Five ways to compute the solid angle subtended by a face of a regular dodecahedron w.r.t. the centroid led to {2 (ArcCot[√15] + 2 ArcCot[√(3(9 + 4 √5))]), ArcCos[7/8] + 2 ArcCos[1/8 + (3 √5)/8], π/3, π/3, 4 (ArcTan[(-√5 + √6) Cot[1/4 (π + ArcSec[-3])]] + ArcTan[Root[1 - 220 #1^2 + 582 #1^4 - 220 #1^6 + #1^8 &, 5]] + ArcTan[Root[1 - 6 #1 + 2 #1^2 + 6 #1^3 + #1^4 &, 4] Tan[1/4 ArcSec[3/√5]]])} (Recall that) these should all be 4π/12 = π/3. Can FullSimplify manage? FullSimplify[ArcTan[Print[ByteCount /@ #]; #] &@TrigExpand@Tan@%] {33224,10912,128,128,81747368} I.e., the intermediate byte counts are 33K, 10K, 128, 128, and 81M! The 33K takes several minutes the first time. I don't expect to see it finish the 81M. I actually don't know a way to symbolically prove case 5. MinimalPolynomial[Tan[...]] seems to have the best shot to answer within a human lifespan, but not a span of hours. --rwg Apropos the timezone map, everybody knows about the 3½ hr discontinuity, right?