Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$
 A: Let $\theta=\pi/7$. Note that 
$$\cos(4\theta)=-\cos(3\theta).\tag{1}$$ 
Let $x=\cos\theta$. By the double-angle identity for cosine, we have $\cos(2\theta)=2x^2-1$, so $$\cos(4\theta)=2(2x^2-1)^2-1.$$ 
By the triple-angle identity for cosine we have
$$\cos(3\theta)=4x^3-3x.$$
Now using Equation (1) we obtain 
$$2(2x^2-1)^2-1=-(4x^3-3x).$$
This is a quartic equation. However, it has the obvious root $x=-1$. Divide  the polynomial $2(2x^2-1)^2-1+4x^3-3x$ by $x+1$, and we obtain the cubic of the question.
To finish, verify that if $\varphi=3\pi/7$ or $5\pi/7$, then $\cos(4\phi)=-3\cos(3\phi)$.
A: Let $z=e^{\frac{i\pi}{7}}$. Then $\cos (\frac{\pi}{7})=\frac{z+z^{-1}}{2}$, $\cos (\frac{2\pi}{7})=\frac{z^2+z^{-2}}{2}$, $\cos (\frac{3\pi}{7})=\frac{z^3+z^{-3}}{2}$. Also $0=\frac{z^7-1}{z-1}=1+z+z^2+z^3+z^4+z^5+z^6$. As $z\ne0,$ divide both sides by $z^3$ to get $$z^3+\frac1{z^3}+z^2+\frac1{z^2}+z+\frac1z+1=0.$$
Now use $$\displaystyle z^2+\frac1{z^2}=\left(z+\frac1z\right)^2-2,\quad z^3+\frac1{z^3}=\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)$$
to obtain a cubic polynomial with the above roots.
A: If $\sin4x=\sin3x,$
$4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$  $$4\cos x\cos2x=3-4\sin^2x\iff4\cos x(2\cos^2x-1)=3-4(1-\cos^2x)$$
$$\iff8\cos^3x-4\cos^2x-4\cos x+1=0$$
Again if $\sin4x=\sin3x,$
$4x=n\pi+(-1)^n3x$ where $n$ is any integer
For even $n=2m$(say), $x=2m\pi\implies \sin x=?$ where $m$ is any integer
For odd $n=2m+1$(say),  $7x=(2m+1)\pi$ where $m$ is any integer
$\implies x=\dfrac{(2m+1)\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$
But for $m\equiv3\pmod7,\sin x=0$
So, the roots of $$8t^3-4t^2-4t+1=0$$ are $\cos\dfrac{(2m+1)\pi}7$ where $m\equiv0,1,2\pmod7$ 
as $\cos(-y)=\cos y,\cos\dfrac\pi7=\cos\dfrac{\pi\{2(-1)+1\}}7$ etc.
