rotating 2D coordinates I've tried googling this, but I always end up somewhere that just says it's easy.
Anyhow, I have a coordinate system, where I need to rotate a bunch of points.
It's all 2D.
Coordinates varies and so does the angles, but the point to rotate around, is always 0,0.
I've look a bit at this post, but I haven't really been able to make it work.
https://stackoverflow.com/questions/2259476/rotating-a-point-about-another-point-2d
It mentions subtracting the pivot point, but should I subtract the distance to the pivot point?
Since my pivot point is $(0,0)$ it sounds too easy, to just subtract 0 (and also doesn't give me the results I expect).
As an example, I have a point in $ (2328.30755138590, 1653.74059364716) $ (very accurate, I know).
I need to rotate it $ 5.50540590872993 $ degrees around $(0,0)$.
I would expect it to end in 2339.68319170805, 1878.18099075262(based on a rotation in my CAD software) but I don't really see how I can get it to do that.
Actually, I need to rotate it around $(0, 1884.25802838571)$. Sorry. I have gotten some coordinate systems mixed.

Let's say you want to rotate a point $P = (a, b)$ w.r.t. to point $Q = (c, d)$ by angle $\theta_0$.
You need to first calculate the co-ordinate of point $P$ taking $Q$ as origin, which will be $(a-c, b-d)$. 
Convert the representation of P from cartesian co-ordinate system to polar co-ordinate system. Now the polar co-ordinates of P(taking Q as origin) will be 

$$\left (\sqrt{{(a-c)^2}+(b-d)^2}, sin^{-1}\frac{c-d}{\sqrt{{(a-c)^2}+(b-d)^2} }  \right )$$ 

All you have to do now is to increase the angle by $\theta_0$. Lastly convert the polar co-odinate back to cartesian co-ordinate using this $(r, \theta)$ ~ $(r\sin\theta, r\cos\theta)$.
But note that your co-ordinates are taking point Q as the origin. To find the co-ordinate w.r.t. to the real origin, just add $c$ and $d$ to the x and y co-ordinates respectively.
Note- To rotate a point $(a,b)$ in cartesian co-ordinate by angle $\phi$ you just need to multiply matrix 
$\left[
  \begin{array}{ c c }
     \cos\phi & -\sin\phi \\
     \sin\phi & \cos \phi
  \end{array} \right]
$and
$ 
\left[
  \begin{array}{ c c }
     a \\
     b
  \end{array} \right]$, hence you can apply this matrix multiplication to 
$\left[
  \begin{array}{ c c }
     a-c \\
     b-d
  \end{array} \right]
$ and then add $c$ and $d$ to the $x$ and $y$ co-ordinate.
