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

My approach
$$A^2 = \begin{pmatrix}
\cos2\theta  & \sin2\theta\\ 
 -\sin 2\theta&  \cos 2\theta
\end{pmatrix} $$
$2010$ 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
$$\begin{pmatrix}
\cos\theta  & \sin\theta\\ 
 -\sin \theta&  \cos \theta
\end{pmatrix} *
\begin{pmatrix}
\cos\phi  & \sin\phi\\ 
 -\sin \phi&  \cos \phi
\end{pmatrix}
=
\begin{pmatrix}
\cos(\theta+\phi)  & \sin(\theta+\phi)\\ 
 -\sin (\theta+\phi)&  \cos (\theta+\phi)
\end{pmatrix}.$$
(You can show this with the multiangle formulae for $\sin$ and $\cos$.) So we have
$$A^{67} = \begin{pmatrix}
\cos67\theta  & \sin67\theta\\ 
 -\sin 67\theta&  \cos 67\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos2\pi  & \sin2\pi\\ 
 -\sin 2\pi&  \cos 2\pi
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1 
\end{pmatrix} = I.$$
$$\therefore A^{2010} = (A^{67})^{30} = I.$$
(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
$$M_\theta = \begin{pmatrix}
\cos\theta  & \sin\theta\\ 
 -\sin \theta&  \cos \theta
\end{pmatrix}$$
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! :)
