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For which $n\in \mathbb{N}$, can we find (reps. find explicitly) $n+1$ integers $0 < k_1 < k_2 <\cdots < k_n < q<2^{2n}$ such that $$\prod_{i=1}^{n} \sin\left(\frac{k_i \pi}{q} \right) =\frac{1}{2^n} $$

P.S.: $n=2$ is obvious answer, $n=6 $ is less obvious but for instance we have $k_1 = 1$, $k_2 = 67$, $k_3 = 69$, $k_4 = 73$, $k_5 = 81$, $k_6 = 97$, and $q=130$.

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Are you sure that the tag [cv.complex-variables] is appropriate? All the terms in your question, apart from $\pi$ (which is not a complex variable), are integers. – Alex B. Oct 30 '10 at 2:46
Maybe he suspects switching to the complex exponential form might help? – J. M. Oct 30 '10 at 2:58
Alex, I think it is appropriate. The second example could be obtained via the so-called $\Gamma$ function (using complex variable analysis) – Portland Oct 30 '10 at 3:25
You might be interested in a paper I published some years ago, Rational products of sines of rational angles. – Gerry Myerson Oct 30 '10 at 10:13
Indeed, thanks Gerry. – Portland Oct 30 '10 at 15:01
up vote 12 down vote accepted

Consider the identity (as quoted by drvitek): $$\prod_{k=1}^{n} \sin \left(\frac{(2k-1) \pi}{2n}\right) = \frac{2}{2^n},$$ This is completely correct but doesn't quite answer the question because the RHS is not $1/2^n$ $\text{---}$ this is an issue related to the fact that $\zeta - \zeta^{-1}$ is not a unit if $\zeta$ is a root of unity of prime power order. Replace $n$ by $3n$, and take the ratio of the corresponding products. Then one finds that $$\prod_{(k,6) = 1}^{k < 6m} \sin \left(\frac{k \pi}{6m}\right)= \sin \left(\frac{\pi}{6m}\right) \sin \left(\frac{5 \pi}{6m}\right) \sin \left(\frac{7 \pi}{6m}\right) \cdots \sin \left(\frac{(6m-1) \pi}{6m}\right) = \frac{1}{2^{2m}}.$$ This provides the identity you request for $n = 2m$, since $q = 6m < 2^{4m} = 2^{2n}$ is true for all $m \ge 1$. There are other "obvious" identities that can be written down, but they tend to have length $\phi(r)$ for some integer $r$, and $\phi(r)$ is always even (if $r > 2$).

For odd $n$, note the "exotic" identity: $$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{15 \pi}{42}\right) \sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8}.$$ Since $42 < 64$, this is an identity of the required form for $n = 3$. On the other hand, none of the rational numbers $1/21$, $5/14$, $8/21$ can be written in the form $k/6m$ where $(k,6) = 1$. Hence $$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{15 \pi}{42}\right) \sin \left(\frac{16 \pi}{42}\right) \prod_{(k,6) = 1}^{k < 6m} \sin \left(\frac{k \pi}{6m}\right) = \frac{1}{2^{2m+3}},$$ when written under the common denominator $q = \mathrm{lcm}(42,6m)$, consists of distinct fractional multiples $k_i/q$ of $\pi$ with $0 < k_i < q$, and is thus an identity of the required form for $n = 2m + 3$, after checking that $$q = \mathrm{lcm}(42,6m) \le 42m \le 2^{4m+6} = 2^{2n}.$$ Thus the answer to your question is that such an identity holds for all $n > 1$. (It trivially does not hold for $n = 1$.) The first few identities constructed in this way are: $$\sin \left(\frac{\pi}{6}\right) \sin \left(\frac{5 \pi}{6}\right) = \frac{1}{4},$$ $$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{15 \pi}{42}\right) \sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8},$$ $$\sin \left(\frac{\pi}{12}\right) \sin \left(\frac{5 \pi}{12}\right) \sin \left(\frac{7 \pi}{12}\right) \sin \left(\frac{11 \pi}{12}\right) = \frac{1}{16},$$ $$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{7 \pi}{42}\right) \sin \left(\frac{15 \pi}{42}\right) \sin \left(\frac{16 \pi}{42}\right) \sin \left(\frac{35 \pi}{42}\right) = \frac{1}{32},$$ &. &.

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@Ace: Thank you for taking the time to check my answer, find that it was off by one, and do the hard work in extending it! Have an upvote! – dvitek Nov 1 '10 at 1:55

From Andreescu and Andrica, Complex Numbers from A to Z p. 48.

$$\prod_{1 \le k \le n} \sin{\frac{(2k-1)\pi}{2n}} = \frac{1}{2^{n-1}}.$$

Ibid., p. 50.

"The following identities hold:

a) $$\prod_{1\le k \le n-1; \gcd{(k,n)} = 1}\sin{\frac{k\pi}{n}} = \frac{1}{2^{\phi(n)}}$$ whenever $n$ is not a power of a prime.

b) $$\prod_{1\le k \le n-1; \gcd{(k,n)} = 1}\cos{\frac{k\pi}{n}} = \frac{(-1)^{\frac{\phi(n)}{2}}}{2^{\phi(n)}}$$ for all odd positive integers $n$.

The first one completely answers your question; the next two are just for fun!

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@drvitek I think Portland is asking for a collection of $n$ points $k_1<k_2<k_3,\dots<k_n$ and another point $q>k_n$ which are between $[0,2^{2n}]$. So your answers are partial answers. Of course the identity a) seems to indicate a way to answer the original question. – Vagabond Oct 30 '10 at 6:11
@Vagabond: Gah! A fencepost error! (To be honest, though, it was late, I knew where some semi-relevant formulas were, and I regurgitated them.) – dvitek Nov 1 '10 at 1:52

I couldn't resist... A different construction, which is "irreducible" also for odd $n$ unlike the construction of Ace of Base, is already implied by the very first example the OP gives. (Did you find it by computer and didn't notice?) Look at the differences of the numerators... It can be straight away generalized to $$\boxed{\sin\left(\dfrac{\pi}{2^{n+1}+2} \right)\prod\limits_{k=1}^{n-1}\sin\left(\dfrac{2^n+2^k+1}{2^{n+1}+2}\pi \right)=\dfrac1{2^n}}.$$ Sure enough, denoting $\sin\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ and $\cos\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ by $s_k$ and $c_k$ respectively, we have $$LHS=s_1\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{2c_1}s_2\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{4c_1}s_4\prod\limits_{k=2}^{n-1}c_{2^k} =\cdots=\frac{s_{2^{n}}}{2^nc_1}=\frac1{2^n}.$$

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