Limit of a recursiv trigonometric function So while doing my math homework I recently stumbled upon this little thing:
It seems that $y_n = \sin(y_{n-1}) + k$ with arbitrary $y_0$ and $k$ converges to a definite value  for $n \to \infty $.
 
Here is a Mathematica application for little clarification if needed:

f[start_, depth_, const_] := 
    N /@ NestList[Sin[#] + const &, start, depth];
Grenz[start_, const_] := N[Nest[Sin[#] + const &, start, 10000], 10];
Manipulate[
   Show[ListPlot[f[start, depth, const], 
     PlotRange -> {{0, depth + 2}, {0, 
        Max@f[start, depth, const] + 0.1}}], 
    Plot[Grenz[start, const], {x, 0, depth + 2}, 
     PlotStyle -> Red]], {{start, 2}, 0.5, 10}, {{depth, 5}, 1, 100, 
    1}, {const, 0, 10}]

The question now is whether there is a way to exactly calculate the limit for any k and y_0 or at least some sort of general rule etc. that could explain that behavior?
My first approach was to find a x such that sin(x) + k == sin(sin(x)+k)+k but even that seems to be above my head or just not non numerically solvable.
(Btw. replace sin with sin^2 and the situation gets 6 times worse!:
)
I really appreciate any help you can provide.
 A: For a recursive function $y_n=f(y_{n-1})$, a limit/fix point would exist, if something like $y_n=y_{n-1}$ happened. In your case, this would be the solution of the equation $\sin(x)+k=x$, i.e. the intersection point of $f_1(x)=\sin(x)+k$ and $f_2(x)=x$.
If I'm not mistaken, this applies to every recursive function (could someone confirm this? I'm not 100% sure.).
EDIT:
In this specific case, since the maximum slope of function $\sin(x)+k$ is 1 and this maximum slope appears only at points (and not for a wider range), there should be always a unique solution to equation $\sin(x)+k=x$, i.e. there should be always a unique fix point for this recursive function, which is not dependant on the starting value $x_0$, but only on $k$.
EDIT2:
Besides the existance of a limit, apparently it's also important that for any random $x$, the recursive function will 'lead' towards the fix point $x_l$. So a relation like $|f(x)-x_l|<|x-x_l| \;\;\forall \,x\backslash\{x_l\}$ is needed additionally, I think. Not sure for all values of $k$, but at least for $k=0$, the Sinus-function has slopes < 1 in the area around the fix point (0), so the condition is fulfilled here. (Btw, I think something like $|f(x)-x_l|<|x-x_l| \Leftrightarrow \frac{|f(x)-f(x_l)|}{|x-x_l|} < 1$ applies (i.e. kinda the slope definition) because of $f(x_l)=x_l$, therefore the rule with the slopes, I guess?)
