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This question already has an answer here:

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?

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marked as duplicate by Simply Beautiful Art calculus Oct 13 '17 at 1:34

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

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