Question

For all $\theta$ in $[0,\frac{\pi }{2}]$ Prove that $\cos \left(\sin \theta \right)>\sin \left(\cos \theta \right)$

Answer 1

$\theta ϵ[0,\frac{\pi }{2}]$
Now $\cos \left(\sin \theta \right)>\sin \left(\cos \theta \right)$
$\Rightarrow$ $\sin (\pi /2-\sin \theta )>\sin \left(\cos \theta \right)$
Now we have to prove $\frac{\pi }{2}-\sin \theta >\cos \theta$ $\forall \theta [0,\pi /2]$
Let $f\left(\theta \right)=\pi /2-\sin \theta -\cos \theta$ be a function. Then
$f\left(\theta \right)=\frac{\pi }{2}-\sqrt{2}(\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta )$
$f\left(\theta \right)=\frac{\pi }{2}-\sqrt{2}\sin (\frac{\pi }{4}+\theta )$
$\because$ $0\le \theta \le \frac{\pi }{2}$
$\therefore$ $\frac{\pi }{4}\le (\theta +\frac{\pi }{4})\le \frac{3\pi }{4}$
or $\frac{1}{\sqrt{2}}\le \sin (\theta +\frac{\pi }{4})\le 1$
or $f\left(\theta \right)=\frac{\pi }{2}-\sqrt{2}\sin (\frac{\pi }{4}+\theta )>0$ $\forall \theta ϵ[0,\frac{\pi }{2}]$
or $\frac{\pi }{2}-\sin \theta -\cos \theta >0$
or $\frac{\pi }{2}-\sin \theta >\cos \theta$
or $\sin (\frac{\pi }{2}-\sin \theta )>\sin \left(\cos \theta \right)$
or $\cos \left(\sin \theta \right)>\sin \left(\cos \theta \right)$ $\theta ϵ[0,\frac{\pi }{2}]$
Ans: 5