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)sin(3)sin(5)sin(89)

  2. 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 AB=(sin1sin89)(sin2sin88)(sin44sin46)sin45=(sin1cos1)(sin2cos2)(sin44cos44)sin45=1244[sin(2)sin(4)sin(88)sin(90)]12=12892B

Which implies that B(A2892)=0

So A=2892 because B0.

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
n1k=1sin(kπn)=n2n1
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

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