(In reply to
re: Interesting observation by Ady TZIDON)
Good point, Ady.
20 minutes effort and I'm not seeing how I can manipulate the tangents with different denominators.
Call P(n) = Prod{i from 1 to n} tan(iπ/(2n+1))
where n is the number of terms (6 in the example in the problem)
Clearly true for n=1 since tan(π/3) = √3.
Assume true for n, then prove it for n+1, or:
Prove it for P(n+1) = Prod{i from 1 to n+1} tan(iπ/(2n+3))
Seems like a formula for tan(a*b) would help.
I'm stuck, so I'll pass for now

Posted by Larry
on 20140817 11:19:09 