# How can I prove that $\cos(\sin x)>\sin(\cos x)$ when $0\leq x\leq \pi$ [duplicate]

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I had a midterm a few days ago,.. and my question goes like this. Prove $\cos(\sin x)>\sin(\cos x)$ where $0\leq x\leq \pi$

Also, you can use sinx>x when x>0 without proving it.

How can I prove it strictly?

## marked as duplicate by Simply Beautiful Art calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 13 '17 at 1:34

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• $sin x$ is smaller then $x$ on for $x$ greater then 0 – Muzi Sep 30 '17 at 5:00
• – user222031 Sep 30 '17 at 5:03
• In particular, see the first answer mentioned in user222031's link. – Simply Beautiful Art Oct 13 '17 at 1:35

## 1 Answer

$$\cos\sin{x}-\sin\cos{x}=\sin\left(\frac{\pi}{2}-\sin{x}\right)-\sin\cos{x}=$$ $$=2\sin\frac{\frac{\pi}{2}-\sin{x}-\cos{x}}{2}\cos\frac{\frac{\pi}{2}-\sin{x}+\cos{x}}{2}>0$$ because by C-S $$\sin{x}\pm\cos{x}\leq\sqrt{(1^2+1^2)(\sin^2x+\cos^2x)}=\sqrt2<\frac{\pi}{2}$$