# For all $x \in [0,90]$ show that $\cos(\sin(x))>\sin(\cos(x))$ diff from the one which I posted earlier

For all $x \in [0,90]$ show that $\cos(\sin(x))>\sin(\cos(x))$

I understood the solution given in my book which said

$\cos(x)+\sin(x)≤\sqrt{2}<pi/2$

$\cos(x)<pi/2−\sin(x)$. 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.

• I strongly suggest using radians ($\sqrt{2} < \pi/2\;$ looks OK but$\sqrt{2} < 90\;$ is somewhat ridiculous). Further: the direction of the inequality is only preserved if you apply an increasing function (e.q. $\sin$ in $0..\pi/2$). But $\cos$ is decreasing. – gammatester Jul 29 '14 at 12:57
• Ok thanks @gammatester is it that always we need to take apply an increasing function whichever sign the inequality is? – geek101 Jul 29 '14 at 13:31
• No, you can use decreasing functions but the relations must be reversed. E.g. $f(x)=\frac{1}{x}$ is dereasing for $x>0$ and therefore $$2 < 5 \Longrightarrow \frac{1}{2}=f(2) > f(5) = \frac{1}{5}$$ – gammatester Jul 29 '14 at 13:38
• Ok @gammatester I understood but if there is an equality sign then we can use either of them right? – geek101 Jul 29 '14 at 13:40
• Yes if course. If $a=b$ then $f(a)=f(b)$ for all functions, but in most cases this not a very useful result. – gammatester Jul 29 '14 at 13:41

$\cos x$ is a decreasing function when $x$ is between 0 and $\pi/2$, so if $a<b$, then $\cos a>\cos b$