tan(3pi/14), cot(pi/7), and -tan(pi/14) are the roots of x^3 - sqrt(7) x^2+ x + 1/sqrt(7) and so their sum is sqrt(7) and their product is -1/sqrt(7) and the sum of their products in pairs is 1. I found the polynomial after the fact, so no proof here. I have no idea how Mathematica simplifies the sum. On Wed, Jul 29, 2009 at 10:33 PM, <rwg@sdf.lonestar.org> wrote:
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun