For all $x \in [0,90]$ show that $\cos(\sin(x))>\sin(\cos(x))$
I understood the solution given in my book which said
$\cos(x)<pi/2−\sin(x)$. Over here if we take $\sin$ of both sides we get the answer.
But if $\sin(x)<pi/2−\cos(x)$,then when we take $\cos$ of both sides, we get two different and opposite answers.
Please explain to me where I have gone wrong.
I have already posted this question but it is an edited version of it and it was wrongly interpreted. Made a few changes.