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?

By induction,
$$ A^n = A^{n-1}A = \begin{pmatrix} \cos{(n-1)\theta} & \sin{(n-1)\theta} \\
 -\sin{(n-1)\theta} & \cos{(n-1)\theta} \end{pmatrix} \begin{pmatrix} \cos{\theta} & \sin{\theta} \\
 -\sin{\theta} & \cos{\theta} \end{pmatrix} = \begin{pmatrix} \cos{(n-1)\theta}\cos{\theta} - \sin{\theta}\sin{(n-1)\theta}  & \sin{(n-1)\theta}\cos{\theta} + \cos{(n-1)\theta}\sin{\theta} \\
 -\sin{(n-1)\theta} & \cos{(n-1)\theta}\cos{\theta} - \sin{\theta}\sin{(n-1)\theta} \end{pmatrix} \\
= \begin{pmatrix} \cos{n\theta} & \sin{n\theta} \\
 -\sin{n\theta} & \cos{n\theta} \end{pmatrix}. $$
The prosthaphaeresis formulae are
$$ \sin{A}+\sin{B} = 2\sin{\tfrac{1}{2}(A+B)}\cos{\tfrac{1}{2}(A-B)} \\
 \sin{A}-\sin{B} = 2\cos{\tfrac{1}{2}(A+B)}\sin{\tfrac{1}{2}(A-B)}\\
 \cos{A}+\cos{B} = 2\cos{\tfrac{1}{2}(A+B)}\cos{\tfrac{1}{2}(A-B)}\\
 \cos{A}-\cos{B} = -2\sin{\tfrac{1}{2}(A+B)}\sin{\tfrac{1}{2}(A-B)}. $$
Thus
$$ \cos{4\theta}-\cos{3\theta} = -2\sin{\tfrac{7}{2}\theta}\sin{\tfrac{1}{2}\theta}, \\
\cos{2\theta}-\cos{\theta} = -2\sin{\tfrac{3}{2}\theta}\sin{\tfrac{1}{2}\theta}, $$
and adding gives
$$ \cos{4\theta}-\cos{3\theta} + \cos{2\theta}-\cos{\theta} = -2\sin{\tfrac{1}{2}\theta} ( \sin{\tfrac{7}{2}\theta} + \sin{\tfrac{3}{2}\theta} ) \\
= -4\sin{\tfrac{1}{2}\theta}\sin{\tfrac{5}{2}\theta}\cos{\theta}. $$
Similarly,
$$ \sin{4\theta}-\sin{3\theta} + \sin{2\theta}-\sin{\theta} = 4\sin{\tfrac{1}{2}\theta}\cos{\tfrac{5}{2}\theta}\cos{\theta}, $$
so
$$ B = 4\sin{\tfrac{1}{2}\theta}\cos{\theta} \begin{pmatrix} -\sin{\tfrac{5}{2}\theta} & -\cos{\tfrac{5}{2}\theta} \\ \cos{\tfrac{5}{2}\theta} & -\sin{\tfrac{5}{2}\theta} \end{pmatrix}. $$
