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

Printable View

- Aug 5th 2013, 07:54 AMKristen111111111111111111Limits
Can you please tell me a geometrical method to prove that 0 < |sin(x)| < √2(1-cos(x)) < |x| ?

- Aug 5th 2013, 09:01 AMShakarriRe: 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 - Aug 5th 2013, 10:20 AMHallsofIvyRe: 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.

- Aug 5th 2013, 03:39 PMShakarriRe: Limits
Your inequality signs should be less than or equal. At x=0 you get 0<0<0<0

- Aug 5th 2013, 05:09 PMKristen111111111111111111Re: 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)) - Aug 5th 2013, 06:12 PMShakarriRe: Limits
Well I don't know a geometric proof but to prove |sin(x)| < √(2(1-cos(x))) you can do it this way

$\displaystyle |sin(x)|<\sqrt{2(1-cos(x)}$

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

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

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

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