$4^\text{th}$ 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.
 A: 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 $B^2=A^4$.
A: All you need to do here is to recognise that $A$ represents an anti-clockwise rotation about the origin by $x$. If you square the matrix then you repeat the rotation a second time, i.e. rotate by $2x$. Similarly, the cube $A^3$ can be thought of as performing the rotation three times. Hopefully you can see that $A^4$ is an anti-clockwise rotation about the origin by $4x$. As a matrix, this looks like
$$A^4 = \left[ \begin{array} 1\cos(4x) & -\sin(4x) \\ \sin(4x) & \cos(4x) \end{array}\right]$$
If you want $ad+bc$ then you have 
$$\cos(4x)\cos(4x)-\sin(4x)\sin(4x) \equiv \cos(4x+4x) \equiv \cos(8x)$$
 Here I used the double angle formula. If, however, you actually wanted the determinant, i.e. $ad-bc$ then you have 
$$\cos(4x)\cos(4x)+\sin(4x)\sin(4x) \equiv \cos(4x-4x) \equiv 1$$
Note that the last result is obvious. A rotation preserves volume and so its matrix must have determinant one. (If the determinant had been $2$, say, then all areas would double.)
A: Here's my way of looking at this:  let $J$ be the $2 \times 2$ matrix given by
$J = \begin{bmatrix}
0 & -1 \\
1 & 0 \end{bmatrix}$;
then $J^2 = -I$, where $I$ is the $2 \times 2$ identity matrix:
$I = \begin{bmatrix}
1 & 0 \\
0 & 1 \end{bmatrix}$.
Now observe that 
$A(x) = \cos x I + \sin x J$,
and that since $J^2 = -I$, $J$ behaves algebraically exactly like $i = \sqrt {-1} \in \mathbb{C}$, $\mathbb{C}$ being the ordinary complex numbers.  In particular we have, for $a, b$ real
$(aI +  bJ)^2 = (a^2 - b^2)I + 2abJ$,
and if $c,d$ are real as well,
$(aI + bJ)(cI + dJ) = (ac - bd)I + (ad + bc)J$.
These formulas are easy to verify simply using elementary matrix algebra, one really doesn't even need to look at specific matrix entries, just maneuver expressions as if they were polynomials in $J$, and reduce using $J^2 = -I$.  Furthermore, they show that de Moivre's classic formula 
$(\cos \theta + i\sin \theta)^n = \cos n \theta + i\sin n \theta$
holds for  matrices of the form  $\cos \theta I + \sin \theta J$:
$(\cos \theta I + \sin\theta J)^n = \cos n \theta I + \sin n \theta J$.
Proving the above equation merely relies on a simple induction on the exponent $n \in \Bbb{Z}$, $n \ge 0$: if 
$(\cos \theta I + \sin\theta J)^k = \cos k \theta I + \sin k \theta J$,
then multiplying through by $\cos \theta I + \sin \theta J$ yields
$(\cos \theta I + \sin\theta J)^{k + 1} = (\cos \theta I + \sin \theta J)(\cos k \theta I + \sin k \theta J)$,
and we have using the above formulas 
$(\cos \theta I + \sin \theta J)(\cos k \theta I + \sin k \theta J)$
$= (\cos \theta \cos k \theta - \sin \theta \sin k \theta)I + (\sin \theta \cos k \theta + \cos \theta \sin k \theta)J$
$=\cos(k + 1) \theta I + \sin(k + 1) \theta J$,
this last equality relying on the standard addition formulas for $\sin (x + y)$ and $\cos (x + y)$.  For more information on this derivation, see this wikipedia page.
The OP's specific concern is neatly wrapped up by observing that since
$A = \cos x I + \sin x J$,
it follows from what we have seen that
$A^4 = \cos 4x I + \sin 4x J$,
and we have $ad + bc$ (using now our OP guest's $a, b, c, d$) as
$ad + bc = \cos^2 4x - \sin^2 4x = \cos 8x$,
as our friend Fly by Night pointed out in his answer to this question.
What I like about the present approach is, that once we have
$(\cos \theta I + \sin\theta J)^n = \cos n \theta I + \sin n \theta J$,
it is really not much extra work to calculate all sorts of stuff about $A^n$ for any $n \in \Bbb{Z}$, $n \ge 0$; for example
$A^{17} = \cos 17 x I + \sin 17 x J$,
and we have
$ad + bc = \cos 34 x$
in this case.
Hope I haven't been too long winded, or gone too far afield.  Cheeer-i-o, Ladies and Gents!
A: \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
$$
A^{n} = {\rm e}^{-{\rm i}\,n\,x\,\sigma_{y}}
=\left(%
\begin{array}{rr}
\cos\left(nx\right) & -\sin\left(nx\right)
\\
\sin\left(nx\right) & \cos\left(nx\right)
\end{array}
\right)\,,
\quad
A\ \mbox{is a Rotation Matrix}
$$
