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.
 A: Use the sine's definition in terms of complex exponentials:
\begin{align}
& 8x^3-4x^2-4x+1 \\[10pt]
= {} & 8\left(\frac{e^{i\pi/14}-e^{-i\pi/14}}{2i}\right)^3 - 4\left(\frac{e^{i\pi/14}-e^{-i\pi/14}}{2i}\right)^2 -4 \frac{e^{i\pi/14}-e^{-i\pi/14}}{2i} +1
\end{align}
Then recall that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ and $(a+b)^2 = a^2+2ab+b^2$.
And $(2i)^3 = -8i$ and $(2i)^2=-4$, and $1/i=-i$.
Then turn the crank.
A: $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle\cos4x=-\cos3x=\cos(\pi+3x)\implies4x=2n\pi\pm(\pi+3x)$ where $n$ is any integer
'+' $\displaystyle\implies x=(2n+1)\pi\equiv\pi\pmod{2\pi}\implies\cos x=-1$
'-' $\displaystyle\implies x=\frac{(2n-1)\pi}7$
Now $\displaystyle\cos4x=-\cos3x$
Using Multiple angle formula,
$\iff8c^4-8c^2+1=-(4c^3-3c)\iff8c^4+4c^3-8c^2-3c+1=0\ \ \  \ (1)$ where $c=\cos x$
Clearly, the roots of $(1)$ are $\cos x,$ where $\displaystyle x=\pi, \frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$
So, the equation whose roots are $\displaystyle\cos\frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$ is  $$\frac{8c^4+4c^3-8c^2-3c+1}{c+1}=0$$
Here $n=2$
A: $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
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 $?$
A: 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)
