can anybody solve this trignometrical identity?

Question

sin pie/7 sin 2pie/7 sin 3pie/7 = (sqrt7)/8

Answer 1

We need to find the value

Eq.(1): x = sin(π/7) sin(2π/7) sin(3π/7).

Consider another value

Eq.(2): y = cos(π/7) cos(2π/7) cos(3π/7).

Multiply both sides of Eq.(2) on sin(π/7).
Take into account that cos(3π/7)=-cos(4π/7) and
sin(π/7) cos(π/7) = (1/2) sin(2π/7),
sin(2π/7) cos(2π/7) = (1/2) sin(4π/7),
sin(4π/7) cos(4π/7) = (1/2) sin(8π/7) = -(1/2) sin(π/7).
Hence,

Eq.(3): y = 1/8.

Write Eqs.(1), (2) as sums of sines and cosines.
The product of two first terms in Eq.(1) is
sin(π/7) sin(2π/7) = (1/2)( cos(π/7) - cos(3π/7) ).
Multiplying it on sin(3π/7) gives
x = (1/4) ( sin(4π/7) + sin(2π/7) - sin(6π/7) ).
Since sin(π-a) = sin(a), we have

Eq.(4): x = (1/4) ( - sin(π/7) + sin(2π/7) + sin(3π/7) ).

Similarly (using cos(π-a) = -cos(a) ), we get

Eq.(5): y = (1/4) (1- cos(π/7) + cos(2π/7) - cos(3π/7) ).

Calculate x^2 + y^2. This is a little bit lengthy, but rather straightforward. Squares of sines and cosines are summed to make unities. You also have terms of the type cos(...)cos(...) and sin(...)sin(...), which are simplified after adding together. For example, cos(π/7) cos(3π/7)-sin(π/7) sin(3π/7) =cos(4π/7)=-cos(3π/7). Remaining linear terms are the same as in the right side of Eq.(5). So, the result is

Eq.(6): x^2 + y^2 = y.

Then x^2 = y(1-y) = (1/8) (7/8), or

Eq.(7): x = √7/8.

Answer 2

I ran this through my math lab and I came up with no solution.
Sorry I couldnt do more for you.