sequence $x_{n+1} = x_{n} + \sin x_{n}$ There is a sequence which satisfies
$$x_{1} = a$$
$$x_{n+1} = x_{n} + \sin x_{n}$$
where a = 1 . 
Why does $$\lim_{n \rightarrow \infty} x_{n} = \pi$$ hold ??
(the first version of question was with 2 misprints!!)
 A: If $f(x)=x+\sin(x)$ then $f$ is strictly increasing on $[0,\pi)$ and $f((0,\pi))\subset(0,\pi)$, so if $x_1=a\in(0,\pi)$ then $x_n\in(0,\pi)$. Moreover, it is clear that $f(x)>x$ for $x\in(0,\pi)$. That is
$$\forall\,x\in(0,\pi),\qquad 0<x<f(x)<\pi$$
So, if $x_1=a\in(0,\pi)$ then by an easy induction we get
$$\forall\,n\geq1,\qquad 0<x_n<x_{n+1}<\pi$$
This proves that $(x_n)_{n\geq1}$ is bounded and increasing. So, it must convege to some
limit $\ell\in(0,\pi]$ with $f(\ell)=\ell$. this implies that $\ell=\pi$ and we are done.
Remark. The convergence of this sequence is remarkably fast. Indeed, if $e_n=\pi-x_n$ then
$$0\leq e_{n+1}=e_n-\sin e_n\leq \frac{e_n^3}{6}$$
So the rate of convergence is cubic. For instance, when $x_1=1$ we have $e_6=5.6 \times 10^{-17}$, while for $x_1=3$ we have $e_4<10^{-16}$.
$$
0<\pi-\left(
3+\sin 3+\sin (3+\sin  3)+\sin (3+\sin  3+\sin (3+\sin  3))\right)<10^{-16}.$$
A: If we assume that the limit exists, let $\lim_{n\to\infty} x_n = x$.  Then:
$$\left(\lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}x_n+\sin(x_n)\right) \implies x=x+\sin(x)$$
Thus:
$$0=\sin(x) \implies x = k\pi, \quad k\in\Bbb{Z}$$
Thus, your stated equality does not hold.
A: It doesn't necessarily hold for all a. For example
$$x_{1} = {n\pi} \implies x_{n} = x_{1} \forall n$$ 
A: Do you mean $x_{n+1}=x_n+\cos x_n$?
We know $\sin y<y$ when $0<y<\pi/2$, so substitute $x=\pi/2-y$ and get $\cos x< \pi/2-x$ when $0<x<\pi/2$.
So adding $\cos x_n$ will not quite reach $\pi/2$.
On the other hand, $\sin y$ and $y$ get very close when $x$ is small, so adding this term gets very close to $\pi/2$.
