<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] let
$$
r = \cos \frac{2\pi}7+i\sin \frac{2\pi}7
$$
so $r$ is a primitive seventh root of unity and
$$
2 \cos \frac{2\pi}7 = r + r^6 = a$$
$$
2 \cos \frac{4\pi}7 = r^2 + r^5 = b$$
$$
2 \cos \frac{6\pi}7 = r^3 + r^4 = c
$$
and so if
$$
H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}
$$
then
$$
\frac{H}2 = \frac1a +\frac1b + \frac1c = \frac{bc+ca+ab}{abc}
$$
by simple drudgery, using $\sum_{k=0}^6 r^k = 0$ (sum of roots of $x^7 = 1$)
$$
bc+ca+ab = (r^2+r^5)(r^3+r^4) + (r^3+r^4)(r^1+r^6) + (r^1+r^6)(r^2+r^5) = -2
$$
and
$$
abc = (r^1+r^6)(r^2+r^5)(r^3+r^4) = 1
$$
from which $H= -4$
</s>