TITLE: Show that function is strictly monotone increasing
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: The derivative is $$f'(x)=\frac{2(1-\cos(x))-x \sin(x)}{(1-\cos(x))^2}=\frac{4\sin^2 \frac{x}{2}-2x\cos \frac{x}{2} \sin \frac{x}{2}}{(1-\cos(x))^2}=\frac{2\sin \frac{x}{2}(2\sin \frac{x}{2}-x \cos \frac{x}{2})}{(1-\cos(x))^2}$$
All terms here are non-negative on $(0,2\pi)$, except possibly $2 \sin \frac{x}{2}-x \cos \frac{x}{2}$. We will prove that it is nonnegative as well. i.e. $$2 \sin \frac{x}{2} \geq x \cos \frac{x}{2}$$
For $x \in[\pi,2\pi)$ is it obvious, because the RHS is non-positive, white the LHS is non-negative.
For $x \in (0,\pi)$, divide by the cosine, to get $2 \tan \frac{x}{2}-x\geq0$, which follows from the identity $\tan t \geq t$ for $t \in(0,\frac{\pi}{2})$.
In summary, we proved $f' \geq 0$, as required.