Let $\Phi_n(x)$ denote the cyclotomic polynomial of order $n$. Show that if $n>1$ is odd, then $\Phi_{2n}(x)=\Phi_n(-x)$
Here: $\Phi_n(x)=\prod\limits_{\zeta}(x-\zeta)$ where $\zeta$ are the primitive $n^{th}$ roots
Upon checking an explicit example, for $n=3$, I observed that the primitive roots arranged in the opposite quadrants so that the given equation is satisfied.
This happened basically because,
$\zeta$ for $n^{th}$ root are powers of form $e^{\frac{2\pi i}{n}}$ and
$\zeta$ for $2n^{th}$ root are powers of the form $e^{\frac{\pi i}{n}}$, which is the half angle of the previous one.
But how should I obtain the exact result for any odd $n$.
Appreciate your help
Let $\zeta$ be a primitive $n$th root of unity, i.e., $\zeta^n=1$ whereas $\zeta^d\ne1$ for all $0<d<n$. Then $\zeta^d\ne -1$ for all $0<d<n$ as otherwise $\zeta^{2d}=1$ if $2d<n$ and $\zeta^{2d-n}=1$ if $n<2d<2n$ (whereas $2d=n$ cannot occur with $n$ odd), contradicting primitivity of $\zeta$.
Now from $(-\zeta)^{2n}=(-1)^{2n}(\zeta^n)^2=1$, we see that $-\zeta$ is a $2n$th root of unity. If $0< d<n$, then $\zeta^d\ne\pm1$, hence $(-\zeta)^d\ne \pm1$. If $d=n$, then $(-\zeta)^d=(-1)^n\zeta^n=-1\ne 1$. If $n<d<2n$, then $0<2n-d<n$ and as just seen, $(-\zeta)^d=\frac1{\zeta^{2n-d}}\ne\frac1{\pm1}$. We conclude that $\-\zeta$ is a primitive $2n$th root.
It follows that $$\Phi_n(-x)\mid \Phi_{2n}(x).$$ From $\phi(2n)=\phi(n)$ for odd $n$, we know that both polynomials have the same degree. Therefore, $$\Phi_n(-x)=c \Phi_{2n}(x)$$ for some non-zero $c$. Finally, as $\phi(n)$ is even for odd $n>1$, we see from the leading monomials that $c=1$.