Basically, I'd like to add $n$ random vectors in a 2 dimensional space of unit length and of angle $\theta$ relative to a global axis. The probability density function of the angle $\theta$ is a uniform distribution over the interval $[0..2\pi)$. I've come to the equation to find the following vector sum for $n$ vectors:
$$ d=\sqrt{(\sum_{i=1}^{n}{\sin(\theta_i)})^2+(\sum_{i=1}^{n}\cos(\theta_i))^2} $$
Where might I go from here to find the probability distribution of $d$? I'm mostly curious about probability distribution of $d$ in the case where $n \rightarrow \infty$.
Thanks!
Note that $D_n=|Z_1+\cdots+Z_n|$ where $Z_k=\mathrm e^{\mathrm i\Theta_k}$ and the random variables $(\Theta_k)$ are i.i.d. and uniform on $[0,1]$. The increments $Z_k$ are centered hence the central limit theorem implies that $(Z_1+\cdots+Z_n)/\sqrt{n}$ converges in distribution to a centered complex normal random variable $U+\mathrm iV$ of covariance matrix $\Sigma=\mathrm{Cov}(\Re Z_1,\Im Z_1)$, that is, $\Sigma=\mathrm{Cov}(\cos\Theta_1,\sin\Theta_1)$.
Now $E(\cos^2\Theta_1)=E(\sin^2\Theta_1)=\frac12$ and $E(\cos\Theta_1\sin\Theta_1)=0$ hence $\Sigma=\frac12I$, that is, $U^2+V^2=\frac12(X^2+Y^2)$ where $(X,Y)$ is i.i.d. standard normal.
In particular, $D_n/\sqrt{n}\to W$ in distribution, where $W=\sqrt{\frac12(X^2+Y^2)}$. The distribution of $W$ is called a $\chi$ distribution and its density is $w\mathrm e^{-w^2/2}$ on $w\geqslant0$. Thus, the complementary CDF of $W$ is such that $P(W\gt w)=\mathrm e^{-w^2/2}$ for every $w\geqslant0$. Finally, $W=\sqrt{-2\log \Xi}$ where $\Xi$ is uniform on $[0,1]$.