Let $\theta=\frac{2 \pi}{67}$ consider the rotation matrix $A$. What is $A^{2010}$? Let $\theta=\frac{2 \pi}{67}$. Consider the matrix
  $$A = \begin{pmatrix}
\cos\theta  &amp; \sin\theta\\ 
 -\sin \theta&amp;  \cos \theta
\end{pmatrix} $$
  Then the matrix $A^{2010}$ is?


My approach
$$A^2 = \begin{pmatrix}
\cos2\theta  &amp; \sin2\theta\\ 
 -\sin 2\theta&amp;  \cos 2\theta
\end{pmatrix} $$

$2010$ is a multiple of $67$. So I'm trying to convert in such format by taking $A$ matrix as power of $2$ and multiplying. But $2048$ is the nearest $2$ power term. So help please. I dont know if you can understand what I did!! Sorry for being not explaining properly
Observe that when you multiply the matrices
$$\begin{pmatrix}
\cos\theta  &amp; \sin\theta\\ 
 -\sin \theta&amp;  \cos \theta
\end{pmatrix} *
\begin{pmatrix}
\cos\phi  &amp; \sin\phi\\ 
 -\sin \phi&amp;  \cos \phi
\end{pmatrix}
=
\begin{pmatrix}
\cos(\theta+\phi)  &amp; \sin(\theta+\phi)\\ 
 -\sin (\theta+\phi)&amp;  \cos (\theta+\phi)
\end{pmatrix}.$$

(You can show this with the multiangle formulae for $\sin$ and $\cos$.) So we have

$$A^{67} = \begin{pmatrix}
\cos67\theta  &amp; \sin67\theta\\ 
 -\sin 67\theta&amp;  \cos 67\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos2\pi  &amp; \sin2\pi\\ 
 -\sin 2\pi&amp;  \cos 2\pi
\end{pmatrix}
=
\begin{pmatrix}
1 &amp; 0 \\
0 &amp; 1 
\end{pmatrix} = I.$$
$$\therefore A^{2010} = (A^{67})^{30} = I.$$
(I tried to put the end bit as a 'spoiler' which you have to mouse over to view, but I couldn't work out how to!)

As mentioned by other people, applying the matrix
$$M_\theta = \begin{pmatrix}
\cos\theta  &amp; \sin\theta\\ 
 -\sin \theta&amp;  \cos \theta
\end{pmatrix}$$
has the action of rotation a vector (anticlockwise) by $\theta$. Thus doing it repeatedly simply means that they sum: $M_\theta * M_\phi = M_{\theta + \phi}$.

Hopefully this helps! :)
Building on what was said, you can view $R(\theta)$ as a rotation matrix in the complex plane. Indeed, for a vector written as $v=x+iy$ , then the rotation matrix is simply the operator $e^{i\theta}$. 
This is shown by Euler's formula.
From this, you know that ${(e^{i\theta})}^n=(e^{ni\theta})$ which links back to what was shown.