# How to simplify $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$?

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?

• 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 $\sec A+\sec 2A+\sec 4A=−4$". – Blue Jul 28 at 19:52
• @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? =) – Chris Steinbeck Bell Jul 28 at 21:44
• $\cos(\pi-\theta) = \cos(\pi+\theta)$. (See, for instance, this answer.) For this situation, $\theta=\pi/7$. – Blue Jul 28 at 21:49
• 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? – Chris Steinbeck Bell Jul 28 at 21:54
• 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.) – Blue Jul 28 at 21:59

Here is a more accessible evaluation, based only on familiar trigonometric identities and free from any complex variables.

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,

$$H=\frac{\cos4\theta \cos6\theta + \cos6\theta \cos2\theta + \cos2\theta \cos4\theta}{\cos2\theta \cos4\theta \cos6\theta}$$

Furthermore, simplify $$H$$ as

$$H =\frac{\cos2\theta + \cos4\theta + \cos6\theta}{\cos2\theta \cos4\theta \cos6\theta} \tag{1}$$

where we applied the identity $$\cos(x+y)+\cos(x-y)=2\cos x\cos y$$ to each of the three products in the numerator, and recognized the relationships $$\cos 4\theta = \cos 10\theta$$ and $$\cos 6\theta = \cos 8\theta$$.

Now, we compute the numerator and denominator separately. Applying the identity $$\sin 2x = 2\sin x \cos x$$ to the denominator three times, we have

$$\cos2\theta \cos4\theta \cos6\theta = \frac{\sin 4\theta \cos 4\theta\cos 6\theta}{2\sin 2\theta} = \frac{\sin 8\theta \cos 8\theta }{4\sin 2\theta}= \frac{\sin 16\theta}{8\sin 2\theta} = \frac{1}{8} \tag{2}$$

To compute the numerator, we write it equivalently as

$$\frac{1}{\sin 2\theta} \left({\sin 2\theta\cos2\theta + \sin 2\theta\cos4\theta + \sin 2\theta\cos6\theta} \right)$$

Then, applying the identity $$\sin(x+y)+\sin(x-y)=2\sin x\cos y$$ to each of the three terms in the parenthesis, we get

$$\cos2\theta + \cos4\theta + \cos6\theta = \frac{\sin 4\theta + (\sin 6\theta - \sin 2\theta) + (\sin 8\theta - \sin 4\theta)}{2\sin 2\theta}$$

After some cancellation due to identical terms and $$\sin 6\theta = -\sin 8\theta$$, the numerator is simply

$$\cos2\theta + \cos4\theta + \cos6\theta = -\frac{1}{2}\tag{3}$$

Finally, plugging (2) and (3) into (1), we arrive at

$$H=-4$$

• Very clear answer; I upvoted. – user2661923 Aug 1 at 6:03
• @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. – Chris Steinbeck Bell Aug 2 at 4:16
• @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 $\sin 6\omega = \sin(2\pi - 8\omega) = -\sin 8\omega$. – Quanto Aug 2 at 14:55
• @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?. – Chris Steinbeck Bell Aug 2 at 20:43
• @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. – Quanto Aug 2 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 $$\sec\left(\frac{2\pi}7\right)\sec\left(\frac{4\pi}7\right)\sec\left(\frac{6\pi}7\right)=8\tag6$$

• I believe that multiple-angle formulas and Vieta's formulas are well within algebra-precalculus. – robjohn Aug 2 at 23:36
• 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 $\sin\left(\frac{2\pi}7\right)=\sin\left(\frac{16\pi}7\right)$. – robjohn Aug 5 at 15:04