1. Limits

Can you please tell me a geometrical method to prove that 0 < |sin(x)| < √2(1-cos(x)) < |x| ?

2. Re: Limits

The inequality it false for x=2. The easiest way to check an inequality before attempting to prove it is to google search the function. I google searched sqrt(2)*(1-cos(x))-abs(x) to check that part of the inequality and found a value for x when it was not less than zero.
Also, this does not belong in the new users forum

3. Re: Limits

Do you mean $\displaystyle \sqrt{2}(1- cos(x))$ or $\displaystyle \sqrt{2(1- cos(x))$? What you wrote is the first but, as Shakarri said, that is simply not true.

4. Re: Limits

Your inequality signs should be less than or equal. At x=0 you get 0<0<0<0

5. Re: Limits

Yeah I'm really sorry I forgot to mention that x belongs to (-pi/2 , pi/2 ) \ {0}
And what I meant was \sqrt{2(1- cos(x))

6. Re: Limits

Well I don't know a geometric proof but to prove |sin(x)| < √(2(1-cos(x))) you can do it this way
$\displaystyle |sin(x)|<\sqrt{2(1-cos(x)}$

$\displaystyle sin^2(x)<2(1-cos(x))$

$\displaystyle 0< 1+1-sin^2(x)-2cos(x)$

$\displaystyle 0< 1+cos^2(x)-2cos(x)$

$\displaystyle 0< (1-cos(x))^2$