Verify: $$(z^2-z_0^2)\prod_{k=1}^{n-1}(z^2-z_k^2)=z^{2n}-z_0^{2n}$$
where $z_k=\cos \left(x+\frac{k\pi}{n}\right)+i\sin\left(x+\frac{k\pi}{n}\right)$ , $n$ = natural number, $n\geq2$ and $k\in\{1,2,\ldots,n-1\}$, $z$ = root of unity , $x$ = real number.
I don't know how to prove that equality because I need that to understand this product: $$\prod_{k=0}^{n-1}\sin\left(x+\frac{k\pi}{n}\right)$$
On
Let $z^2=z,z_0^2=y_0$
$$\left(\dfrac y{y_0}\right)^n=1$$
$$\implies\dfrac y{y_0}=e^{2m\pi i/n}\text{ where }0\le m<n$$
So, $\left(\dfrac y{y_0}\right)^n-1=\prod_{m=0}^{n-1}\left(\dfrac y{y_0}-e^{2m\pi i/n}\right)$
$$y^n-y_0^n=\prod_{m=0}^{n-1}\left(y-e^{2m\pi i/n}\cdot y_0\right)$$
Now $\displaystyle e^{2m\pi i/n}\cdot y_0=e^{2m\pi i/n}\cdot e^{2ix}=e^{2i(x+m\pi/n)}=y_m$
Both sides are polynomials of degree $2n$ in $z$ and with leading coefficient $1$. You know $2n$ roots of the polynomial on the right, namely $\pm z_k$. If yo can show that these are also roots of the right hand side, you are done (why?).