HMMT February

Consider the polynomial $P(z)=z^{7}-1$. Let $z=e^{i x}=\cos x+i \sin x$. Then $$\begin{aligned} z^{7}-1= & \left(\cos ^{7} x-\binom{7}{2} \cos ^{5} x \sin ^{2} x+\binom{7}{4} \cos ^{3} x \sin ^{4} x-\binom{7}{6} \cos x \sin ^{6} x-1\right) \\ & +i\left(-\sin ^{7} x+\binom{7}{2} \sin ^{5} x \cos ^{2} x-\binom{7}{4} \sin ^{3} x \cos ^{4} x+\binom{7}{6} \sin x \cos 6 x\right) \end{aligned}$$ Consider the real part of this equation. We may simplify it to $64 \cos ^{7} x-\ldots-1$, where the middle terms are irrelevant. The roots of $P$ are $x=\frac{2 \pi}{7}, \frac{4 \pi}{7}, \ldots$, so $\prod_{k=1}^{7} \cos \left(\frac{2 \pi k}{7}\right)=\frac{1}{64}$. But $$\prod_{k=1}^{7} \cos \left(\frac{2 \pi k}{7}\right)=\left(\prod_{k=1}^{3} \cos \left(\frac{k \pi}{7}\right)\right)^{2}$$ so $\prod_{k=1}^{3} \cos \left(\frac{k \pi}{7}\right)=\frac{1}{8}$. Now consider the imaginary part of this equation. We may simplify it to $-64 \sin ^{11} x+\ldots+7 \sin x$, where again the middle terms are irrelevant. We can factor out $\sin x$ to get $-64 \sin ^{10} x+\ldots+7$, and this polynomial has roots $x=\frac{2 \pi}{7}, \ldots, \frac{12 \pi}{7}$ (but not 0 ). Hence $\prod_{k=1}^{6} \sin \left(\frac{2 \pi k}{7}\right)=-\frac{7}{64}$. But, like before, we have $$\prod_{k=1}^{6} \sin \left(\frac{2 \pi k}{7}\right)=-\left(\prod_{k=1}^{3} \sin \left(\frac{2 \pi k}{7}\right)\right)^{2}$$ hence $\prod_{k=1}^{3} \sin \left(\frac{k \pi}{7}\right)=\frac{\sqrt{7}}{8}$. As a result, our final answer is $\frac{\frac{\sqrt{7}}{8}}{\frac{1}{8}}=\sqrt{7}$.