Method to find $\sin (2\pi/7)$ I just thought a way to find $\sin\frac{2π}{7}$. 
Considering the equation $x^7=1$
$⇒(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0$
$⇒(x-1)[(x+\frac1 x)^3+(x+\frac1 x)^2-2(x+\frac1 x)-1]=0$
We can then get the 7 solutions of x, but the steps will be very complicated, especially when solving cubic equation, and expressing x as a+bi. The imaginary part of the second root of x will be $\sin\frac{2π}{7}$. 
Besides this troublesome way, are any other approach? Thank you.

Just for laughs, we can at least in principle compute $\cos{(\pi/7)}$ by observing that
$$\sin{\frac{3 \pi}{7}} = \sin{\frac{4 \pi}{7}}$$
Using a combination of double-angle forumlae, we end up with a cubic equation for $\cos{(\pi/7)}$:
$$8 \cos^3{\frac{\pi}{7}} - 4 \cos^2{\frac{\pi}{7}} - 4 \cos{\frac{\pi}{7}}+1=0$$
This equation has one real solution which is $\cos{(\pi/7)}$.  The bad news is that the expression is unwieldy at best:
$$\cos{\frac{\pi}{7}}=\frac{1}{6} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 i\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}\right)$$
The imaginary part of this expression is of course zero.  The real part, however, ends up being expressed in terms of a sine and cosine of another angle, and I think the point of an exercise like this is to not do that.  Anyway, I hope this adds to the discussion above.
