You can get various infinite families of these values by starting with a known one of them and repeatedly applying a half-angle formula.
– coffeemathMar 25 '21 at 19:53
$\cos n\theta+i\sin n\theta=(\cos \theta+i\sin \theta)^n$ together with binomial expansion gives you multiple angle formulas for $\cos n\theta$ and $\sin n\theta.$ But (as far as I know) there is no algebraic formula for $\sin(\theta/n)$ in terms of $\sin \theta, \cos\theta.$
– BumblebeeMar 25 '21 at 20:01