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What is the simplest way we can find which one of $\cos(\cos(1))$ and $\cos(\cos(\cos(1)))$ [in radians] is greater without using a calculator [pen and paper approach]? I thought of using some inequality relating $\cos(x)$ and $x$, but do not know anything helpful. We can use basic calculus. Please help.

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    $\begingroup$ Sorry, I deleted my comment because it was unhelpful, but I can try again. Since $1$ is somewhere like $\pi/3$, $\cos 1$ is approximately $1/2$. Then $1/2$ is kind of like $\pi/6$, so $\cos 1/2$ is about $\sqrt 3/2$. I think that's close enough to solve it. $\endgroup$ – Ian Coley Feb 19 '14 at 5:19
  • $\begingroup$ If you draw the graphs of $y=\cos x$ and $y=x$ on the same pair of axes, you will immediately see an inequality relating $\cos x$ and $x$, at least for certain values of $x$. $\endgroup$ – David Feb 19 '14 at 5:20
  • $\begingroup$ PS I don't think there should be any need to use calculus. $\endgroup$ – David Feb 19 '14 at 5:21
  • $\begingroup$ @David Yes, I know that there is a root of cos x=x, but to find it is very difficult, and then using the obtained inequality to solve this would surely be an overkill. Some simple solution? $\endgroup$ – Sawarnik Feb 19 '14 at 5:22
  • $\begingroup$ @IanColey I do not think that it may be rigorous enough. Can you make it more detailed? $\endgroup$ – Sawarnik Feb 19 '14 at 5:26
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Since $0\le\cos(x)\le1$ for all $0\le x\le \pi/2$, and cos is a decreasing function on that region, you have $0\le\cos(1)\le1\Rightarrow0\le\cos(1)\le\cos(\cos(1))\le1\Rightarrow0\le\cos(\cos(\cos(1)))\le\cos(\cos(1))\le1$.

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    $\begingroup$ Not quite; note that $\cos(\cos(1))>\cos(1)$ $\endgroup$ – Omnomnomnom Feb 19 '14 at 5:34
  • $\begingroup$ Sorry about that, it might be the local time. $\endgroup$ – blues66 Feb 19 '14 at 5:40
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Here's an idea: we can write $$ \cos(x_0) - \cos(\cos(x_0)) = -2\sin\left(\frac{x_0-\cos(x_0)}{2}\right) \sin\left(\frac{\cos(x_0)+x_0}{2}\right) $$ Since $\sin(x)>0$ when $x>0$ and $\sin(x)<0$ when $x<0$ (for all $x$ within a certain range of $0$), we can deduce that the result will have the opposite sign of $x_0-\cos(x_0)$.

Now, we can set $x_0 = \cos(\cdots(\cos(x))\cdots)$, where $\cos$ is iterated any number of times. We deduce that the sign of the sequence $$ \{1 - \cos(1), \cos(1) - \cos(\cos(1)),\cos(\cos(1))-\cos(\cos(\cos(1))),\dots\} $$ switches sign each time. Proceeding inductively, we have our base case $$ \cos(1) < 1 $$ Which means that $$ \cos(\cos(1))>\cos(1) $$ Which means that $$ \cos(\cos(\cos(1))) < \cos(\cos(1)) $$ and so on.

We can deduce that all iterations of this process will stay in a range for which this trick will work since $\cos([0,\pi/2]) \subset [0,\pi/2]$. We can deduce that the sequence of iterates will converge by noting that since $\cos(x)$ has a derivative whose absolute value strictly less than $1$ on $[0,1]$ and therefore defines a contractive mapping.

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