Of course, if we're allowed to use arccos, then the solid angle formula is equivalent to omega = arccos(u*v) + arccos(v*w) + arccos(w*u) - pi where u, v, w are the unit vectors in R^3 of the vertices, and y*z denotes the dot product of vectors y and z. And something similar for hyperbolic triangles, since in both cases areas are in absolute value equal to the spherical excess. —Dan
On Feb 22, 2016, at 12:18 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Folke Eriksson: On the measure of solid angles, Mathematics Magazine 63,3 (June 1990) 184-187. http://www.maa.org/sites/default/files/Eriksson14108673.pdf <http://www.maa.org/sites/default/files/Eriksson14108673.pdf>
Eriksson's formula can be re-written as follows. Let the coordinates of the vertices of the triangle (drawn on unit sphere) be the rows of 3x3 matrix X. Then NonEuclideanSignedArea = 2 * arctan( det(X) / (1+M12+M23+M13) ) where M=X Xtranspose and for spherical geometry K=+1.
This formula also works to give the signed area of a hyperbolic nonEuclidean triangle where the first two coordinates of each unit-length 3-vector now are imaginary, and we change arctan to argtanh.
And of course in 1 dimension lower (X now a 2x2 matrix) the corresponding formula is Measure = arccos( M12 ) or for hyperbolic use argcosh.
But I do not believe there is any formula anywhere near as simple as these if we go to one dimension higher. Specifically mere arctrig will not suffice. Dilogarithms also are needed.