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?
 A: Here is the evaluation based on familiar trigonometric identities. Let $\theta = \pi/7$ and express $H$ in terms of cosine functions
$$H=\frac{1}{\cos2\theta} + \frac{1}{\cos4\theta} + \frac{1}{\cos6\theta}$$
or, in the form of common denominator,
$$H =\frac{\cos2\theta + \cos4\theta + \cos6\theta}{\cos2\theta \cos4\theta \cos6\theta}=\frac ND \tag{1}$$
where we used $\cos(x+y)+\cos(x-y)=2\cos x\cos y$ and the relationships $\cos 4\theta = \cos 10\theta$ and $\cos 6\theta = \cos 8\theta$. To compute the denominator, apply  $\sin 2x = 2\sin x \cos x$ to the denominator three times,
$$ D = \frac{\sin 4\theta \cos 4\theta\cos 6\theta}{2\sin 2\theta} = \frac{\sin 8\theta \cos 8\theta }{4\sin 2\theta}= \frac{\sin 16\theta}{8\sin 2\theta} = \frac{1}{8} \tag{2} $$ 
To compute the numerator, write it as
$$ N =\frac{1}{\sin 2\theta} \left({\sin 2\theta\cos2\theta + \sin 2\theta\cos4\theta + \sin 2\theta\cos6\theta} \right)$$
and apply $\sin(x+y)+\sin(x-y)=2\sin x\cos y$ and $\sin 6\theta = -\sin 8\theta$ to get
$$ N = \frac{\sin 4\theta + (\sin 6\theta - \sin 2\theta) + (\sin 8\theta - \sin 4\theta)}{2\sin 2\theta}=-\frac12\tag 3$$
Finally, plugging (2) and (3) into (1), we arrive at
$$H=-4$$
A: 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 $H= -4$
A: If $z=e^{2\pi i/7}$, then
$$
\cos\frac{2n\pi}{7}=\frac{z^n+z^{-n}}{2}=\frac{z^{2n}+1}{2z^n}
$$
so your expression becomes
$$
\frac{2z}{z^2+1}+\frac{2z^2}{z^4+1}+\frac{2z^3}{z^6+1}
$$
We get the numerator
$$
2z(z^{10}+z^4+z^6+1+z^9+z^3+z^7+z+z^8+z^4+z^6+z^2)
$$
Now we can note that $z^7=1$ and $z^6+z^5+z^4+z^3+z^2+z+1=0$, so the expression becomes
$$
4z(z^6+z^4+z^3+z^2+z+1)=-4z^6
$$
The denominator is
\begin{align}
(z^2+1)(z^{10}+z^6+z^4+1)
&=z^{12}+z^8+z^6+z^2+z^{10}+z^6+z^4+1\\
&=z^5+z+z^6+z^2+z^3+z^6+z^4+1\\
&=z^6
\end{align}
A: 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
$$
\sec\left(\frac{2\pi}7\right)\sec\left(\frac{4\pi}7\right)\sec\left(\frac{6\pi}7\right)=8\tag6
$$
