# Why are some equal expressions more equal than others?

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.

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

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] *)

• Excellent work around, thank you. Dec 22, 2012 at 5:37
• @Phillip why note vote for his answer in that case? Dec 22, 2012 at 11:04
• Ok, how do I vote for an answer? Dec 22, 2012 at 16:23