Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$

Question

Prove that $$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$

My attempt

\begin{align} \text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\ &=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\ &=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1 \end{align} Now, please help me to complete the proof.

@Theophile, yeah you are right. Please check the edited version. — pi-π, Jul 15 '16 at 15:20
Please edit and include whatever is in your image in the post itself using $\LaTeX$, and then discard the picture. Formatting mathematics as pictures is highly discouraged. Remember there are blind people. It should be your responsibility as a side interested in getting free help to at least provide decently formatted questions. Not my downvote yet. — dbanet, Jul 15 '16 at 15:21
I have posted a solution of this on this question: (Strictly speaking, it is not the same question, but my solution covers this case as well) http://math.stackexchange.com/questions/1772998/if-alpha-frac2-pi7-prove-that-sin-alpha-sin2-alpha-sin4-alpha-frac/1773070#1773070 — Ege Erdil, Jul 15 '16 at 15:38
@starfall, your solution is out of my mind. I am just a tenth grader. — pi-π, Jul 15 '16 at 15:54

seavoyage · Accepted Answer · 2016-07-15T15:58:56.383

2

$cos(2\pi/7)$+$cos(4\pi/7)$+$cos(8\pi/7)$

= $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(6\pi/7)$ (angles add to give $2\pi$, thus one is $2\pi$ minus the other)

At this point, we'll make an observation

$cos(2\pi/7)$$sin(\pi/7)$ = $\frac{sin(3\pi/7) - sin(\pi/7)}{2}$ ..... (A)

$cos(4\pi/7)$$sin(\pi/7)$ = $\frac{sin(5\pi/7) - sin(3\pi/7)}{2}$ ..... (B)

$cos(6\pi/7)$$sin(\pi/7)$ = $\frac{sin(7\pi/7) - sin(5\pi/7)}{2}$ ..... (C)

Now, add (A), (B) and (C) to get

$sin(\pi/7)*(cos(2\pi/7)+cos(4\pi/7)+cos(6\pi/7))$ = $\frac{sin(7\pi/7) - sin(\pi/7)}{2}$ = -$sin(\pi/7)/2$

The $sin(\pi/7)$ cancels out from both sides to give you your answer.

edited Jul 15 '16 at 15:58

answered Jul 15 '16 at 15:41

seavoyage

556

Please elaborate a bit more. – pi-π Jul 15 '16 at 15:42
at what step exactly? – seavoyage Jul 15 '16 at 15:44
Don't the angles you refer to as differing by $2\pi$ actually differ by $\frac{2\pi}{7}$?...Although they do in fact have the same cosine value, but because of their symmetry with respect to $\pi$. – paw88789 Jul 15 '16 at 15:46
Actually what is meant by cosine formula in A.P? I am just a 10 grade student. – pi-π Jul 15 '16 at 15:48
They add to give you $14\pi/7$, which is $2\pi$...oh....they add to give you $2\pi$....i'll make an edit there.... – seavoyage Jul 15 '16 at 15:48
@user354073 I'm sorry...I wasn't aware of that....i'll add the details of that step then – seavoyage Jul 15 '16 at 15:53
@user354073 I hope it is clear now! :) – seavoyage Jul 15 '16 at 16:00
Yeah. It was nice talking to you Mr. Bajaj. – pi-π Jul 15 '16 at 16:01

score 2 · Answer 2 · answered Jul 16 '16 at 04:47

Recall the identity $$\sin \alpha \cos \beta = \frac{1}{2} \left( \sin(\alpha+\beta) + \sin(\alpha - \beta) \right).$$ Then

$$\begin{align*} \sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right) &= \frac{1}{2} \left( \sin \frac{4\pi}{7} + \sin \frac{6\pi}{7} + \sin \frac{-2\pi}{7} + \sin \frac{10\pi}{7} + \sin \frac{-6\pi}{7} \right), \\ \end{align*}$$ and observing that $$\sin \frac{10\pi}{7} = \sin \left(2\pi - \frac{4\pi}{7}\right) = \sin \frac{-4\pi}{7} = -\sin \frac{4\pi}{7}, \\ \sin \frac{-2\pi}{7} = - \sin \frac{2\pi}{7}, \\ \sin \frac{-6\pi}{7} = - \sin \frac{6\pi}{7},$$ we have $$\sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right) = -\frac{1}{2} \sin \frac{2\pi}{7},$$ where upon dividing both sides by $\sin 2\pi/7$, we obtain the desired result.

score 1 · Answer 3 · answered Jul 15 '16 at 16:17

1

Consider the polynomial $x^6+x^5+\dots+x+1$. This has roots at $e^{i2n\pi/7} = cos(2n\pi/7)+ isin(2n\pi/7)$ for $n = 1,2,\dots,7$.

Since $cos(x) = cos(2\pi - x)$ the sum you want is just half of the real part of the sum of the roots of this polynomial. Now by Vieta's formula the sum of the roots is $-1$, hence the sum is $-1/2$.

answered Jul 15 '16 at 16:17

Nate

11,206

Quick note on why are the roots at that for those $n$ pls? – dbanet Jul 15 '16 at 19:53
This polynomial is $(x^7-1)/(x-1)$ so it's roots are all of the roots of unity of order $7$ other than $1$. – Nate Jul 16 '16 at 15:48

score 0 · Answer 4 · answered Jul 16 '16 at 02:01

Here a solution solving a cubic equation in order to extend a bit the statement of the post. $$\cos x+\cos 2x+\cos 4x=\cos x+(2\cos 3x\cos x)=\cos x(1+2\cos 3x)$$ When $x=\frac{2\pi}{7}$ one has $$\cos 3x=\cos(\pi-\frac{\pi}{7})=-\cos(\frac{\pi}{7})$$ so the equality becomes $$\cos(\frac{2\pi}{7})(1-2\cos(\frac{\pi}{7}))=-\frac 12\\(2\cos^2(\frac{\pi}{7})-1)(1-2\cos(\frac{\pi}{7}))=-\frac 12$$ Puting now $X=\cos x$ we get the equation $$8X^3-4X^2-4X+1=0$$ The roots are $$X_1=-0.623489801819=\cos(\frac{5\pi}{7})\\X_2=0.222520933956=\cos(\frac{3\pi}{7})\\\color{red}{X_3=0.900968867902=\cos(\frac{\pi}{7})}$$ It follows that we also could have put, instead of the given equality $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ the two following ones $$\cos\frac {5\pi}{7}+ \cos\frac {10\pi}{7}+ \cos\frac {20\pi}{7}=-\frac{1}{2}\\\cos\frac {3\pi}{7}+ \cos\frac {6\pi}{7}+ \cos\frac {12\pi}{7}=-\frac{1}{2}$$ (Or even put $-X_i+n\pi; i=1,2,3$ instead of $X_i$)

Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$

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