On 2016-02-20 11:11, Eugene Salamin via math-fun wrote:
I came across an interesting formula in an optics book for the solid angle of a spherical triangle. Let a, b, c, be the unit vectors from the center to the vertices of a spherical triangle on the unit sphere. Then the solid angle Omega, which equals the area of the triangle, is given by
tan(Omega/2) = [a,b,c] / (1 + b.c + a.c + a.b),
where a.b etc. are dot products, and [a,b,c] is the scalar triple product. This paper is the cited reference.
http://www.maa.org/sites/default/files/Eriksson14108673.pdf
-- Gene
Yow, converting to angles between the vectors, and using the formula 52d8effb9069c8b7454fc03683e6f1ae.png <https://upload.wikimedia.org/math/5/2/d/52d8effb9069c8b7454fc03683e6f1ae.png> makes a *fifth* formula for solid angle in terms of vertex angles: solidAngle[a_, b_, c_] := {2 ArcTan[Sqrt[ 1 - Cos[a]^2 - Cos[b]^2 + 2 Cos[a] Cos[b] Cos[c] - Cos[c]^2]/( 1 + Cos[a] + Cos[b] + Cos[c])], ArcCos[-1 + (1 + Cos[a] + Cos[b] + Cos[c])^2/((1 + Cos[a]) (1 + Cos[b]) (1 + Cos[c]))], 2 ArcCos[1/ 4 (1 + Cos[a] + Cos[b] + Cos[c]) Sec[a/2] Sec[b/2] Sec[c/2]], 2 ArcSin[1/(4 Sqrt[2]) Sqrt[-1 - Cos[2 a] - Cos[2 b] + 4 Cos[a] Cos[b] Cos[c] - Cos[2 c]] Sec[a/2] Sec[b/2] Sec[c/2]], 4 ArcTan[\[Sqrt](Tan[1/4 (a + b - c)] Tan[1/4 (a - b + c)] Tan[ 1/4 (-a + b + c)] Tan[1/4 (a + b + c)])]} DanA's would make a sixth, but I can't make sense of it. --rwg