You have a linear transformation from one polar coordinate system to another: \(\displaystyle r'= 1- r\) and \(\displaystyle \theta'= sin^2(\theta/2)\). I would start by using the "half angle formula":

\(\displaystyle sin(\theta/2)= \sqrt{(1/2)(1- cos(\theta)}\) so that \(\displaystyle \theta'= sin^2(\theta/2)= \frac{1}{2}(1- cos(\theta)\). Now use the fact that \(\displaystyle \theta= arctan\left(\frac{y}{x}\right)\), \(\displaystyle r= \sqrt{x^2+ y^2}\), \(\displaystyle \theta'= arctan\left(\frac{y'}{x'}\right)\), and \(\displaystyle r'= \sqrt{x'^2+ y'^2}\)