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?

  • 6
    $\begingroup$ Noting that $\sec\frac{6\pi}{7} = \sec\frac{8\pi}{7}$, this question becomes identical to "If $A=2\cdot\pi/7$ then show that $\sec A+\sec 2A+\sec 4A=−4$". $\endgroup$
    – Blue
    Jul 28, 2019 at 19:52
  • $\begingroup$ @Blue Sorry. I'm still stuck on how does $\sec \frac{6\pi}{7}=\sec\frac{8\pi}{7}$?. I thought that the trigonometric function remains the same unless you sum it by $2\pi$. How can I prove what you had just commented? =) $\endgroup$ Jul 28, 2019 at 21:44
  • 4
    $\begingroup$ $\cos(\pi-\theta) = \cos(\pi+\theta)$. (See, for instance, this answer.) For this situation, $\theta=\pi/7$. $\endgroup$
    – Blue
    Jul 28, 2019 at 21:49
  • $\begingroup$ If the angle lies i.e in the fourth quadrant and I add half turn ($\pi$) it will land in the second quadrant (the opposite side) hence in that zone cosine will have a negative value. This part is where I'm confused. If I subtract half the turn it will yield the same result. Can you help me to clear out this doubt? $\endgroup$ Jul 28, 2019 at 21:54
  • 2
    $\begingroup$ It's not a question of adding half-turns to an angle, it's a matter of adding or subtracting some angle to or from a half-turn. Here, $\frac{6\pi}{7} = \pi - \frac{\pi}{7}$ is in the second quadrant, and $\frac{8\pi}{7}=\pi+\frac{\pi}{7}$ is in the third quadrant. Each has wiggled the terminal end of angle $\pi$ by $\pi/7$ one way or the other, so their cosines will match. (This is not unlike how $\cos\theta = \cos(-\theta)$, or $\sin(\frac{\pi}{2}+\theta)=\sin(\frac{\pi}{2}-\theta)$, etc.) $\endgroup$
    – Blue
    Jul 28, 2019 at 21:59

4 Answers 4


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


  • 1
    $\begingroup$ Very clear answer; I upvoted. $\endgroup$ Aug 1, 2019 at 6:03
  • $\begingroup$ @Quanto Hi. I'm still confused on why $\cos 4 \omega = \cos 10 \omega$ and $\cos 6 \omega = \cos 8 \omega$ the same goes for the second identity you had posted which says on $\sin 6 \omega = - \sin 8 \omega$ perhaps you could include this missing step on the justification of both because i am still stuck on these. $\endgroup$ Aug 2, 2019 at 4:16
  • $\begingroup$ @ChrisSteinbeckBell Here is the justification. Since $\omega=\pi/7$, then $4\omega + 10\omega = 2\pi$. As a result, $\cos 4\omega = \cos(2\pi - 10\omega) = \cos 10\omega$. Similarly, $\cos 6\omega = \cos(2\pi - 8\omega) = \cos 8\omega$ and $\sin 6\omega = \sin(2\pi - 8\omega) = -\sin 8\omega$. $\endgroup$
    – Quanto
    Aug 2, 2019 at 14:55
  • $\begingroup$ @Quanto This only applies to that particular problem but not a generalization for all cases right?. What if I didn't know the angle omega beforehand?. Initially I was trying to understand if $\cos 4\omega = \cos 10 \omega$ on all cases?. Can this be generalized?. $\endgroup$ Aug 2, 2019 at 20:43
  • $\begingroup$ @ChrisSteinbeckBell - You’re correct. The derivation presented here, as well as the result $H = -4$, is only applicable to the case of $ω=π/7$, which may not be generalized for other $ω$’s. $\endgroup$
    – Quanto
    Aug 2, 2019 at 22:46

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 $H= -4$


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}


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 $$ \sec\left(\frac{2\pi}7\right)\sec\left(\frac{4\pi}7\right)\sec\left(\frac{6\pi}7\right)=8\tag6 $$

  • $\begingroup$ I believe that multiple-angle formulas and Vieta's formulas are well within algebra-precalculus. $\endgroup$
    – robjohn
    Aug 2, 2019 at 23:36
  • $\begingroup$ An alternate derivation of $(6)$ follows from $$\begin{align} \sin\left(\frac{2\pi}7\right)\cos\left(\frac{2\pi}7\right)\cos\left(\frac{4\pi}7\right)\cos\left(\frac{8\pi}7\right) &=\frac12\sin\left(\frac{4\pi}7\right)\cos\left(\frac{4\pi}7\right)\cos\left(\frac{8\pi}7\right)\\ &=\frac14\sin\left(\frac{8\pi}7\right)\cos\left(\frac{8\pi}7\right)\\ &=\frac18\sin\left(\frac{16\pi}7\right) \end{align} $$ then dividing by $\sin\left(\frac{2\pi}7\right)=\sin\left(\frac{16\pi}7\right)$. $\endgroup$
    – robjohn
    Aug 5, 2019 at 15:04

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