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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.

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    $\begingroup$ There is a simple formula for a product of sines $\prod_{k=1}^n \sin\left(2^{k-1}x\right)=\frac{\sin 2^nx}{2^n\cos x}$ provided $x\ne \frac \pi2+\pi n$, $n\in \Bbb Z$. :-) $\endgroup$ Jun 29, 2017 at 18:34

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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.$\qquad\square$

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  • $\begingroup$ Doesn't $e^{i(2n+1)\theta}=-1$? $\endgroup$
    – D.R.
    Jun 29, 2017 at 2:08
  • $\begingroup$ @D.R. I think it is a typographic error, he means $e^{2i(2n+1)\theta}=1$. $\endgroup$ Jun 29, 2017 at 6:26

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