TITLE: For all real $\theta$ prove that $ \cos(\sin\theta) \gt \sin(\cos\theta)$
QUESTION [12 upvotes]: How do I prove this? Im not able to even start it. Help please!

REPLY [3 votes]: The inequality $\cos(\sin\theta)\gt\sin(\cos\theta)$ for all $\theta$ really boils down to the inequality $x\gt\sin x$ for $x\gt0$ and the fact that $\cos x$ is decreasing on the interval $0\le x\le\pi$.  I'll take those as known.  (They are easy enough to prove.)
Let's start by observing that we need only verify the inequality $\cos(\sin\theta)\gt\sin(\cos\theta)$ for $0\lt\theta\lt\pi/2$:
By the periodicity of sine and cosine, it suffices to prove the inequality for $-\pi\le\theta\le\pi$.  Since cosine is an even function (and sine is odd), it suffices to prove it for $0\le\theta\le\pi$.  The inequality is clearly satisfied on since $\pi/2\lt\theta\le\pi$, $\sin(\cos\theta)$ is negative there while $\cos(\sin\theta)$ is positive.  Finally, it's easily checked for $\theta=0$ and $\pi/2$.  So that leaves $0\lt\theta\lt\pi/2$
Since we're now on an inteval where $\cos\theta\gt0$, we can let $x=\cos\theta$ and get $\cos\theta\gt\sin(\cos\theta)$.  But we also have $\pi\gt\theta\gt\sin\theta\gt0$ on this interval, which, by the fact that the cosine function in decreasing there, gives us $\cos(\sin\theta)\gt\cos\theta$.  Putting it all together gives the desired inequality.