# Value of sin(2∘)⋅sin(4∘)⋅sin(6∘)⋯sin(90∘)\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

Now

Which implies that

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?

Putting $n=180$, it gives
The proof of identity $(1)$ can be found at the end of this pdf.