Maximum of $\frac{1 -\cos(lx)}{1 - \cos(x)}$ on $[\frac{\pi}{l}, 2\pi - \frac{\pi}{l}$]

Question

Given a positive integer $l$, I am pretty certain that the maximum of the function $$ f(x) = \frac{1 - \cos(lx)}{1 - \cos(x)} $$ on the interval $[\frac{\pi}{l}, 2\pi - \frac{\pi}{l}]$ is attained precisely in the endpoints, and nowhere else. This is clear from making a few plots for different values of $l$, but it is surprisingly hard to prove. I have tried manipulating the equalities $$ \frac{x^2}{2} \leq 1 - \cos(x) \leq \frac{x^2}{2} - \frac{x^4}{24} $$ but to no avail; they are too local in nature, it would seem.

Answer 1

By the symmetry $f(x) = f(2\pi -x)$, it is sufficient to look at the interval $[\frac{\pi}{l}, \pi]$. For $x \in ]\frac{\pi}{l}, \pi]$ we have

$$ f(1) = \frac{2}{1-\cos(\frac{\pi}{l})} > \frac{2}{1 - \cos(x)} \geq f(x). $$