# Thread: Chain of Sine and Cosine

1. ## Chain of Sine and Cosine

As I was messing around in Maple (yes I'm a nerd), I noticed something.

It appears as if the infinite chain $\displaystyle \sin(\cos(\sin(\cos(\sin(\cos(.....(\sin(\cos(x))) ))))))$ converges for all x.

It appears to be converging to a number very close to $\displaystyle \ln(2)$, but I don't think it actually does, unless it's really, really slowly.

The number (calculated for 7 sine-cosine pairs) is approximately $\displaystyle 0.69481969$.

Also, if you let the first term be cosine instead of sine, this chain gets close to $\displaystyle 0.768169156736795977462$.

2. Originally Posted by redsoxfan325
As I was messing around in Maple (yes I'm a nerd), I noticed something.

It appears as if the infinite chain $\displaystyle \sin(\cos(\sin(\cos(\sin(\cos(.....(\sin(\cos(x))) ))))))$ converges for all x.

It appears to be converging to a number very close to $\displaystyle \ln(2)$, but I don't think it actually does, unless it's really, really slowly.

The number (calculated for 7 sine-cosine pairs) is approximately $\displaystyle 0.69481969$.

Also, if you let the first term be cosine instead of sine, this chain gets close to $\displaystyle 0.768169156736795977462$.

This kind of thing is called fixed point iteration. Starting with $\displaystyle x_0$ being whatever, you let $\displaystyle x_1=\sin(\cos x_0)$ and more generally $\displaystyle x_{n+1}=\sin(\cos x_n)$ for every $\displaystyle n\geq 0$.
There are various assumptions that ensure that the sequence converges (for instance, here the function $\displaystyle f(x)=\sin\cos x$ is contracting: $\displaystyle |f(x)-f(y)|\leq k|x-y|$ where $\displaystyle k\in(0,1)$, hence Banach's fixed point theorem applies). Then, if $\displaystyle x_n\to\ell$, necessarily we have $\displaystyle \ell=\sin\cos\ell$ (taking the limit in $\displaystyle x_{n+1}=\sin\cos x_n$).
So that your constants are just the fixed points of $\displaystyle \sin\cos$ and $\displaystyle \cos\sin$: the numbers $\displaystyle c_1,c_2$ such that $\displaystyle \sin\cos c_1=c_1$ and $\displaystyle \cos\sin c_2=c_2$ (they are unique). I don't think there exits a simple explicit formula for them.