Let t = tan x. Let C(x,y) denote the combinatoric function x!/(y!*(x-y)!). Then the general multi-angle formula for tan(n*x) can be expressed as:
Sigma{k=0 to floor((n-1)/2)} (-1)^k*C(n,2k+1)*t^(2k+1)
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Sigma{k=0 to ceil((n-1)/2)} (-1)^k*C(n,2k)*t^(2k)
Specifically for this problem tan(13x) then equals:
13t - 286t^3 + 1287t^5 - 1716t^7 + 715t^9 - 78t^11 + t^13
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1 - 78t^2 + 715t^4 - 1716t^6 + 1287t^8 - 286t^10 + 13t^12
For all the angles x=n*pi/13 with n equals any integer, the value of tan(13x) = 0. Therefore the values of the tangents of all thirteen angles are the roots of the polynomial:
13t - 286t^3 + 1287t^5 - 1716t^7 + 715t^9 - 78t^11 + t^13
Factoring out the trivial root t=0, corresponding to x=0, the product of the remaining 12 roots equals 13.
Note that tan(n*pi/13)=-tan((13-n)pi/13). Then the product of the 12 roots equals the square of the product of the first six roots - the expression we want to evaluate.
Therefore tan(pi/13) * tan(2pi/13) * tan(3pi/13) * tan(4pi/13) * tan(5pi/13) * tan(6pi/13) = sqrt(13).
This is easily adaptable to prove the more general equation suggested in the comments:
Prod{k=1 to n} tan(k*pi/(2n+1)) = sqrt(2n+1).
Solution
