Since $\cos (2x) = 2 \cos^2 (x) - 1$, I wonder about the iteration $(x,y) \mapsto (2x\mod 2\pi, 2y^2-1)$. Will it converge to the cosine graph? I tried it in mathematica and got quite a few points, if I cheat and give it an initial value.
x = {1, 0.540302306}; step[{a_, b_}] := {Mod[2 a, 2 Pi], 2 b^2 - 1} Table[(x = step[x]; x), {k, 1, 30}]; ListPlot[%, AspectRatio -> Automatic]
Here's a variation based on the sum of angles formula gives nice convergenc to the graph.
x = {1, 0.540302306}; step[{a_, b_}] := (c = Mod[a + 1, 2 Pi]; {c , b Cos[1] - If[c < Pi, 1, -1] Sin[1] Sqrt[1 - b^2]}) Table[(x = step[x]; x), {k, 1, 500}]; ListPlot[%, AspectRatio -> Automatic]
These two algorithms have difference tolerances for error. In the first algorithm, if I start with a wrong initial value like {1, 0.4} I get a chaotic "cloud", but in the second case, it still converges to the cosine graph.
I guess I would like to know, does my original iteration have any other fixed points besides the cosine graph?