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 :-)

By Taylor, we have
$$\csc \left(\frac{\pi  n}{6 (n-2)}\right)=2-\frac{2 \pi }{\sqrt{3} n}+\frac{\frac{7 \pi ^2}{9}-\frac{4 \pi
   }{\sqrt{3}}}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\csc^2 \left(\frac{\pi  n}{6 (n-2)}\right)=4-\frac{8 \pi }{\sqrt{3} n}+\frac{8 \pi  \left(5 \pi -6 \sqrt{3}\right)}{9
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$4-\csc^2 \left(\frac{\pi  n}{6 (n-2)}\right)=\frac{8 \pi }{\sqrt{3} n}-\frac{8 \left(\pi  \left(5 \pi -6
   \sqrt{3}\right)\right)}{9 n^2}+O\left(\frac{1}{n^3}\right)$$
$$4-\csc^2 \left(\frac{\pi  n}{6 (n-2)}\right)=\frac{8 \pi }{\sqrt{3} n}+O\left(\frac{1}{n^2}\right)$$
Edit
Another way to do it using @Dzoooks' hint.
$$\frac{\pi  n}{6 (n-2)}=\frac{\pi (n-2)+2\pi}{6 (n-2)}=\frac \pi 6+\frac{\pi }{3 (n-2)}=\frac \pi 6+a$$
$$\csc \left(a+\frac{\pi }{6}\right)=2-2 \sqrt{3} a+7 a^2-\frac{23 a^3}{\sqrt{3}}+O\left(a^4\right)$$
$$\csc^2\left(a+\frac{\pi }{6}\right)=4-8 \sqrt{3} a+40 a^2+\left(-\frac{92}{\sqrt{3}}-28 \sqrt{3}\right)
   a^3+O\left(a^4\right)$$
$$4-\csc^2\left(a+\frac{\pi }{6}\right)=8 \sqrt{3} a-40 a^2+\left(\frac{92}{\sqrt{3}}+28 \sqrt{3}\right)
   a^3+O\left(a^4\right)$$ Now, let $a=\frac{\pi }{3 (n-2)}$ and, continuing with Taylor series, the same result.
