[Maxima] [Fwd: Trig puzzle]

Richard Fateman fateman at cs.berkeley.edu
Sun Aug 2 17:52:39 CDT 2009

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-------- Original Message --------
Subject: 	Trig puzzle
Date: 	Thu, 30 Jul 2009 03:33:26 +0000 (UTC)
From: 	rwg at sdf.lonestar.org
To: 	math-fun <math-fun at mailman.xmission.com>
References:
<mailman.35692.1248585699.4407.math-fun at mailman.xmission.com>
<p06240801c69436da184d@[192.168.1.105]>

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

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