TITLE: Evaluate $\prod_{i=1}^{89} \sin (i)$
QUESTION [6 upvotes]: 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?

REPLY [6 votes]: 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.