<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] Like my other answers $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Using this, $$\cos\frac{\pi}7-\cos\frac{2\pi}7+\cos\frac{3\pi}7=0$$
Now $\displaystyle\cos\frac{\pi}7=\cos\left(\pi-\frac{6\pi}7\right)=-\cos\left(2\cdot\frac{3\pi}7\right)$ (use Double Angle formula $\displaystyle\cos2A=2\cos^2A-1$)
$\displaystyle\cos\frac{2\pi}7=-\cos\left(\pi+\frac{2\pi}7\right)=-\cos\left(3\cdot\frac{3\pi}7\right)$ (use $\displaystyle\cos3A=4\cos^3A-3\cos A$ formula)
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