Recursive sequence with $\sin x$ and Taylor series

Question

I have been trying to solve that problem for two days but no answer yet. There it is:
there is a recursive sequence:
$a_{1}=1$
$a_{n+1}=\sin(a_{n})$
I am to proof that:
$$\lim_{n \to \infty} \sqrt{n}a_{n}=\sqrt{3}$$
What I know is that:
for $x>0: \sin(x) < x$
Thus:
$a_{2}=\sin(a_{1})<a_{1}<1$
$a_{3}=\sin(a_{2})<a_{2}<a_{1}<1$
$...$
$a_{n+1}=\sin(a_{2})<a_{2}<a_{1}<1$
Now we know that $a_{n}$ is decreasing. And from that and Banach's theorem we know that:
$\lim\limits_{n \to \infty}a_{n}=0$
However I have no idea how to proof the other limit.