Why (correct) expressions like
Assuming[p > 0, 2 ArcTan[Sinh[p]] == Pi - 2 ArcTan[Csch[p]] // FullSimplify]
Are not correctly evaluated to: True? What is the best approach in these cases
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Your have to help a bit to FoolSimplify[]
f[e_] := 100 Count[e, _Gudermannian | _Csch, {0, Infinity}] +
LeafCount[e]
Assuming[p > 0,
FullSimplify[2 ArcTan[Sinh[p]] == Pi - 2 ArcTan[Csch[p]],
ComplexityFunction -> f]]
(True)
Assuming[p > 0, 2 ArcTan[p] == Pi - 2 ArcTan[1/p] // FullSimplify]does evaluate toTrue. The problem seems to be that Mathematica doesn't recognize $\sinh x = 1/\mathrm{csch}\, x$ in this context. $\endgroup$ – Michael Seifert Mar 14 '19 at 13:57Seriesshows that the difference between the LHS and RHS is zero, i.e.,Assuming[p > 0, Series[ 2 ArcTan[Sinh[p]] - (Pi - 2 ArcTan[Csch[p]]), {p, 0, 50}]] // Normalevaluates to0$\endgroup$ – Bob Hanlon Nov 14 '19 at 1:51