Value of $\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ) $
Solution 1:
Using the identity $$\prod_{k=1}^{n-1}\sin \left( \frac{k\pi}{n}\right) = \frac{n}{2^{n-1}} \tag{1}$$ Putting $n=180$, it gives $$ \left(\sin (1^\circ)\sin (2^\circ)\dots \sin (89^\circ)\right)^2= \frac{180}{2^{179}} \tag{2}$$ The value of $$\boxed{ \sin(2^{\circ})\sin(4^{\circ}) \dots \sin(90^{\circ}) = \sqrt{\frac{180}{2^{179}}}\div \sqrt{\frac 1 {2^{89}}} = \sqrt{\frac{180}{2^{90}}}} \tag{3}$$
The required value seems to be in agreement with calculated value. The proof of identity $(1)$ can be found at the end of this pdf.