<s>[INST] Prove that $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$. 
Prove that
  $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.

What I've tried doing :
If $\theta=\frac{\pi}{7}:$
$$
3\theta+4\theta=\pi
$$
This allowed me to prove that :
$$
\tan^2\frac{\pi}{7}+\tan^2\frac{2\pi}{7}+\tan^2\frac{3\pi}{7}=21
\\
\cot^2\frac{\pi}{7}+\cot^2\frac{2\pi}{7}+\cot^2\frac{3\pi}{7}=5
$$
Is my reasoning wrong or is this entirely the wrong way to approach this question ? 
 [/INST] You can show that $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ are the three roots of $64x^3-112x^2+56x-7 = 0$, as follows:
Take $\zeta_7$ a $7$'th root of unity (a root of $(x^7-1)/(x-1) = x^6+x^5+x^4+x^3+x^2+x+1 = 0$), then $2 \cos(k 2\pi/7) = \zeta_7^k + \zeta_7^{-k}$, and with $\sin^2+\cos^2=1$ you can now express $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ in terms of $\zeta_7$. After that you can verify that they are roots of $64x^3-112x^2+56x-7 = 0$.
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