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I know I asked a similar question sometime before, and the thing is I need them for a proof. So, please help, I promise this is the last one.

What is the simplest way we can find which one of $\sin(\sin(\sin(1)))$and $\cos(\cos(\cos(1)))$ [in radians] is greater without using a calculator [pen and paper approach]? We can use basic calculus.

Any approximation which can be reasonably done with paper and pen is also welcome.

EDIT: On Stevens recommendation, I will tell the background. What I did was that I proved that the minimum of $f=\cos (\cos (\cos (\cos(x))))$ is $\cos(\cos(\cos(1)))$ and maximum of $g=\sin (\sin (\sin (\sin (x))))$ is $\sin(\sin(\sin(1)))$ . So, I just needed this to prove that the maximum of $g$ is smaller than the minimum of $f$, to show that there is no root of the equation $f=g$.

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    $\begingroup$ Out of curiosity, (a) what sort of proof do you need this for, and (b) why is a calculator approach inappropriate? $\endgroup$ – Steven Stadnicki Feb 19 '14 at 8:09
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    $\begingroup$ @StevenStadnicki a) You can see the chat. I have disturbed others for this thing. b) Well, because I want it to be a non calculator approach. $\endgroup$ – Sawarnik Feb 19 '14 at 8:11
  • $\begingroup$ Have you considered Taylor series? $\endgroup$ – Billie Feb 19 '14 at 8:53
  • $\begingroup$ @Billie I have not studied that yet :(, but would that give an answer? $\endgroup$ – Sawarnik Feb 19 '14 at 8:53
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    $\begingroup$ Note that the two values are $0.65429$ and $0.67843$. Therefore it will not be easy at all to solve this problem "from first principles". $\endgroup$ – Christian Blatter Feb 19 '14 at 9:25
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For small positive $x$ one has $$p(x):=1-{x^2\over2}<\cos x<1-{x^2\over2}+{x^4\over24}=:q(x)\ .$$ It follows that $\cos1<q(1)={13\over24}$ and therefore $$\cos(\cos 1)>\cos\biggl({13\over24}\biggr)>p\biggl({13\over24}\biggr)={983\over1152}\ .$$ This in turn implies that $$\cos\bigl(\cos(\cos 1)\bigr)<\cos{983\over1152}<q\biggl({983\over1152}\biggr) ={27814243110433\over42268920643584}\doteq 0.658\ .$$ Similarly, for small positive $x$ one has $$\sin x>1-{x^3\over6}=:s(x)\ .$$ It follows that $\sin 1>s(1)={5\over6}$ and therefore $$\sin(\sin 1)>\sin{5\over 6}>s\biggl({5\over6}\biggr)={955\over1296}\ .$$ This in turn implies that $$\sin\bigl(\sin(\sin 1)\bigr)>\sin{955\over1296}>s\biggl({955\over1296}\biggr)= {8753215805\over13060694016}\doteq 0.670\ .$$

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