explain why irrational cosine values added and multiplied equal a rational number

Question

why does

[cos(2π/7)+cos(π/7)]*[cos(2π/7)-1]*[cos(12π/7)+cos(π/7)]=-(7/8)

hint;

cos(7x)=64cos⁷x-112cos⁵x+56cos³x-7cosx

Answer 1

The beauty of cosine (and sine) is that all cos(pπ/q) numbers are

algebraic. p and q integers, q≠0.

let z(x)=cos(x)

know z(7x)=0 when x=(2n+1)π/14,,,,,n=0,1,...,6

same with f(z(x))=64z⁷-112z⁵+56z³-7z

f(z(x))=64[(z-0)(z-cos(π/14))(z-cos(3π/14))(cos(5π/14))(z-cos(9π/14))(z-cos(11π/14))(z-cos(13π/14))

(z-0)=(z-cos(7π/14))

algebraic numbers are roots of polynomial equation with integer coefficients

we know for seven degree polynomial,.the coefficient for z term equals multiplying he first 6 roots

(where the first root is the z=0 one), which is zero, then adding last 6 roots multiplied.

-7/64=[cos(π/14)]*[cos(3π/14)]*[cos(5π/14)]*[cos(9π/14)]*[cos(11π/14)]*[cos(13π/14)]

[cos(π/14)]*[cos(3π/14)]=(1/2)[cos(2π/7)+cos(π/7)]

[cos(5π/14)]*[cos(9π/14)]=(1/2)[-1+cos(2π/7)]

[cos(11π/14)]*cos(13π/14)]=(1/2)[cos(12π/7)+cos(π/7)]

since cos(12π/7)=cos(2π/7)

I should have written the identity

-7/8=(cos(2π/7)+cos(π/7))²(cos(2π/7)-1)

Answer 2

Follow the hint.

Write out the hint with x = 2 pi/7, pi/7 and 12 pi/7. (a)

Multiply out the product in the first line. (b)

Relate a to b.

The key point is that cos pi, cos 2 pi, and cos 12 pi are all integers (and rational).