0
$\begingroup$

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: $$ n \delta = {2\pi}. $$

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

enter image description here

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 $(\pi-\delta)/2$.

By the Spherical Law of Cosines for Angles, we have

$$\begin{align} \cos A &= -\cos P \cos Q + \sin P \sin Q \cos\alpha \\[4pt] 0 &= -\cos\frac{\pi}{n}\sin\frac\delta2+\sin\frac{\pi}{n}\cos\frac\delta2\cos\alpha \\[4pt] \tan\frac{\delta}{2} &= \tan\frac{\pi}{n}\;\cos\alpha \end{align}$$ as desired. $\square$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .