Which is greater, $\cos(\cos(1))$ or $\cos(\cos(\cos(1)))$? 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. 
 A: 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 $0\le\cos(1)\le1\Rightarrow0\le\cos(1)\le\cos(\cos(1))\le1\Rightarrow0\le\cos(\cos(\cos(1)))\le\cos(\cos(1))\le1$.
A: 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.
