Product of Sines and Sums of Squares of Tangents There is a nice formula for products of cosines, found by multiplying by the complementary products of sines and using the double angle sine formula (as I asked in my question here): $$\prod_{k=1}^n \cos\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^n}$$
First question: Is there a formula for products of sines of the form $\prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)$? Or any other interesting formula for products of sines? 
Second question, I recently saw on another post here, that there is a formula for the sums of squares of tangents:
$$\sum_{k=1}^n \tan^2\left(\frac{k\pi}{2n+1}\right)=2n^2+n$$
How would you prove this formula? From this post here, they use De Moivre's Theorem, root finding and Vieta's formulas, but that seems inefficient when dealing with general $n$ in the above formula. 
 A: Let us define $\theta=\frac{\pi}{2n+1}$ and $~\xi_k=\tan(k\theta)$ for $-n\le k\le n$. Clearly
$$\xi_k=\frac{1}{i}\frac{e^{2ik\theta}-1}{e^{2ik\theta}+1}\iff\frac{1+ i\xi_k}{1-i\xi_k}=e^{2ik\theta}$$
and since $e^{i2(2n+1)\theta}=1$ we conclude that
$\{\xi_k:-n\le k\le n\}$ are zeros of the polynomial $P(X)$ defined by
$P(X)=(X+i)^{2n+1}+(X-i)^{2n+1}$, but $\deg P=2n+1$ and the coefficient of $X^{2n+1}$ in $P$ is $2$. So
$$P(X)=2\prod_{k=-n}^n(X-\xi_k)=2X \prod_{k=1}^n(X^2-\xi_k^2)$$
because $\xi_{-k}=-\xi_k$. On the other hand
$$\eqalign{P(X)&=\sum_{\ell=0}^{2n+1}\binom{2n+1}{\ell}X^\ell i^{2n+1-\ell}(1-(-1)^{\ell})\cr
&=2X\sum_{\ell=0}^{n}\binom{2n+1}{2\ell+1}X^{2\ell} (-1)^{n-\ell}\cr
}$$
It follows that
$$\prod_{k=1}^n(X^2-\xi_k^2)=
\sum_{\ell=0}^{n}\binom{2n+1}{2\ell+1}X^{2\ell} (-1)^{n-\ell}$$
Or
$$\prod_{k=1}^n(t-\xi_k^2)=
\sum_{\ell=0}^{n}\binom{2n+1}{2\ell+1}t^{\ell} (-1)^{n-\ell}\tag{1}$$
Now comparing the coefficients of $t^{n-1}$ on both sides we get
$$\sum_{k=1}^n\xi_k^2=\binom{2n+1}{2n-1}=\binom{2n+1}{2}=n(2n+1)$$
or 
$$\sum_{k=1}^n\tan^2\frac{\pi k}{2n+1}=n(2n+1)\tag{2}$$
Again, comparing the constant terms in $(1)$ we get
$$\prod_{k=1}^n\xi_k^2=\binom{2n+1}{1}=2n+1$$
and since $\prod_{k=1}^n\xi_k>0$ we get
$$\prod_{k=1}^n\tan\frac{\pi k}{2n+1}=\sqrt{2n+1}\tag{3}$$
This implies also that
$$\prod_{k=1}^n\sin\frac{\pi k}{2n+1}=
\prod_{k=1}^n\tan\frac{\pi k}{2n+1}\prod_{k=1}^n\cos\frac{\pi k}{2n+1}=
\frac{\sqrt{2n+1}}{2^n}\tag{4}$$
and we are done.$\qquad\square$
