4
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n1 = 1/3 (2 + (-Sqrt[3] + 2 Sin[(2 π)/9])/(Sqrt[3] Cos[π/9] - Sin[π/9]));
n2 = 1/2 + Sin[π/18]/(3 Cos[(2 π)/9] + Sqrt[3] Sin[(2 π)/9]);
n3 = 2/3 (1 - Cos[4 π/9]);

These three expressions are numerically equivalent. How can I coax Mathematica to simplify the first two into the third? FullSimplify only returns the original forms.

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  • 1
    $\begingroup$ It's at least worth pointing out that the form produced by RootReduce is the same for all three: RootReduce /@ {n1, n2, n3} $\endgroup$
    – Mr.Wizard
    Dec 22, 2012 at 5:02
  • 2
    $\begingroup$ Is that a political question ;-) ? $\endgroup$
    – Yves Klett
    Dec 22, 2012 at 7:18
  • 1
    $\begingroup$ @Yves The title, while amusing, doesn't describe the problem. Perhaps we should change it. $\endgroup$
    – Mr.Wizard
    Dec 22, 2012 at 8:19

1 Answer 1

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$\begingroup$

Note that MMA aggressively evaluates n3 to 2/3 (1 - Sin[\[Pi]/18]), so I assume this is n3.

This is the best I could do:

ExpToTrig[ToRadicals[Root[MinimalPolynomial[n1, x], 2]] /. {
    1 + I Sqrt[3] :> 2 (-1)^(1/3), 1 - I Sqrt[3] -> -2 (-1)^(2/3)}]

(* 2/3 - 2/3 Sin[π/18] *)

and

ExpToTrig[ToRadicals[Root[MinimalPolynomial[n2, x], 2]] /. {
    1 + I Sqrt[3] :> 2 (-1)^(1/3), 1 - I Sqrt[3] -> -2 (-1)^(2/3)}]

(* 2/3 - 2/3 Sin[π/18] *)
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3
  • $\begingroup$ Excellent work around, thank you. $\endgroup$ Dec 22, 2012 at 5:37
  • $\begingroup$ @Phillip why note vote for his answer in that case? $\endgroup$
    – Mr.Wizard
    Dec 22, 2012 at 11:04
  • $\begingroup$ Ok, how do I vote for an answer? $\endgroup$ Dec 22, 2012 at 16:23

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