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In the result of solving the equation or the integral often get something like k*ArcTan[x], I prefer the ArcTan[x] form.

For example, given list1, how do I get list2? FullSimplify doesn't work, my current approach is too limited, is there a more general way?

list1={2 ArcTan[1/2 (1+Sqrt[5])],2 ArcTan[Sqrt[5]-2],2 ArcTan[1/5 (-3+Sqrt[34])],3 ArcTan[Sqrt[1/6]],4 ArcTan[1/3]};

list2={Pi-ArcTan[2],ArcTan[1/2],ArcTan[5/3],ArcTan[17/(3 Sqrt[6])],ArcTan[24/7]};

FullSimplify[list1]
FullSimplify[list1==list2]

list1 /. 2 ArcTan[x_] :> ArcTan[Simplify[(2 x)/(1 - x^2)]] /; -1 < x < 1

$\left\{2 \tan ^{-1}\left(\frac{1}{2} \left(\sqrt{5}+1\right)\right),-2 \tan ^{-1}\left(2-\sqrt{5}\right),2 \tan ^{-1}\left(\frac{1}{5} \left(\sqrt{34}-3\right)\right),3 \cot ^{-1}\left(\sqrt{6}\right),4 \cot ^{-1}(3)\right\}$

True

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1 Answer 1

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FunctionExpand + FullSimplify seems to work:

ArcTan @@ FullSimplify[FunctionExpand[{Cos[list1], Sin[list1]}]]
(* {π - ArcTan[2], ArcTan[1/2], ArcTan[5/3], ArcTan[17/(3 Sqrt[6])], ArcTan[24/7]} *)
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    $\begingroup$ +1. TrigExpand works too: ArcTan @@ TrigExpand@Through[{Cos, Sin}[list1]] // Simplify $\endgroup$
    – Michael E2
    Commented Jun 3, 2022 at 17:46

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