For rotation matrix $A$, find $B = A^4- A^3 + A^2 - A$ 
Find the value of $B = A^4- A^3 + A^2 - A$ where $A$ is the matrix given below $$ A=
    \left [ \begin{matrix}
    \cos\alpha & \sin \alpha  \\
    -\sin\alpha & \cos\alpha 
    \end{matrix} \right ]
$$

It's actually quite simple when you look at it. Find $A$, $A^2$, $A^3$, $A^4$ by doing some matrix multiplication and then add and subtract following the question.
I did all of the tedious work mentioned above. 
But the matrix $B$ I obtain is filled with garbage. I assuming either we must use some trigno mumbo jumbo identities, or there's something beautiful that I missed from the question.
Edit: Could the sum of a geometric series help?

Some hints...you could do some simplification by considering $$(A+I)B=A^5-A=A(A^4-I)$$
$R(\theta)$ is the standard rotation matrix $\left(\begin{matrix}\cos\theta &-\sin\theta\\ \sin\theta&\cos\theta\end{matrix}\right)$ which rotates by angle $\theta$ anticlockwise about the origin. Note that $R(\theta)R(\phi)=R(\theta+\phi)$.
Therefore $A=R(-\alpha)$ represents  clockwise rotation by angle $\alpha$,$$A+I=\left(\begin{matrix}\cos\alpha+1 &\sin\alpha\\ -\sin\alpha&\cos\alpha+1\end{matrix}\right)=2\cos\frac{\alpha}{2}R(-\frac{\alpha}{2})$$
$$\implies(A+I)^{-1}=\frac{1}{2\cos\frac{\alpha}{2}}R(\frac{\alpha}{2})$$
Then $$B=(A+I)^{-1}A(A^4-I)$$
