Inspired by John Muradeli

Let $x$ be a an arbitrary real number and $(x_{n})_n$ be a sequence defined by the recursion $x_{n+1}=\cos x_{n},$ where $x_{1}=\cos x$ . Obviously, we have that $|x_{n}| \leq 1$ for any natural number $n$ . On the other hand, using the mean value theorem , we have that $\cos a- \cos b = \sin (c) (a-b)$ for a certain number $c$ in between $a$ and $b$ . Therefore if $a$ and $b$ are in the interval $[-1, 1]$ the following inequality holds $|\cos a -\cos b|\leq r|a-b|\:\:\:\:\:\:(*)$ where $r= \sin(1)$ and it is obvious that $0<r<1$ . Now, using mathematical induction we can prove that $|x_{n+1}-x_{n}|\leq r^{n-1}| x_{2}-x_{1}|.$ Indeed, this is true for $n=1$ and now if we assume the inequality is true when $n=k,$ then using $(*)$ we obtain that $|x_{k+2}-x_{k+1}|=|\cos x_{k+1}-\cos x_{k}|\leq r |x_{k+1}-x_{k}|\leq r^{k}|x_{2}-x_{1}|.$

Now, we can also see the following: $|x_{n+m}-x_{n}|\leq |x_{n+m}-x_{n+m-1}|+...+|x_{n+2}-x_{n+1}|+|x_{n+1}-x_{n}|\leq (r^{n+m-2}+...+r^{n-1})|x_{2}-x_{1}|$ $\leq r^{n-1}(r^{m-1}+r^{m-2}+...+1)|x_{2}-x_{1}|<\frac{r^{n-1}}{1-r}|x_{2}-x_{1}|.$ The latter inequality implies that $x_{n}$ is a Cauchy sequence and therefore is convergent. So $\lim_{n \to \infty}{x_{n}}$ exists and is a finite number that we can denote by $l$ . Now taking limits in both sides of the equations $x_{n+1}=\cos x_{n},$ and using the continuity of cosine we get that $l=\cos l.$ Solving this equation approximately we get that $l\approx 0.739.$ And this is the answer of the question.

Assume that limit of the expression exist and it is $x$ . Then, we have:

$\begin{aligned} x & = \lim_{n \to \infty} {(\cos{(\cos{(\cos{(...)})})})} \\ & = \cos{\left( \lim_{n \to \infty} {(\cos{(\cos{(\cos{(...)})})})}\right)} \\ & = \cos{x} \end{aligned}$

Using numerical method, we get $x=\boxed{0.739}$

Assuming that x is such that

0<cosx<1,

then:

cos1<cos(cosx)<1

cos1<cos(cos(cosx))<cos(cos1))

cos(cos(cos1))<cos(cos(cos(cosx)))<cos(cos1))

and so on. These cosines vary less and less, alternating up and tending the limits at every step.It can be shown that after a certain number of cases we have, to three decimal places, the threshold value 0739.