1
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$Assumptions = Element[t, Reals] && Element[ω, Reals] && ω > 0 
&& t > 0;
ArcTan[8.54596*10^7 (1. + 1. t^2 ω^2) (1.2214 + E^ω Cos[
2 ArcTan[t ω]]), -1.70919*10^8 E^ω t ω] + 
ArcTan[8.54596*10^7 (1. + 1. t^2 ω^2) (1.2214 + E^ω 
Cos[2ArcTan[t ω]]),1.70919*10^8 E^ω t ω]//FullSimplify

Mathematica doesn't simplify this seemingly ArcTan[x,y]+ArcTan[x,-y] expression. I plotted the 3D-graph, it shows the expression is zero for all values of t and \[Omega].

I understand one underlying reason might be Mathematica isn't sure about the positivity of Cos term in x, and hence, x. Am I right? Any ideas to simplify this term in code, instead of manually, would be very much appreciated. I'm relatively new to the language.

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  • $\begingroup$ Maybe Simplify[expr, Cos[2 ArcTan[t \[Omega]]] > 0], where expr is your expression involving Arctan? You can use the second argument of Simplify to submit assumptions. $\endgroup$ – Henrik Schumacher Apr 30 '18 at 10:16
  • $\begingroup$ The Cos term does not necessarily have to be greater than zero. And, I have tried out inserting the global assumptions inside Simplify and FullSimply but it still doesn't work. $\endgroup$ – naeema18 Apr 30 '18 at 11:32
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    $\begingroup$ If (1.2214 + E^\[Omega] Cos[2 ArcTan[t \[Omega]]]) is not always greater than zero, then it's not necessarily true that the ArcTan terms sum to zero. (Consider x < 0, y == 0.) $\endgroup$ – Michael E2 Apr 30 '18 at 11:43
1
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First you should avoid using inexact numbers if possible.

expr = ArcTan[ α (1 +  t^2 ω^2) (γ +E^ω Cos[2 ArcTan[t ω]]), -β E^ω t ω] + 
  ArcTan[α(1 +  t^2 ω^2) (γ + E^ω Cos[2 ArcTan[t ω]]), β E^ω t ω]

Now, with assumptions inside , the simplification evaluates as you expect

Simplify[expr, 
Assumptions -> {ω > 0,t > 0 , α > 0, β > 0, γ >0,Cos[2 ArcTan[t ω]] > 0,Elements[{ω, t, α, β,γ}, Reals]}]
(* 0 *)
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  • $\begingroup$ Your solution does simplify it to 0, because you're assuming Cos[2 ArcTan[t ω]] > 0. I don't see any good reason to assume so. If you do, please let me know. $\endgroup$ – naeema18 Apr 30 '18 at 11:46
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    $\begingroup$ @ naeema18: In the question I read ... about the positivity of Cos term.. as an assumption. $\endgroup$ – Ulrich Neumann Apr 30 '18 at 11:53
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$Assumptions = ω > 0 && t > 0;

Note that if a variable is positive it is redundant to add that it is real.

If you want exact results you must use exact numbers. Use Rationalize

expr = ArcTan[
     8.54596*10^7 (1. + 1. t^2 ω^2) (1.2214 + 
        E^ω Cos[
          2 ArcTan[t ω]]), -1.70919*10^8 E^ω t ω] + 
    ArcTan[8.54596*10^7 (1. + 1. t^2 ω^2) (1.2214 + 
        E^ω Cos[2 ArcTan[t ω]]), 
     1.70919*10^8 E^ω t ω] // Rationalize;

TrigToExp is what you need

expr // TrigToExp // Simplify

(* 0 *)
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  • $\begingroup$ If you plot Plot3D[TrigToExp[ArcTan[x, y] + ArcTan[x, -y]], {x, -1, 1}, {y, -1, 1}] you can see that the comment from Michael E2 concerning the first argument x stll holds. $\endgroup$ – Ulrich Neumann May 1 '18 at 9:40

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