[math-fun] OK, so where did Betsy Ross get cos(2π/13)?

4 Jul 2019


      She probably neither knew nor cared that
Out[87]= 1/24 (-2 + (
   2 ((-13)^(2/3) 2^(1/3) + (-13 - 195 I Sqrt[3])^(1/3)))/(-5 +
     3 I Sqrt[3])^(1/3) + Sqrt[
   6 (26 + 2^(2/3) (1703 - 195 I Sqrt[3])^(1/3) +
      2^(2/3) (1703 + 195 I Sqrt[3])^(1/3))])

In[89]:= ArcCos@%% // FullSimplify

Out[89]= (2 \[Pi])/13

In[53]:= MinimalPolynomial[Cos[2 \[Pi]/13]]

Out[53]= -1 + 6 #1 + 24 #1^2 - 32 #1^3 - 80 #1^4 + 32 #1^5 + 64 #1^6 &
These are guaranteed to solve in radicals.  (Why?
Well trivially,
In[27]:= ToRadicals[Cos[2 \[Pi]/13]]
Out[27]= -(1/2) (-1)^(11/13) (1 + (-1)^(4/13)))
A great source of tests for Julian's and my sextic solver.
—Bill

[math-fun] OK, so where did Betsy Ross get cos(2π/13)?

Bill Gosper