Help Evaluating Complex Numbers

[VertexOperator]

Homework Statement

Evaluate:

Homework Equations

sin\frac{\pi }{7}.sin\frac{2\pi }{7}.sin\frac{3\pi }{7}

The Attempt at a Solution

Using z^{7}-1 got:
cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}

 

[micromass]

Seems correct!

 

[VertexOperator]

Seems correct!

But my solution is for the cosine product. How can I find the solution to the question from what I got?

 

[micromass]

Can't you repeat much of the same proof? Where does it fail?

 

[VertexOperator]

The way I got: cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8} is by solving the equation z^{7}-1=0
z=1,cis\pm\frac{2\pi }{7},cis\pm\frac{4\pi }{7},cis\pm\frac{6\pi }{7}
w+w^{-1}=cis\frac{2\pi }{7}+cis\frac{-2\pi }{7}=2cos\frac{2\pi }{7}
Similarly;
w^{2}+w^{-2}=-2cos\frac{3\pi }{7}
w^{3}+w^{-3}=-2cos\frac{\pi }{7}
(w+w^{-1})(w^{2}+w^{-2})(w^{3}+w^{-3})=
w^{6}+1+w^{2}+w^{3}+w^{4}+w^{5}+1+w=1
, since w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1=0 \therefore cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}
So I don't think I can do the same to get the produce for sine.

 

[micromass]

It might be easier to determine

\prod_{k=0}^6 \sin(k\pi/7)

Do this by expressing the sine as complex exponentials. Also try to make use of the following formula

\frac{z^7-1}{z-1}=\prod_{k=1}^6 (1- \zeta^k)

with \zeta = \cos(\pi/7) + i\sin(\pi/7). Let z go to 1.