Evaluate $\prod_{i=1}^{89} \sin (i)$

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.