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Six tangents (Posted on 2014-08-17) Difficulty: 3 of 5
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tan(pi/13)*tan(2pi/13)*tan (3pi/13)*tan(4pi/13)*tan(5pi/13)*tan (6pi/13) = 13^1/2.

No Solution Yet Submitted by Danish Ahmed Khan    
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re(2): Interesting observation | Comment 3 of 8 |
(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 2014-08-17 11:19:09

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