Product of sine 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 $q=130$.
 A: 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!
A: 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:
$$\sin \left(\frac{\pi}{6}\right) \sin \left(\frac{5 \pi}{6}\right) = 
\frac{1}{4},$$
$$\sin \left(\frac{2 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8},$$
$$\sin \left(\frac{\pi}{12}\right) \sin \left(\frac{5 \pi}{12}\right)      
\sin \left(\frac{7 \pi}{12}\right) \sin \left(\frac{11 \pi}{12}\right) =     
\frac{1}{16},$$
$$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{7 \pi}{42}\right) 
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right)
\sin \left(\frac{35 \pi}{42}\right)  = \frac{1}{32},$$
&. &.
A: 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 $$LHS=s_1\prod\limits_{k=1}^{n-1}c_{2^k}  
=\frac1{2c_1}s_2\prod\limits_{k=1}^{n-1}c_{2^k}
=\frac1{4c_1}s_4\prod\limits_{k=2}^{n-1}c_{2^k} =\cdots=\frac{s_{2^{n}}}{2^nc_1}=\frac1{2^n}.$$ 
