Method to find $\sin (2\pi/7)$

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.