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
 A: 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! :)
A: The matrix $A$ is rotation by $2\pi/67$. Take the vector $(1,0)$. If you rotate it $2010$ times the total angle is $2010*2\pi/67=60\pi$. So it comes back to $(1,0)$.Similar argument shows that the $(0,1)$ also remains at $(0,1)$. So, this is the identity matrix.
A: By noticing that $67|2010$ you are actually done. This means that the rotation $A^{2010}$ corresponds to an integer number of full rotations. If $A(\theta)$ is the matrix $A$ as above with any angle, then notice that
$$(A(\theta))^k = A(k\theta)$$
So
$$(A(\frac{2\pi}{67}))^{2010} = A(\frac{2\pi}{67}\cdot 2010) = A(2\pi \cdot 30)$$
Now notice that $A(\theta + 2\pi) = A(\theta)$ to see that $A(\frac{2\pi}{67})^{2010} = A(0) = I_2$
A: 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. 
