Trigonometric solution of a 6 degree polynomial

Question

How do i prove that $\sin^2 \frac{\pi}{13}$ is a root of the equation $$2^{12}.x^6-13(2^{10}.x^5-5.2^8.x^4+3.2^8.x^3-7.2^5.x^2+7.2^2x-1)$$?

Any hints/answers would be appreciated.

Answer 1

HINT:

Let $13x=\pi\iff 7x=\pi-6x\implies\sin7x=\sin(\pi-6x)=\sin6x$

$(1):$ Using de Moivre's Identity, find $\displaystyle\sin6x,\sin7x$ in terms of $\sin x$

$(2):$ Use Prosthaphaeresis formula, $\displaystyle\sin(n+2)x-\sin nx=2\sin x\cos nx$ $\iff\sin(n+2)x=\sin x+2\sin x\cos nx$ (Reduction formula$\#1$)

$\displaystyle\cos(n+2)x-\cos nx=?,\iff\cos(n+2)x=\cos nx-\sin x\sin nx$ (Reduction formula$\#2$)

$(3):$ or $\displaystyle\sin7x=\sin(4x+3x)=\sin4x\cos3x+\cos4x\sin3x$

and $\displaystyle\sin6x=\sin3(2x)=3\sin2x-4\sin^32x=\cdots$

Answer 2

Chebyshev polynomials of first kind.