From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Radical expressions Date: 24 Feb 1995 16:00:27 GMT In article <793557375snz@wincoll.demon.co.uk>, Thomas O. Womack wrote: >Can anyone tell me how to solve these polynomials? In particular, what >is cos(2pi/7)? Heh heh. You wanted to solve x^7-1=0 with x not 1. That is, you want a root of x^6+x^5+...+x+1=0. Here's the trick for such 'symmetric' polynomials. Divide by x^3 and let u=(x+1/x); you'll see u satisfies u^3+u^2-2u-1=0. All right, how do you solve a cubic? Let v= u + (1/3) to clear the quadratic term: v satisfies v^3-(7/3)v-(7/27)=0. Well, cubics in this normal form are solved by a 'well-known' formula, which is derived roughly like this: you look for solutions v of the form v = w + C/w for some C and w. Choosing C = -(1/3)(-7/3) makes some terms clear so we need only solve (w^3) + C / (w^3) - 7/27 = 0. Setting z=w^3 we get a quadratic in z which we solve as z = (7/54)(1 + 3 sqrt(-3) ). Now work your way backwards: given this (complex) z, extract any cube root to get w, find v= w + 7/(9w), u = v- 1/3; then you can solve for x from u using the quadratic formula, although your particular question only asked for cos(2pi/7) which is precisely (x+1/x) / 2 for one of the roots x, that is, cos(2pi/7) = (1/2) ( - 1/3 + w + (7/9) / w ) for one of those three cube roots w. What's that you say, you've got too many complex numbers? Indeed you do, and that's the so-called 'casus irreduciblus' (sp?) of the cubic equation. When all three roots are real, the formula for the roots expresses them as sums of cube roots of complex numbers. If you attempt to solve (x+iy)^3=a+bi for x and y (with a and b known) you'll find you have to solve precisely the cubic equation you started with. So you're kind of stuck if you insist on real calculations. Oh, there is a way out: one can express the solutions to the last problem as x = r cos(theta), y = r sin(theta) where r = (a^2+b^2)^(1/3) and theta is any of three certain angles, but I'll bet you can see where this is going: you'll end up expressing cos(2pi/7) in terms of cos(2pi/7) ! dave (who wishes they still taught 'Theory of Equations')