For all $\theta$ in $[0,\pi/2]$ I need to show that $\cos (\sin \theta)>\sin (\cos \theta)$.
In my book it is done like $cos (\theta)<\pi/2- sin (\theta) $.Then they took sine on both sides ? But I doubt this approach.Since $\sin $ is not an increasing function everywhere is it correct to directly take sine on both sides of inequality ? And can someone provide alternate solution to this problem too ? Thanks.