Prove that 2Cos(2pi/7) is a root to x^3+x^2-2x-1

[Johan Lans]

I need help with this problem:
Prove that 2Cos(2pi/7) is a root to x^3+x^2-2x-1.
My thought is that I will prove this by showing that
8Cos^3(2pi/7)+4Cos^2(2pi/7)-4Cos(2pi/7)-1 equals zero, but I just can't
do the math. Any tips?

[Stan Brown]

Post by Johan Lans
Prove that 2Cos(2pi/7) is a root to x^3+x^2-2x-1.
My thought is that I will prove this by showing that
8Cos^3(2pi/7)+4Cos^2(2pi/7)-4Cos(2pi/7)-1 equals zero, but I just can't
do the math. Any tips?

I don't immediately see a useful factorization of this. I suspect the
answer is to use some double-angle formulas and perhaps good old
sin^2(theta) = 1 - cos^2(theta).

[Alex. Lupas]

I need help with this problem:.............

Let P(z):=z^3+z^2-2z-1 and
T_3(z)=cos(3*arccos(z))=4z^3-3z
T_4(z)=cos(4*arccos(z))=8z^4-8z^2+1 .
Please verify the equalities

(1) Q(z):=(T_4(z)-T_3(z))/(z-1)=

=8z^3+4z^2-4z-1=P(2z) .
Therefore P(z)=Q(z/2). But this means that if
P(z_j)=0 and Q(y_j)=0, then z_j=2y_j .

From (1) it may be seen that y_j=cos(2j*pi/7)

Therefore z_j=2*cos(2*j*pi/7) , j in {1,2,3}.

[Badger]

Post by Johan Lans
Prove that 2Cos(2pi/7) is a root to x^3+x^2-2x-1.
My thought is that I will prove this by showing that
8Cos^3(2pi/7)+4Cos^2(2pi/7)-4Cos(2pi/7)-1 equals zero, but I just can't
do the math. Any tips?

Can you take as given the fact that cos(2 pi/7) is the real part of
a root of the 7th cyclotomic polynomial? That is,

x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0

If so, then consider

cos(6y) + cos(5y) + ... + cos(y) + 1 = 0

For the specific value y = 2 pi/7,

cos(6y) = cos(y)
cos(5y) = cos(2y)
cos(4y) = cos(3y)
cos(3y) = 4 (cos y)^3 - 3 cos(y)
cos(2y) = 2 (cos y)^2 - 1

Letting x = cos(2 pi/7),

x + (2 x^2 - 1) + 2 (4 x^3 - 3x) + (2 x^2 - 1) + x + 1 = 0

Collecting like terms,

8 x^3 + 4 x^2 - 4x - 1 = 0

So (2 pi/7) is indeed a root of your equation, and thus also of the
minimal polynomial x^3 + x^2 - 2x - 1.


References:
----------
<http://mathworld.wolfram.com/CyclotomicPolynomial.html>
<http://mathworld.wolfram.com/TrigonometryAnglesPi7.html>


Bonus:
-----
In their paper "Homogeneous Polynomials and the Minimal Polynomial of
cos(2 pi/n)", Surowski and McCombs give an explicit formula with proof
for the minimal polynomial of 2 cos(2 pi / p) for p an odd prime.

Here's a link to the preprint (PDF format):
<http://www.math.ksu.edu/~dbski/cosine-mo-jo.pdf>