Find the limit of: cos(cos(...cos1)

Question

Find the limit of cosine composing with itself n times:

$\lim\limits_{n \to \infty}\left ( cos(cos(\cos(...\cos1 \right )$

I have worked on it a bit: The sequence is: 0,1,cos(1),cos(cos1),... I identified the function mapping [0,1] to [0,1] and the function is decreasing in this interval. Then I saw that if even terms sequence is increasing the odd ones is decreasing.

$${a_{2n}}\le{a_{2n+1}}.$$ Then compose of two decreasing functions is increased on [0,1]. I stuck here :( Would it be enough to show that $${a_{2n+1}-a_{2n}}\le{k}(a_{2n-1}-a_{2n-2})$$ for some k: 0<k<1

I have saw an answer on this site but I could not understand. It says that from Mean Value Theorem there exist a number x of the segment [a2n-1,a2n-2] which is on [0,1] so that: (a2n-1 - a2n-2)*sin(cosx)*sinx

0 < sin(cosx)*sinx < sin(cos1)*sin1= k < 1

and this:

$${a_{2n+1}-a_{2n}}\le{k}(a_{2n-1}-a_{2n-2})$$

This wasn't clear for me, from mean value theorem. Did he used that for function or for the sequence? should we take an=f(n) and then (an-1 - a2n-2)*sin(cosx)*sinx= then for [x1,x2]; f (f(x1)-f(x2))=f'(an)(x2-x1) ? Could we use that in this way? Please can someone clarify the part from mean value theorem becuase I couldn't understand it.