TITLE: Why does $\cos (\pi\cos (\pi \cos (\log (20+\pi)))) \approx -1$
QUESTION [6 upvotes]: I read on Wikipedia that 
$$\cos (\pi\cos (\pi \cos (\log (20+\pi)))) \approx -1$$
to a high degree of accuracy.  Why is this true?  Is this pure coincidence or is there some mathematical background?

REPLY [13 votes]: It is a well known coincidence that 
$$e^{\pi}-\pi \approx 20$$
Using this, we find
$$e^{\pi}-\pi \approx 20 \implies \pi\approx \log ( 20+\pi)$$
then
$$-1 =\cos (\pi) \approx \cos(\log ( 20+\pi))$$
$\cos (-\pi)=-1$, so a closer approximation of $-1$ can be found with
$$-1 =\cos(\pi\cos (\pi)) \approx \cos(\pi\cos(\log ( 20+\pi)))$$
and again
$$-1 =\cos(\pi \cos(\pi\cos (\pi))) \approx \cos(\pi\cos(\pi\cos(\log ( 20+\pi))))$$

In fact, if $x_0 \approx -1$ and $x_n=\cos (\pi x_{n-1})$ then 
$$\lim_{n \to \infty}x_n=-1$$