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Mathematica provides the following function to compute the arctangent of a number, preserving quadrant information:

ArcTan[x,y]

for real $x$ and $y$, when $y = 0$ and $x \ne 0$, $\arctan(y/x) = 0$. However, Mathematica doesn't perform this simplification:

FullSimplify[ArcTan[x,0], Assumptions->{x \[Element] Reals}]
(* ArcTan[x, 0] *)

Is this an error on my part or is there an underlying reason why this simplification isn't performed?

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It is because M does not know the actual numerical value of x (or rather its sign).

Let say x was 1

ArcTan[1,0]

Mathematica graphics

But what if x was -1 ?

ArcTan[-1,0]

Mathematica graphics

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  • $\begingroup$ Ah, a mathematical error on my end then. I suppose I was overthinking it. Thanks! $\endgroup$ – Billy Kalfus Nov 21 '17 at 22:43
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Assuming[0 < x < π, Simplify@ArcTan[x, 0]]
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ArcTan simplifications are somehow limited:

Simplify[ArcTan[x, y], {x == y, 0 < y, 0 < x}]

Sadly only gives:

ArcTan[y, y]
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