I want to show that $$ f(x)=\dfrac{x-\sin(x)}{1-\cos(x)} $$ is strictly increasing in $(0,2 \pi) $. Unforunately, this is not that easy for me , as the derivative is not very manageable and trigonometric identities appear to be not very helpful, too.$$$$
Are there any further ways to do this?
Best Answer
let $f(x)=\dfrac{x-\sin(x)}{1-\cos(x)}$
$f'(x)=\dfrac{(x-\sin x)(-\sin x)}{(1-\cos x)^2}+\dfrac{1-\cos x}{1-\cos x}=\dfrac{\sin^2x-x\sin x+1-2\cos x+\cos^2 x}{(1-\cos x)^2}=\dfrac{2(1-\cos x)-x\sin x}{(1-\cos x)^2}=\dfrac{2\cdot 2\sin^2 \dfrac{x}{2}-2x\sin\dfrac{x}{2}\cos{x}{2}}{(1-\cos x)^2}=\dfrac{2\cdot sin \dfrac{x}{2}}{(1-\cos x)^2}\cdot (2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}) $
$\dfrac{2\cdot \sin \dfrac{x}{2}}{(1-\cos x)^2}>0 $ when $0<x<2\pi$,
so we only check $(2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}) $.
When $0< x < \pi, \tan\dfrac{x}{2}>\dfrac{x}{2},$ we have $ 2\sin\dfrac{x}{2}>x\cos x \dfrac{x}{2}$, ie. $2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}>0$.
When $\pi \leq x < 2\pi$, $\cos\dfrac{x}{2} \leq 0, -x \cos\dfrac{x}{2} \ge 0,$ and $2\sin\dfrac{x}{2}>0$, ie. $2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}>0$
so $f'(x)>0$ when $0<x<2\pi$, QED