Closed form of $\arccos\left(\frac{2 \pi}{2^N}\right)$ Is there a closed form for $\arccos\left(\dfrac{2 \pi}{2^N}\right)$ in terms of $N \in \mathbb{Z}, N \ge 3$?
I'm not super optimistic, but I'm not sure how to really start exploring the problem, either.

$$\arccos z= \frac {\pi} {2} - \left( z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \cdots\ \right) 
= $$
$$=\frac {\pi} {2} - \sum_{n=0}^\infty \frac {\binom{2n} n z^{2n+1}} {4^n (2n+1)}; \qquad | z | \le 1 $$
