BCCee Dick Askey replied: Here are two similar identities. tan(3pi/11) + 4sin(2pi/11) = 11^(1/2) sin(2pi/7) + sin(4pi/7) + sin(8pi/7) = (1/2)*7^(1/2). These are in Chapter 7 of Hobson's "A Treatise on Plane and Advanced Trigonometry". In case you do not know this book, it has a number of gems. Dover has it back in print. Dick Mma 7.0 FullSimplify does these, too. Maybe it just guesses them numerically? No, it has a ferocious algebraic number simplifier: In[39]:= ToRadicals[Tan[3*Pi/11] + 4*Sin[2*Pi/11]] Out[39]= 5/11 7/22 4/11 I (1 + (-1) ) -2 (-1) (-1 + (-1) ) + ---------------- 5/11 -1 + (-1) In[40]:= RootReduce[%] Out[40]= Sqrt[11] --rwg
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