Proving the Inequality: sin(x) < x for x > 0

[nos]

Hello all,

I want to prove the following inequality.
sin(x)<x for all x>0.

Now I figured that I put a function f(x)=x-sin(x), and show that it is increasing for all x>0. But this alone doesn't prove it. I need to show we have inequality from the start. I can't show that lim f(x) as x->0 is positive cause this limit equals 0. I can show that we have equality at x=0, and only at x=0. Therefore, sin(x) <=x for all x>=0, and we only have equality at x=0. So sin(x)<x for all x>0. This doesn't seem the right way to do it though.

Thanks.

 

[pasmith]

Hello all,

I want to prove the following inequality.
sin(x)<x for all x>0.

Now I figured that I put a function f(x)=x-sin(x), and show that it is increasing for all x>0. But this alone doesn't prove it. I need to show we have inequality from the start. I can't show that lim f(x) as x->0 is positive cause this limit equals 0. I can show that we have equality at x=0, and only at x=0. Therefore, sin(x) <=x for all x>=0, and we only have equality at x=0. So sin(x)<x for all x>0. This doesn't seem the right way to do it though.

No, you are correct: x - \sin x is zero at x = 0 and is thereafter strictly increasing, so x - \sin x can't be zero or negative for x &gt; 0.

 

[Ssnow]

You can visualize it comparing the graphs of $y(x)=x$ and $y(x)=\sin{(x)}$

 

[zinq]

Hint: sin(0) = 0, and 0 < d(sin(x))/dx < 1 for 0 < x <= π/2 (why?).