<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] \begin{eqnarray*}
A
& = &
\left(%
\begin{array}{rr}
\cos\left(x\right) & -\sin\left(x\right)
\\
\sin\left(x\right) & \cos\left(x\right)
\end{array}
\right)
=
\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}
\\
A^{2}
& = &
\cos^{2}\left(x\right) - \sin^{2}\left(x\right)
-
2{\rm i}\sin\left(x\right)\cos\left(x\right)\,\sigma_{y}
=
\cos\left(2x\right) - {\rm i}\sin\left(2x\right)\,\sigma_{y}
\\&&\mbox{Then}
\\
A^{4}
& = &
\cos\left(4x\right) - {\rm i}\sin\left(4x\right)\,\sigma_{y}
=
\left(%
\begin{array}{rr}
\cos\left(4x\right) & -\sin\left(4x\right)
\\
\sin\left(4x\right) & \cos\left(4x\right)
\end{array}
\right)
\quad\Longrightarrow\quad
{\rm det}\,\left(A^{4}\right) = 1
\\[5mm]&&\mbox{}
\end{eqnarray*}
$$
ad + bc = \cos^{2}\left(4x\right) - \sin^{2}\left(4x\right) = \cos\left(8x\right)
$$
$\sigma_{y}$ is a Pauli Matrix: http://en.wikipedia.org/wiki/Pauli_matrices
Indeed, ${\rm det}\,\left(A^{4}\right) =\left({\rm det}\,A\right)^{4} = 1^{4} = 1$.
Also,
$$
A'
=
-\sin\left(x\right) - {\rm i}\cos\left(x\right)\,\sigma_{y}
=
-{\rm i}\left\lbrack\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}\right\rbrack
=
-{\rm i}A
\ \Longrightarrow\ A = {\rm e}^{-{\rm i}\,x\,\sigma_{y}}
$$
since $A'' + A = 0$ with $\left.A\right\vert_{x\ =\ 0} = 1$
and $\left.A'\right\vert_{x\ =\ 0} = -{\rm i}\,\sigma_{y}$. That means
$$
A^{n} = {\rm e}^{-{\rm i}\,n\,x\,\sigma_{y}}
=\left(%
\begin{array}{rr}
\cos\left(nx\right) & -\sin\left(nx\right)
\\
\sin\left(nx\right) & \cos\left(nx\right)
\end{array}
\right)\,,
\quad
A\ \mbox{is a Rotation Matrix}
$$
</s>