PDF of a squared L2-norm of a vector that is the difference between two uniformly chosen vectors on unit circle Let $Z = ||X-Y||_2^2$ where $X$ and $Y$ are two points selected uniformly-randomly and independently on the unit  circle. I'm trying to find the PDF of $Z$.
It is very similar to this question but answers over there are too clever for me. So here is what I did:
Since we are on the unit circle, if we switch to polar coordinates, both $X$ and $Y$ can be characterized by the angle. So, $$ X = \begin{bmatrix}cos(\theta_x)\\sin(\theta_x)\end{bmatrix}\ \ \  Y = \begin{bmatrix}cos(\theta_y)\\sin(\theta_y)\end{bmatrix}$$
where $\theta_x$ and $\theta_y$ are both $\sim Unif(0, 2\pi)$. Now the squared $L_2$ norm becomes
\begin{align*}
    ||X-Y||_2^2 & =
    (cos(\theta_x)-cos(\theta_y))^2 + (sin(\theta_x)-sin(\theta_y))^2 \\[7pt]
         & =  2 - 2(cos(\theta_x)cos(\theta_y)+sin(\theta_x)sin(\theta_y)) \\[7pt]
         & =  2 - 2cos(\theta_x - \theta_y)
\end{align*}
I can now formulate the CDF of $Z$:
\begin{align*}
    P(Z\leq z) & = P(||X-Y||_2^2 \leq z) \\[10pt]
               & = P(2 - 2cos(\theta_x - \theta_y) \leq z) \\[7pt]
               & = P(cos(\theta_x - \theta_y) \geq \frac{2-z}{2})
\end{align*}
Now I take $arccos$ of both terms:
\begin{align*}
    P(Z\leq z) & = P(\theta_x - \theta_y \leq arccos(\frac{2-z}{2}))
\end{align*}
And I think this corresponds to the area of the white trapezoid below, where axes represent $\theta_x$ and $\theta_y$, red & blue lines are boundaries for $(0, 2\pi)$ and green line is $\theta_x - \theta_y = arccos(\frac{2-z}{2})$: 

which gives
$$ F_Z(z) = 
\begin{cases}
  0  & \text{ for } z \leq 0\\[7pt]
  4\pi^2 - \frac{1}{2} (2\pi - arccos(\frac{2-z}{2}))^2 & \text{ for } 0 \lt z \lt 4\\[7pt]
  1  & \text{ for } z \geq 4\\
\end{cases}$$
but this can't be true because $F_Z(z)$ is not continous at either end. I suspect I failed with the $arccos$ operation, any thoughts on how can I correct the steps?
 A: Plese note the fact that
$$P(\cos(a)>b)=P(\color{red}{|a|}<\text{arccos}(b))=P(-\text{arccos}(b)<a<\text{arccos}(b))$$
(but in fact this is unimportant for the sequel).
Edit: You have in fact a classical arcsine distribution under a slightly generalized form $a=0,b=4$. Please note that (see the Wikipedia article) that an arcsine distribution is obtained from a uniform RV $U$ on [0,1] by considering $X=\cos(\pi U)$.
A: Not repeating all the steps that you have already done correctly.
$\theta = \theta_x - \theta_y$ is the angle measured between the two randomly chosen points on the circle from its center.
Please note $\theta$ is uniformly distributed between $0$ and $\pi$.
Formulating the CDF of $Z$ -
\begin{align*}
F_Z(z) \equiv P(Z\leq z) & = P(||X-Y||_2^2 \leq z) \\[10pt]
               & = P((2 - 2cos \theta) \leq z) \\[7pt]
               & = P(cos \theta \geq \frac{2-z}{2}) \\[7pt]
               & = P(cos \theta \geq \frac{2-z}{2}) \\[7pt]
               & = P(\theta \leq \arccos\frac{2-z}{2}) \\[7pt]
\end{align*}
We know
$$F_{\theta}(\theta) = 
\begin{cases}
   \frac{\theta}{\pi} & 0 \leq \theta \leq \pi \\[7pt]
  0  & \text{ otherwise } \\
\end{cases}$$
So,
$$F_{Z}(z) = 
\begin{cases}
   \frac{1}{\pi} \arccos\frac{2-z}{2} & 0 \leq z \leq 4 \\[7pt]
  0  & \text{ otherwise } \\
\end{cases}$$
So, pdf $ = \frac{1}{\pi}  \frac{d(\arccos\frac{2-z}{2})}{dz} = \frac{1}{\pi} \frac{1}{\sqrt {4z-z^2}} \text { } (0 \leq z \leq 4)$
