inequality exercise asked by Imad zak in many math groups

Exercise:

Show that, $\cos \left( \sin x \right)>\sin \left( \cos x \right)\,\,,\,\forall x\in \mathbb{R}$

Solution: we have $\cos \left( \sin x \right)>\sin \left( \cos x \right)\Leftrightarrow \cos \left( \sin x \right)-\sin \left( \cos x \right)>0$

Put $f\left( x \right)=\cos \left( \sin x \right)-\sin \left( \cos x \right)$ and let’s prove $f>0$

We know that, \[\cos a-\cos b=-2\sin \left( \frac{a+b}{2} \right)\sin \left( \frac{a-b}{2} \right)\]

So $\cos \left( \sin x \right)-\cos \left( \frac{\pi }{2}-\cos x \right)=-2\sin \left( \frac{1}{2}\left( \sin x+\frac{\pi }{2}-\cos x \right) \right)\sin \left( \frac{1}{2}\left( \sin x-\frac{\pi }{2}+\cos x \right) \right)$

Observe that,

$\sin x+\cos x=\frac{\sqrt{2}}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=\sqrt{2}\left( \cos \left( \frac{\pi }{4} \right)\sin x+\sin \left( \frac{\pi }{4} \right)\cos x \right)=\sqrt{2}\sin \left( \frac{\pi }{4}+x \right)$

And $\sin x-\cos x=\sqrt{2}\sin \left( x-\frac{\pi }{4} \right)$

So $\cos \left( \sin x \right)-\cos \left( \frac{\pi }{2}-\cos x \right)=-2\sin \left( \frac{\sqrt{2}}{2}\sin \left( x-\frac{\pi }{4} \right)+\frac{\pi }{4} \right)\left( \frac{\sqrt{2}}{2}\sin \left( \frac{\pi }{4}+x \right)-\frac{\pi }{4} \right)$

$\left| \sin x-\cos x \right|=\left| \sqrt{2}\sin \left( x-\frac{\pi }{4} \right) \right|=\sqrt{2}\left| \sin \left( x-\frac{\pi }{4} \right) \right|$ but $\left| \sin \left( x-\frac{\pi }{4} \right) \right|\le 1$

Hence $\left| \sin x-\cos x \right|\le \sqrt{2}$ and $\sqrt{2}<\frac{\pi }{2}$ thus $\left| \sin x-\cos x \right|<\frac{\pi }{2}$

Thus $-\frac{\pi }{4}<\frac{\sin x-\cos x}{2}<\frac{\pi }{4}\Leftrightarrow 0<\frac{\sin x-\cos x}{2}+\frac{\pi }{4}<\frac{\pi }{2}$

hence $0<\sin \left( \frac{\sin x-\cos x}{2}+\frac{\pi }{4} \right)<1$

also $\left| \sin x+\cos x \right|=\left| \sqrt{2}\sin \left( \frac{\pi }{4}-x \right) \right|=\sqrt{2}\left| \sin \left( \frac{\pi }{4}-x \right) \right|\le \sqrt{2}<\frac{\pi }{2}$

hence $-\frac{\pi }{2}<\frac{\sin x+\cos x}{2}-\frac{\pi }{4}<0\Leftrightarrow -1<\sin \left( \frac{\sin x+\cos x}{2}-\frac{\pi }{4} \right)<0$

So $\cos \left( \sin x \right)-\sin \left( \cos x \right)>0$ thus $f>0\Leftrightarrow \cos \left( \sin x \right)>\sin \left( \cos x \right)$