How can I calculate the value of

sin(1∘)⋅sin(3∘)⋅sin(5∘)⋯sin(89∘)

sin(2∘)⋅sin(4∘)⋅sin(6∘)⋯sin(90∘)

My solution:Let A=sin(1∘)⋅sin(3∘)⋅sin(5∘)⋯sin(89∘)B=sin(2∘)⋅sin(4∘)⋅sin(6∘)⋯sin(90∘)

Now A⋅B=(sin1∘⋅sin89∘)⋅(sin2∘⋅sin88∘)⋯(sin44∘⋅sin46∘)⋅sin45∘=(sin1∘⋅cos1∘)⋅(sin2∘⋅cos2∘)⋯(sin44∘⋅cos44∘)⋅sin45∘=1244[sin(2∘)⋅sin(4∘)⋯sin(88∘)⋅sin(90∘)]⋅1√2=12892⋅B

Which implies that B⋅(A−2−892)=0

So A=2−892 because B≠0.

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

**Answer**

Using the identity

n−1∏k=1sin(kπn)=n2n−1

Putting n=180, it gives

(sin(1∘)sin(2∘)…sin(89∘))2=1802179

The value of

sin(2∘)sin(4∘)…sin(90∘)=√1802179÷√1289=√180290

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.

**Attribution***Source : Link , Question Author : juantheron , Answer Author : Santosh Linkha*