Prove independence of two random variables. Let two random variables $\xi_1$ and $\xi_2$ be given. They are independent and have a standard normal distribution. Proove that $\frac{\xi_1^2 - \xi_2^2}{\sqrt{\xi_1^2+\xi_2^2}}$ and $\frac{2\xi_1\xi_2}{\sqrt{\xi_1^2+\xi_2^2}}$ are independent.
I was given a little hint that I need to think about the angles and may be it's necessary to go to the polar coordinate system.
Please help to prove it or give me a small hint)
 A: according to
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$
they are independent.
but if you want to continue your way: 
$X=r\cos(\theta)  \hspace{.5cm}
Y=r\sin(\theta) \hspace{.5cm}
X,Y \sim normal(0,1) \Leftrightarrow  \theta \sim  Uniform(0,2\pi) \hspace{.5cm} r^2\sim \chi^2_2$.
$X$ and $Y$ are independent $\Leftrightarrow $ $\theta$ and $r$ are independent.
also $\sin(\theta) \sim \cos(\theta) \sim  \sin(2\theta) \sim  2\sin(\theta) \cos(\theta)
\sim \cos(2\theta) \sim \cos(2\theta) \sim f $ that $ f(z) =\frac{1}{\pi \sqrt(1-z^2)} I_{[-1,1]}(z) $ since
$z=\sin(\theta) \Rightarrow  f(z)=|\frac{d}{dz} \sin^{-1}(z)| f_{\theta}(\sin^{-1}(z)) + |\frac{d}{dz} (\pi-\sin^{-1}(z))| f_{\theta}(\pi -\sin^{-1}(z)) =\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} +\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} =\frac{1}{\pi\sqrt(1-z^2)} $
similar for others.
$\frac{2XY}{\sqrt(X^2+Y^2)}=\frac{2r^2 \cos(\theta) \sin(\theta)}{r}=2r \cos(\theta) \sin(\theta) =r  \sin(2\theta) \sim r \sin(\theta) $ 
$\frac{X^2-Y^2}{\sqrt(X^2+Y^2)}=\frac{r^2(\cos^2(\theta)-\sin^2(\theta))}{r}$
$=r\cos(2\theta)\sim r\cos(\theta)$
note $X=r \sin(\theta) $ and $Y=r\cos(\theta)$ are independent.
