2014/068) which is greater $\cos(\sin\, x)$ or $\sin(\cos\, x)$

we have
$\cos(\sin\, x)-\sin(\cos\, x)$
= $\cos(\sin\, x)-\cos(\dfrac{\pi}{2}- \cos\, x)$
= $\cos(\sin\, x)+\cos(\dfrac{\pi}{2}+ \cos\, x)$
=  $2 \cos \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2} \cos \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$
=  $2 \cos\ t \ \cos \ s$
where $t= \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$
and $s=  \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$

is product of 2 terms and we can show that both t and s are between 0 and $\frac{\pi}{2}$

we have $\sin\, x\pm \cos\, x= \sqrt{2}\sin ( x\pm \dfrac{\pi}{4})$ so between $- \sqrt{2}$ and $\sqrt{2}$

hence s and t both are positive and less than $\frac{\pi}{2}$ so the difference $\cos(\sin\, x)-\sin(\cos\, x)\gt \ 0 $ and hence
$\cos(\sin\, x) \gt \sin(\cos\, x)$