n \theta)$$
whose matrix representation is
$$\begin{bmatrix} \cos(n \theta) & -\sin(n \theta)\\ \sin(n \theta) & \cos(n \theta)\end{bmatrix} = \mathrm R^n (\theta)$$
