<s>[INST] Evaluate $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$ The problem is as follows:
Find the value of $\textrm{H}$ which belongs to a certain vibration coming from a magnet.
$$H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$$
It was easy to spot that each term was related to multiples of two and three of the first angle. So I rewrote that equation like this:
$$H=\sec \frac{2\pi}{7}+\sec \frac{2\times 2\pi}{7}+\sec \frac{3\times 2\pi}{7}$$
One method which I tried was to transform the multiples of each angle into their equivalents as a single one as shown below:
$$\cos^{2}\omega=\frac{1+\cos 2\omega}{2}$$
$$\cos 2\omega= 2 \cos^{2}\omega - 1$$
$$\cos^{3}\omega=\frac{1}{4}\left(3cos\omega+\cos 3\omega \right)$$
$$\cos 3\omega = 4 \cos^{3}\omega - 3 cos\omega$$
Therefore by plugin these expressions into the above equation would become into (provided that secant function is expressed in terms of secant):
$$H=\frac{1}{\cos \frac{2\pi}{7}}+\frac{1}{2\cos^{2}\frac{2\pi}{7}-1}+\frac{1}{4\cos^{3}\frac{2\pi}{7}-3\cos\omega}$$
But from here on it looks convoluted or too algebraic to continue. My second guess was it could be related to sum to product identity but I couldn't find one for the secant.
Does it exist a shortcut or could it be that am I missing something? Can somebody help me to find the answer?
Can this problem be solved without requiring to use Euler's formulas?
 [/INST] Using multiple angle formulas, we get
$$
\begin{align}
\cos(\theta)&=x\\
\cos(2\theta)&=2x^2-1\\
\cos(3\theta)&=4x^3-3x\\
\cos(4\theta)&=8x^4-8x^2+1
\end{align}\tag1
$$
Consider $\cos(3\theta)=\cos(4\theta)$, which happens when $3\theta+4\theta=2k\pi$ for some $k\in\mathbb{Z}$ (it also happens when $3\theta-4\theta=2k\pi$, but those cases are a subset). Thus,
$$
x=\cos\left(\frac{2k\pi}7\right)\implies8x^4-4x^3-8x^2+3x+1=0\tag2
$$
Since $k$ and $7-k$ give the same values for $\cos\left(\frac{2k\pi}7\right)$ and $k=0$ gives $\cos\left(\frac{2k\pi}7\right)=1$, if we divide $(2)$ by $x-1$, we get the polynomial satisfied by $x=\cos\left(\frac{2k\pi}7\right)$ for $k\in\{1,2,3\}$; that is,
$$
8x^3+4x^2-4x-1=0\tag3
$$
The polynomial satisfied by $x=\sec\left(\frac{2k\pi}7\right)$ for $k\in\{1,2,3\}$ is then
$$
x^3+4x^2-4x-8=0\tag4
$$
Vieta's formulas then give that
$$
\bbox[5px,border:2px solid #C0A000]{\sec\left(\frac{2\pi}7\right)+\sec\left(\frac{4\pi}7\right)+\sec\left(\frac{6\pi}7\right)=-4}\tag5
$$
Furthermore, they also give that
$$
\sec\left(\frac{2\pi}7\right)\sec\left(\frac{4\pi}7\right)\sec\left(\frac{6\pi}7\right)=8\tag6
$$
</s>