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Using your Head is Permitted May 2011 riddle
$$
-\prod_{i=0}^{n-1} \cos \left(\pi \frac{2^{i}}{2^{n}-1}\right)=\prod_{i=1}^{n} \cos \left(\pi \frac{i}{2 n+1}\right)
$$
I am happy to report that many readers this month found a simpler solution to the riddle than the one I had in mind. I am presenting this simpler solution here. (The list of solvers is pretty much also the list of those who deserve credit.)
Instead of proving directly that the two sides of the equality equal each other, we will prove independently that each side of the equality is individually equal to $2^{-n}$. In both cases, more or less the only trigonometry we will need is $\sin (2 \alpha)=$ $2 \sin (\alpha) \cos (\alpha)$
To prove the left hand side, let
$$
C S(x)=\prod_{i=0}^{n-1} \cos \left(\pi \frac{x * 2^{i}}{2^{n}-1}\right)
$$
and define
$$
S N(x)=\prod_{i=0}^{n-1} \sin \left(\pi \frac{x * 2^{i}}{2^{n}-1}\right) .
$$
Using the trigonometric equality, we know that
$$
2^{n} * S N(1) * C S(1)=S N(2)=S N(1) * \frac{\sin \left(\pi \frac{2^{n}}{2^{n}-1}\right)}{\sin \left(\pi \frac{1}{2^{n}-1}\right)}=-S N(1),
$$
where the last equality is given by $\sin (\pi+x)=-\sin (x)$. As $S N(1) \neq 0$, we can divide and conclude $C S(1)=-2^{-n}$.
To prove the right hand side, let
$$
C S(x)=\prod_{i=1}^{n} \cos \left(\pi \frac{x * i}{2 n+1}\right)
$$
and define
$$
S N(x)=\prod_{i=1}^{n} \sin \left(\pi \frac{x * i}{2 n+1}\right) .
$$
Now, $2^{n} * S N(1) * C S(1)=S N(2)$, but $S N(2)=S N(1)$ because, using the equality $\sin (\pi-x)=\sin (x)$ we can replace any odd valued $i$ in the calculation of $S N(1)$ with the corresponding even value $2 n+1-i$ that appears in the calculation of $S N(2)$ but not of $S N(1)$. Noting $S N(1) \neq 0$, we conclude that $C S(1)=2^{-n}$. Q.E.D.
I thank Oded Margalit and Dani Vardi for drawing my attention to these beautiful trigonometric equalities. Readers may enjoy pondering this follow-up question, also from Dani Vardi:
Simplify the following trigonometric expression: $\prod_{i=1}^{n} \tan \left(2 \pi \frac{i}{2 n+1}\right)$.
math.stackexchange.com/questions/3556048/proving-that-prod-k-1n- ... 2n1-right-sqrt2n1-us
Building on @WETutorialSchool's hint, start from the identity$$i\tan\frac{\theta}{2}=\frac{2i\sin\frac{\theta}{2}\exp\frac{i\theta}{2}}{2\cos\frac{\theta}{2}\exp\frac{i\theta}{2}}=\frac{\exp i\theta-1}{\exp i\theta+1}.$$Define $z:=\exp\frac{2i\pi}{2n+1}$ so$$\frac{z^k-1}{z^k+1}=i\tan\frac{k\pi}{2n+1},\,\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}=\frac{z^{-k}-1}{z^{-k}+1}=\frac{1-z^k}{1+z^k}=-i\tan\frac{k\pi}{2n+1}.$$Since $\prod_{k=1}^n\tan\frac{k\pi}{2n+1}$ is a product of the positive tangents of $n$ acute angles,$$\prod_{k=1}^n\tan\frac{k\pi}{2n+1}=\sqrt{\prod_{k=1}^ni\tan\frac{k\pi}{2n+1}\cdot-i\tan\frac{k\pi}{2n+1}}=\sqrt{\prod_{k=1}^n\frac{z^k-1}{z^k+1}\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}}=\sqrt{\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}}.$$To prove $\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}=2n+1$, consider the values of $z^k+1,\,0\le k\le 2n$. They are the roots of $(w-1)^{2n+1}-1$, so their product is $-1$ times this polynomial's constant term, i.e. $2$. In other words, $\prod_{k=0}^{2n}(z^k+1)=2$ and $\prod_{k=1}^{2n}(z^k+1)=1$. Similarly, $\prod_{k=1}^{2n}(z^k-1)$ is the product of the roots of $\frac{(w+1)^{2n+1}-1}{w}$, so is equal to its constant term, $\binom{2n+1}{1}$. |
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