<s>[INST] $4^\text{th}$ power of a $2\times 2$ matrix $$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$ 
Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.
I tried to use some trigonometric expressions but it gets very complicated and couldn't solve it.
 [/INST] Here is a hint:
We have
$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$
Use matrix multiplication to compute (I'll do it for you) ...
$B=A^2 = \left(\begin{array}{cc}\cos^2 x -\sin^2 x& -2\sin x \cos x \\ 2\sin x \cos x & \cos^2 x -\sin^2 x\end{array}\right)$
Now simplify using the trigonometry you know, and compute $B^2=A^4$.
</s>