Showing $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$ The expression I have here is part of the upper bound given to the minimum distance of any 2 points out of $n$ points embedded in the unit sphere $\mathbb{S}^2$ in $\mathbb{R}^3$. If you have $n$ points on this unit sphere then we must have 2 points with a distance $d$ such that
$$d \le \sqrt{4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)}$$
This result was proven by Fejes Tóth. What I wish to show is that $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$.
We can see in the limit as $n \to \infty$, $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \to 0^+$. Also using basic numerical simulation for many and large $n$ I can visually verify that this statement seems to be true. The point now would be to show that for any $n$, then there exists some constant $k$ such that
$$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{k}{n}$$
This would be good enough. I'll be working on it here just thought the community may have some ideas :-)

We wish to show that $$f(x)=\left(4-\frac kx\right)\sin^2\frac{\pi x}{6(x-2)}\le1$$ for all positive real $x$. This is evidently the case when $4-k/x\le0$ so we consider $x>k/4$ with $$f'(x)=\frac k{x^2}\sin^2\frac{\pi x}{6(x-2)}-\left(4-\frac kx\right)\frac{24\pi}{(6x-12)^2}\sin\frac{\pi x}{6(x-2)}\cos\frac{\pi x}{6(x-2)}.$$ We claim that $f'(x)>0$ for all $x>k/4$. This is equivalent to showing that $$\tan\frac{\pi x}{6(x-2)}-\frac{2\pi x(4x-k)}{3k(x-2)^2}>0.$$ The second term is at most $8\pi/3k$ so the inequality holds whenever $$\frac x{x-2}>\frac6\pi\arctan\frac{8\pi}{3k}\implies x>2+\frac2{\frac6\pi\arctan\frac{8\pi}{3k}-1}.$$ This is true whenever $k>8\pi/\sqrt3$ so $$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{\frac{8\pi}{\sqrt3}+\epsilon}{n}$$ for any $\epsilon>0$. It follows that $d<3.81n^{-1/2}$.
