Can you please tell me a geometrical method to prove that 0 < |sin(x)| < √2(1-cos(x)) < |x| ?
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
Well I don't know a geometric proof but to prove |sin(x)| < √(2(1-cos(x))) you can do it this way
$\displaystyle |sin(x)|<\sqrt{2(1-cos(x)}$
$\displaystyle sin^2(x)<2(1-cos(x))$
$\displaystyle 0< 1+1-sin^2(x)-2cos(x)$
$\displaystyle 0< 1+cos^2(x)-2cos(x)$
$\displaystyle 0< (1-cos(x))^2$