My attempt:
Since $0\leq\sin^2a_n<1$ which means that $0\leq\arcsin(\sin^2a_n)<\frac{\pi}{2}$, so that the series will decrease when $a_0\in[-1,1]\setminus\{0\}$ without a limit, and that it will stay at $0$ when $a_0=0$.
If i am right (and I highly suspect that I'm not) that the sequence diverges, how would i find $\lim_{n\to\infty}\sqrt{n}a_n$? Should I divide the problem into different cases? If so, I've tried using Stolz-Cesàro when $a_0=0$, but that doesn't seem to lead anywhere.
For the first part we have that if $a_0=0 \implies a_n=0$ and $\sqrt n a_n =0$.
For $a_0 > 0$
$a_{n}>0 $ (by induction)
$a_{n+1}< a_n \iff a_{n+1}- a_n= -\arcsin(\sin^2a_n) < 0$
$a_n \to L$ (monotone sequence theorem)
For $a_0 < 0$ let consider $b_n=\pi+a_n>0$ then
$$a_{n+1}=a_n-\arcsin(\sin^2a_n) \iff b_{n+1}=b_n-\arcsin(\sin^2b_n)$$
$b_{n}>0$ (by induction)
$b_{n+1}< b_n \iff b_{n+1}- b_n= -\arcsin(\sin^2b_n) < 0$
$b_n \to L$ (monotone sequence theorem)
that is $a_n \to -\pi$.
For the second part, for the case $a_0>0$, let consider $na_n^2$ and by Stolz-Cesaro we have
$$na_n^2=\frac n{\frac1{a_n^2}} \implies \frac{n+1-n}{\frac1{a_{n+1}^2}-\frac1{a_n^2}}=\frac{a_{n+1}^2a_n^2}{a_n^2-a_{n+1}^2}\sim \frac{(a_n-a_n^2)^2a_n^2}{a_n^2-a_n+a_n^2}=\frac{(a_n-a_n^2)a_n}{2a_n-1} \to 0$$
and therefore $\sqrt n a_n \to 0$.
By a rough evaluation we can also claim that $a_n\sim \frac 1n$ indeed by $a_n\sim cn^\alpha$
$$c(n+1)^\alpha= cn^\alpha-\arcsin(\sin^2a_n)\sim cn^\alpha-c^2n^{2\alpha}$$
$$\left(1+\frac1n\right)^\alpha -1 \sim-cn^{\alpha} \implies 1+\frac{\alpha}n -1 \sim-cn^{\alpha} \implies c=1,\,\alpha=-1$$
indeed again by Stolz-Cesaro we have
$$na_n=\frac n{\frac1{a_n}} \implies \frac{n+1-n}{\frac1{a_{n+1}}-\frac1{a_n}}=\frac{a_{n+1}a_n}{a_n-a_{n+1}}\sim \frac{(a_n-a_n^2)a_n}{a_n-a_n+a_n^2}=1-a_n \to 1$$
Let $f(x)=x-\arcsin(\sin^2(x))$ then $f'(x)=1-\frac{2\sin(x)\cos(x)}{\sqrt{1-\sin^4(x)}}$ so $f'(x)<1$ for $x\in(0,1]$. Therefore $f(x)<x$ for $x\in(0,1]$ since $f(0)=0$ so your sequence decreases. Also $f(x)>0$ since $f(1)>0$ and $f$ has a local maximum on $(0,1)$. So your sequence is decreasing and bounded and therefore convergent. The limit $L$ must satisfy $L=L-\arcsin(\sin^2(L))\Longleftrightarrow L=\pi z$ with $z\in\mathbb{Z}$ and since $0<a_n<1$ we get $L=0$.
For $x\in[-1,0)$ we have $f'(x)>1$ so again $f(x)<x$ for $x\in[-1,0)$. This also holds for $x\in(-\pi,0)$: Since $f$ is increasing on $(-\frac{\pi}{2},0)$ and has a local maximum on $(-\pi,-\frac{\pi}{2})$ $f$ must have a local minimum at $(-\frac{\pi}{2}\mid-\pi)$ since the function is decreasing before, increasing after and $f'(-\frac{\pi}{2})$ is undefined. Since $f(x)=x\Longleftrightarrow x=\pi z$ with $z\in\mathbb{Z}$ we have $f(x)<x$ for $x\in(-\pi,0)$ and since $f(x)=-\pi\Longleftrightarrow x=-\pi\vee x=-\frac{\pi}{2}$ we have $f(x)\geq-\pi$ for $x\in[-\pi,0]$. So your sequence is decreasing and bounded and therefore convergent again. The limit $L$ must satisfy $L=\pi z$ again and since $-\pi\leq a_n<0$ we get $L=-\pi$.
For $a_0=0$ you don't need Stolz-Cesaro since $a_n=0\forall n\in\mathbb{N}$ so $\lim_{n\to\infty}\sqrt{n}a_n=\lim_{n\to\infty}0=0$
