Doesn't the 2 dimensional rotation preserve the infinitesimal area $dxdy$? Let $R(\theta)$ be the two dimensional rotation in the $xy$ plane. And let $x'$ ad $y'$ the transformed coordinates. Then it is just intuitively clear that $dxdy=dx'dy'$ must hold. But when using the matrix representation for $R(\theta)$ expressed by $sin\;\theta$ and $cos\;\theta$, things just get twisted... Isnt' $dx'=cos\theta dx-sin\;\theta dy$ and $dy'=cos\theta dy+sin\;\theta dx$? I can't see why things go so awry....
 A: We can see this change of coordinates in two ways. 
1) The area element $dS$ in the new coordinates is the standard area element  $dxdy$ multiplied by the jacobian of the transformation and, for a rotation, the Jacobian is
$$J=
\det \begin{pmatrix}
\cos \theta & -\sin \theta\\
\sin \theta&\cos \theta
\end{pmatrix}=1
$$
so $dS'=Jdx'dy'=dx'dy'$
2) we can use as area element the external product of the two vectors $dx'=\cos\theta dx-\sin\theta dy$ and $dy'=\cos\theta dy+\sin\theta dx$:
$$
dx'\wedge dy'=(\cos\theta dx-\sin\theta dy)\wedge(\sin\theta dx+\cos\theta dy)=
\cos^2 \theta dx \wedge dy +\sin^2 \theta dx\wedge dy=dx\wedge dy
$$
where we use the fact that the external ( or wedge) product is an alternating product, so that: 
$dx\wedge dx= dy\wedge dy=0\quad$ and $\quad dx \wedge dy=-dy\wedge dx$

The mistake in your derivation derives from a wrong representation of the area element. If you think at the figure of the rotated elements $dx'$ and $dy'$ than you see that the area defined by them is , in absolute value, $|dS'|=|dx'|\cdot |dy'|$ that is, using the coordinates with respect to $dx,dy$,
$$
dS'=\sqrt{(cos^2 \theta+\sin^2\theta)(sin^2 \theta+\cos^2 \theta) }=1
$$ 
Note that this is the absolute value of the area; the formulas in my answer give, in general, the oriented area (that can be positive or negative). Obviously in this simple case the difference is not immediately appreciated. 
