Value of $\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)$

How can I calculate the value of

1. $\sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)$

2. $\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)$

My solution: Let $$A = \sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)$$

$$B = \sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)$$

Now $$\begin{eqnarray} A\cdot B &=& (\sin 1^\circ \cdot \sin 89^\circ)\cdot (\sin2^\circ\cdot \sin 88^\circ)\cdots(\sin 44^\circ\cdot \sin 46^\circ) \cdot \sin 45^\circ \\ &=& (\sin 1^\circ\cdot \cos1^\circ)\cdot (\sin2^\circ\cdot \cos 2^\circ)\cdots(\sin 44^\circ\cdot \cos 44^\circ)\cdot \sin 45^\circ \\ &=& \frac{1}{2^{44}}\left[\sin(2^\circ)\cdot \sin (4^\circ)\cdots \sin(88^\circ) \cdot \sin (90^\circ)\right]\cdot \frac{1}{\sqrt{2}} \\ &=& \frac{1}{2^{\frac{89}{2}}}\cdot B \end{eqnarray}$$

Which implies that $$B\cdot \left(A-2^{-\frac{89}{2}}\right) = 0$$

So $A = 2^{-\frac{89}{2}}$ because $B\neq 0$.

But I do not understand how can I calculate the value of $B$. Can we calculate these values using complex numbers?

• Is there any reason to believe it has a nice form?
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.