The Wikipedia page for regular icosahedrons says the following:
The locations of the vertices of a regular icosahedron can be described using spherical coordinates, for instance as latitude and longitude. If two vertices are taken to be at the north and south poles (latitude $\pm90°$), then the other ten vertices are at latitude $\pm\arctan(\frac12)\approx\pm26.57°$. These ten vertices are at evenly spaced longitudes ($36°$ apart), alternating between north and south latitudes.
I understand the first and last parts of this. I understand having vertices at latitude $\pm90°$. I also understand putting the other vertices at longitudes $36°$ apart, since there are 5 vertices on each of the pentagonal pyramids ($360° \div 10 = 36$). What I don't understand, is why the latitudes are at $\pm\arctan(\frac12)\approx\pm26.57°$.
Looking at an icosahedron from the side, it certainly looks like the the middle section forms a 2:1 rectangle (depending on orientation). I'm having trouble mathematically proving this, however.
My question is, how did they arrive at $\pm\arctan(\frac12)\approx\pm26.57°$? How was this derived?
If $\gamma = \frac{1}{2}(1 + \sqrt{5})$ is the golden ratio, the twelve points $$ \pm(1, \pm\gamma, 0),\qquad \pm(0, 1, \pm\gamma),\qquad \pm(\pm\gamma, 0, 1) $$ are straightforwardly checked to be vertices of a regular icosahedron. Pick some antipodal pair, say $\pm(1, \gamma, 0)$, to lie at the poles of the sphere. The desired latitude is the angle $\theta'$ in the diagram. Its complementary angle $\theta$ (i.e., the colatitude of the five vertices in the northern hemisphere) is the angle between two "nearby" corners of a golden rectangle: $$ \cos\theta = \frac{(1, \gamma, 0) \cdot (-1, \gamma, 0)}{1 + \gamma^{2}} = \frac{\gamma^{2} - 1}{\gamma^{2} + 1} = \frac{1}{\sqrt{5}}. $$ Consequently $\sin\theta = \frac{2}{\sqrt{5}}$, and $\tan\theta = 2 = \cot\theta'$, i.e., $\theta' = \arctan\frac{1}{2}$.