Showing monotone convergence of recursive relation with $ x_{n+1} = \sin(x_{n}) $ how I can show the following monotonic relation  : $x_{n}-x_{n+1}>=0$ with 
 $x_{n+1}=sin(x_{n})$:
My Idea:
with
$x_{n}-x_{n+1}=x_{n}-sin(x_{n})=....$here I am stuck.
it were  nice if someone can help me at.
greetings

This requires an additional ssumption. For example if $x_1=-\frac {\pi} 2$ then $x_2=-1 >x_1$.
If $x_n \geq 0$ for all $x$ then  this result is true and it follows from the inequality $\sin x \leq x$ for all $x \geq 0$. 
Proof of $\sin x \leq x$ for $x \geq 0$: let $f(x)=x-\sin x$. Then $f(0)=0$ and $f'(x)=1-\cos x \geq 0$. Hence $f$ is montonically increasing so $f(x) \geq f(0)=0$ for $x \geq 0$. 
