# Dihedral in a regular spherical polygon

Planes rotate around a central symmetry axis pass through a sphere centre to intersect on the sphere forming a regular spherical polygon of $$n$$ sides.

A small circle forms as base of cone semi-vertical angle $$\alpha$$ circumscribing the vertices of the polygon.

If $$\delta$$ is the dihedral angle between successive planes then show that $$\tan \dfrac {\delta}{2} =\tan \dfrac {\pi}{n} \;\cos \alpha$$

Special case in Euclidean geometry when $$\alpha\rightarrow 0$$ sum of external angles of a regular polygon: $$n \delta = {2\pi}.$$

The planes carve a regular spherical polygon in the surface of the (unit) sphere, such that the dihedral angles $$\delta$$ between the planes match the exterior angles of the polygon. The semi-vertical angle $$\alpha$$ of the cone gives the (spherical) circumradius of the polygon.
If $$P$$ is the center of the polygon, $$Q$$ one of its vertices, and $$A$$ the midpoint of a side adjacent to $$Q$$, then $$\triangle PQA$$ is a spherical right triangle with hypotenuse $$\alpha$$ and acute angles $$\pi/n$$ and $$(\pi-\delta)/2$$.
\begin{align} \cos A &= -\cos P \cos Q + \sin P \sin Q \cos\alpha \\[4pt] 0 &= -\cos\frac{\pi}{n}\sin\frac\delta2+\sin\frac{\pi}{n}\cos\frac\delta2\cos\alpha \\[4pt] \tan\frac{\delta}{2} &= \tan\frac{\pi}{n}\;\cos\alpha \end{align} as desired. $$\square$$