Evaluate sin(π/7)sin(2π/7)sin(3π/7)

[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.