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