[...] rwg>So, is the solid angle formed by four dihedrals maximal when it inscribes
in a circular cone? (Actually, I should still have the max constraints. But I think the cone part was way messy.) --rwg 3 Sep 2008: "Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0.
[...] maxSolidAngle[a_, b_, c_, d_] := 2*ArcCos[((Cos[d] + Cos[c] + Cos[b] + Cos[a])/(4*Cos[a/2]*Cos[b/2]* Cos[c/2]*Cos[d/2])) - Tan[a/2]*Tan[b/2]*Tan[c/2]*Tan[d/2]]
Browsing Carr's synopsis http://archive.org/stream/asynopsiselemen00carrgoog#page/n216/mode/2up to see if that's where Ramanujan got his Bernoulli notation, I found two more formulæ (appended) for the solid angle given the vertex angles: {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[(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 Sqrt[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)]]]} The equivalence of these gives FullSimplify fits, even for {a,b,c}=π/{2,3,4}. The last one is http://mathworld.wolfram.com/LHuiliersTheorem.html, misprinted in Carr as Llhuillier's by an obviously Welsh typesetter. Amazingly, "huilier" means oiler. (Or cruet.) --rwg