<s>[INST] Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the  whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think the symbols might be different but can be the same. If it is correct, please help me to solve this; if the equation is wrong, then please modify it and solve it.
 [/INST] $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle z^7=-1=\cos\pi+i\sin\pi$
Using this, $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2,3\pmod7$
So, the equation whose roots are  $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2\pmod7$ is $$\frac{z^7+1}{z+1}=0\iff z^6-z^5+z^4-z^3+z^2-z+1=0$$
Like this divide either sides by $z^3\ne0$ to get $$z^3+\frac1{z^3}-\left(z^2+\frac1{z^2}\right)+z+\frac1z-1=0$$ 
$$\iff\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)-\left[\left(z+\frac1z\right)^2-2\right]+z+\frac1z-1=0$$ 
$$\iff \left(z+\frac1z\right)^3-\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)+1=0 $$
Observe that $\displaystyle n\equiv-2,1\pmod7\implies z+\frac1z=2\cos\frac{3\pi}7$
Similarly for $\displaystyle n\equiv-1,0\pmod7; \equiv-3,2\pmod7$ 
So, $\displaystyle2\cos\frac{3\pi}7$ is a root of $$u^3-u^2-2u+1=0 $$
$\displaystyle\implies\cos\frac{3\pi}7$ is a root of $?$
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