Help Evaluating Complex Numbers

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Homework Statement



Evaluate:

Homework Equations



[tex]sin\frac{\pi }{7}.sin\frac{2\pi }{7}.sin\frac{3\pi }{7}[/tex]

The Attempt at a Solution



[tex]Using[/tex] [tex]z^{7}-1[/tex] [tex]got:[/tex]
[tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
 
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micromass said:
Seems correct!

But my solution is for the cosine product. How can I find the solution to the question from what I got?
 
The way I got: [tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex] is by solving the equation [tex]z^{7}-1=0[/tex]
[tex]z=1,cis\pm\frac{2\pi }{7},cis\pm\frac{4\pi }{7},cis\pm\frac{6\pi }{7}[/tex]
[tex]w+w^{-1}=cis\frac{2\pi }{7}+cis\frac{-2\pi }{7}=2cos\frac{2\pi }{7}[/tex]
Similarly;
[tex]w^{2}+w^{-2}=-2cos\frac{3\pi }{7}[/tex]
[tex]w^{3}+w^{-3}=-2cos\frac{\pi }{7}[/tex]
[tex](w+w^{-1})(w^{2}+w^{-2})(w^{3}+w^{-3})=[/tex]
[tex]w^{6}+1+w^{2}+w^{3}+w^{4}+w^{5}+1+w=1[/tex]
, since [tex]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}[/tex]
So I don't think I can do the same to get the produce for sine.