Evaluate $\prod_{i=1}^{89} \sin (i)$ Evaluate: $$\prod_{i=1}^{89} \sin (i)$$
My attempt:We know that 
$\sin \alpha \cdot \sin(90- \alpha)=\sin \alpha \cos \alpha=\frac{1}{2}\sin 2\alpha$
Then we have:
$$
(\sin1\cdot\sin89)(\sin2\cdot\sin88)\cdots (\sin44\cdot\sin46)\sin45
=\left(\frac{1}{2}\sin2\right)\cdots\left(\frac{1}{2}\sin88\right)\frac{\sqrt{2}}{2}
$$
Factorize from $\frac{1}{2}$ then do the same thing again You will get:
$$
\frac{\sqrt{2}}{2^{67}}(\sin4\cdot\sin8\cdots\sin88)
$$
Now what to do?Any hints?
 A: By the symmetry $\sin(180^{\circ}-k^{\circ})=\sin(k^{\circ})$, we have that
$$\prod_{k=1}^{179}\sin (k^{\circ})=\sin(90^{\circ})\left(\prod_{k=1}^{89}\sin (k^{\circ})\right)\left(\prod_{k=1}^{89}\sin (180-k^{\circ})\right)=\left(\prod_{k=1}^{89}\sin (k^{\circ})\right)^2$$
On the other hand by Evaluation of a product of sines (noted by H. H. Rugh and Batominovski).
$$\prod_{k=1}^{179}\sin (k^{\circ})= \frac{180}{2^{179}}$$
Hence
$$\prod_{k=1}^{89}\sin (k^{\circ})=\left(\frac{180}{2^{179}}\right)^{1/2}=\frac{\sqrt{90}}{2^{89}}=\frac{3 \sqrt{10}}{2^{89}}$$
which confirms the comment by Tolaso.
A: Not exactly a closed form, but maybe it is interesting to see the following way. Note that $$\prod_{k=1}^{n}\sin\left(k\right)=\frac{1}{2^{n}i^{n}}\prod_{k=1}^{n}\left(e^{ik}-e^{-ik}\right)=\frac{1}{2^{n}i^{n}}\prod_{k=1}^{n}e^{ik}\prod_{k=1}^{n}\left(1-e^{-2ik}\right)
 $$ now obviously we have $$\prod_{k=1}^{n}e^{ik}=\exp\left(i\sum_{k=1}^{n}k\right)=\exp\left(i\frac{n\left(n+1\right)}{2}\right)
 $$ and $$\prod_{k=1}^{n}1-e^{-2ik}=\left(e^{-2i}\right)_{n}
 $$ where $\left(a,q\right)_{n}
 $ is the $q$-Pochhammer symbol. So $$\prod_{k=1}^{n}\sin\left(k\right)=\frac{\exp\left(i\frac{n\left(n+1\right)}{2}\right)\left(e^{-2i},e^{-2i}\right)_{n}}{2^{n}i^{n}}
 $$ now taking $n=89$ we get $$\prod_{k=1}^{89}\sin\left(k\right)=\color{red}{\frac{\exp\left(i4005\right)\left(e^{-2i},e^{-2i}\right)_{89}}{2^{89}i}}
 $$ now it is interesting to note that $\frac{\exp\left(i4005\right)\left(e^{-2i},e^{-2i}\right)_{89}}{i}$ is real, but at this moment I don't know how to prove it.
