Proving $\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=-88$ and $\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=24$

Question

Prove: $$\begin{align} \sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7} &=-88 \tag{1} \\[6pt] \sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7} &=\phantom{-}24 \tag{2} \end{align}$$

Answer 1

you do need to know how to work with symmetric polynomials. The basic fact is that the roots of $$ x^3 + x^2 - 2x-1 $$ are $$ 2 \cos \frac{2 \pi}{7}, \; \; \; 2 \cos \frac{4 \pi}{7}, \; \; \; 2 \cos \frac{6 \pi}{7}. $$ This tells you the sum $(-1)$, the sum of pairwise products$(-2)$, and the product$(1)$. The observation goes back, essentially, to Gauss. You need to modify by factors of 2,4,8.

Answer 2

Let $\cos\frac{2\pi}{7}=x$, $\cos\frac{4\pi}{7}=y$ and $\cos\frac{4\pi}{7}=z$.

Hence, $$x+y+z=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2};$$ $$xy+xz+yz=\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=$$ $$=\frac{1}{2}\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=-\frac{1}{2}$$ and $$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8}.$$ Thus, $$\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=$$ $$=\frac{x^3y^3+x^3z^3+y^3z^3}{x^3y^3z^3}=\frac{(xy+xz+yz)^3-3(x+y+z)(xy+xz+yz)xyz+3x^2y^2z^2}{x^3y^3z^3}=$$ $$=\frac{\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)\cdot\frac{1}{8}+3\left(\frac{1}{8}\right)^2}{\left(\frac{1}{8}\right)^3}=-88$$ and $$\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=$$ $$=\frac{x^2y^2+x^2z^2+y^2z^2}{x^2y^2z^2}=\frac{(xy+xz+yz)^2-2(x+y+z)xyz}{x^2y^2z^2}=$$ $$=\frac{\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)\cdot\frac{1}{8}}{\left(\frac{1}{8}\right)^2}=24$$