Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


We tried various different trigonometric identities. Still no luck.

Geometric interpretation would be also welcome.

EDIT: Very good answers, I'm clearly impressed. I followed all the answers and they work! I can only accept one answer, the others got my upvote.

share|cite|improve this question
This seems helpful: – Rasmus May 3 '12 at 14:01
Drat. I was trying to find a way to prove $\sum_{k=1}^7\cos\frac{2k\pi-1}7=0$ which could be simplified to double the left hand side plus $\cos \pi$ equals 0. – Mike May 3 '12 at 14:20
+1 for the fantastic reference! – Georges Elencwajg May 3 '12 at 16:32
I noticed interesting reference about the subject. $\cos(\pi/7), \cos(3 \pi/7) , \cos(5 \pi/7)$ are root of the polinom $8x^3-4x^2-4x+1=0$. It is not proof but I can see easily the sum of roots are $-(-4)/8=1/2$ – Mathlover May 4 '12 at 0:47
up vote 22 down vote accepted

Hint: start with $e^{i\pi/7} = \cos(\pi/7) + i\sin(\pi/7)$ and the fact that the lhs is a 7th root of -1.

Let $u = e^{i\pi/7}$, then we want to find $\Re(u + u^3 + u^5)$.

Then we have $u^7 = -1$ so $u^6 - u^5 + u^4 - u^3 + u^2 -u + 1 = 0$.

Re-arranging this we get: $u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$.

If $a = u + u^3 + u^5$ then this becomes $u a + 1 = a$, and rearranging this gives $a(1 - u) = 1$, or $a = 1 / (1 - u)$.

So all we have to do is find $\Re(1 / (1 - u))$.

$1 / (1 - u) = 1 / (1 - \cos(\pi/7) - i \sin(\pi/7)) = (1 - \cos(\pi/7) + i \sin(\pi/7)) / (2 - 2 \cos(\pi/7))$


$\Re(1/(1-u)) = (1 - \cos(\pi/7)) / (2 - 2\cos(\pi/7)) = 1/2 $

share|cite|improve this answer
Very elegant as the calculations are kept to the minimum! – S4M Sep 19 '12 at 19:47

You could obtain this by first multiplying $\cos(\pi/7)+\cos(3\pi/7)+\cos(5\pi/7)$ by $2\sin(\pi/7)$ and then applying the double angle formula for $\sin$ and sum to product formula for $\cos x \sin y$.

Let $a=\pi/7$, $b=3\pi/7$, and $c=5\pi/7$.

Using $$ 2\cos x\sin x =\sin 2x $$ and $$ \cos x\sin y= {\sin(x+y)-\sin(x-y)\over2 } $$ we have $$ \eqalign{ (\cos a+\cos b+\cos c)\cdot 2\sin a &=\color{maroon}{2\cos a\sin a}+\color{darkgreen}{2\cos b\sin a} +\color{darkblue}{2\cos c\sin a}\cr &=\color{maroon}{\sin 2a }+\color{darkgreen}{\sin(b+a)-\sin(b-a)}+\color{darkblue}{ \sin(c+a)-\sin(c-a)}\cr &=\color{teal}{\sin(2\pi/7) }+ {\color{purple}{\sin(4\pi/7)}\color{teal}{-\sin(2\pi/7)}}+ { \sin(6\pi/7)-\color{purple}{\sin(4\pi/7)}}\cr &=\sin(6\pi/7)\cr &=\sin (\pi/7)\cr &=\sin a. } $$ Now divide both sides of $$ (\cos a+\cos b+\cos c)\cdot 2\sin a =\sin a $$ by $2\sin a$.

share|cite|improve this answer

To elaborate on Mathlover's comment, the three numbers $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, and $\cos\frac{5\pi}{7}$ are the three roots of the monic Chebyshev polynomial of the third kind


where $U_n(x)=\frac{\sin((n+1)\arccos\,x)}{\sqrt{1-x^2}}$ is the usual Chebyshev polynomial of the second kind, and the last relationship is derived through the trigonometric identity


The question then is essentially asking for the negative of the coefficient of the $x^2$ term of $\frac18(U_3(x)-U_2(x))$ (Vieta); we can generate the two polynomials using the definition given above, or through an appropriate recursion relation. We then have


and we thus have


which yields the identities

$$\begin{align*} \cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}&=\frac12\\ \cos\frac{\pi}{7}\cos\frac{3\pi}{7}+\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}+\cos\frac{\pi}{7}\cos\frac{5\pi}{7}&=-\frac12\\ \cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}&=-\frac18 \end{align*}$$

share|cite|improve this answer
Thanks a lot for clearance. you shew beauty of Mathematics – Mathlover May 4 '12 at 9:07

Geometric picture: if $N$ is an integer greater than 1, the points $\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right)$, for $0\le j\lt N$, are equally distributed on the unit circle. More precisely, this set of points is unchanged if you rotate the plane by $\frac{2\pi}{N}$ about the origin. If we think of the points as a set of equal masses, then the center of mass has to be a point that is unchanged under rotation. There is only one such point.

With a little bit of work, the sum you have can be connected with this center of mass calculation.

To elaborate on the above: The matrix that rotates a vector by $\frac{2\pi}{N}$ is $$ R=\begin{bmatrix} \cos\frac{2\pi}{N} & -\sin\frac{2\pi}{N}\\\sin\frac{2\pi}{N} & \cos\frac{2\pi}{N} \end{bmatrix}. $$ Try multiplying $R$ by any of the vectors $\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right)$ and then applying addition formulas. You should get $\left(\cos\frac{2\pi (j+1)}{N},\sin\frac{2\pi (j+1)}{N}\right)$. But this means that the sum $$ S=\sum_{j=0}^{N-1}\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right) $$ is unchanged under multiplication by $R$. Can you prove that this implies $S=(0,0)$?

You are trying to prove $$ \frac{1}{2}-\cos\frac{\pi}{7}-\cos\frac{3\pi}{7}-\cos\frac{5\pi}{7}=0. $$ To relate it to the above, multiply both sides by 2, and use $\cos\theta=-\cos(\pi+\theta)=-\cos(\pi-\theta)$ to rewrite the left hand side.

share|cite|improve this answer

The three angles in the formula $\lbrace \pi/7, 3\pi/7, 5\pi/7 \rbrace $, thought of as points on the unit circle, can be completed to a set of seven angles of vertices of a regular heptagon.

These heptagon angles are, up to an integer multiple of $2 \pi$, equal to $\lbrace \pi, \pm \pi/7, \pm 3\pi/7, \pm 5\pi/7 \rbrace $.

The average $x$ coordinate of vertices of the polygon is zero, from the rotational symmetry of the heptagon.

There is also symmetry with respect to reflection in the $x$ axis: $\cos \pm x = \cos x$.

Combining these facts one gets $2S_7 - 1 = 0$ where $S_7$ is the sum of three cosines. Or $S_n = 1/2$ for odd $n \geq 3$.

share|cite|improve this answer
The correct denomination is heptagon. – Did May 8 '12 at 9:48

Let $y = \cos\theta + i \cdot \sin\theta$, where $\theta$ has either of the values $$\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{11\pi}{7} \ \text{and} \ \frac{13\pi}{7}$$ Then $y^{7} = \cos{7\theta} + i \:\sin{7\theta} =-1$. Then you have $$y^{7}+1=0 \Rightarrow (y+1)(y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1)=0$$ Now the root $y=-1$ corresponds to the value $\theta=\pi$. The roots of the equation $$y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1 =0 \qquad \cdots (1)$$ are therefore $\cos\theta + i\: \sin\theta$ where $\theta$ takes other values than $\pi$. Put $2x = y + \frac{1}{y} = 2 \cos\theta$ and note that:

  • $\displaystyle y^{2}+\frac{1}{y^2} = 4x^{2}-2$

  • $\displaystyle y^{3} + \frac{1}{y^3} = 8x^{3}-6x$.

Now divide $(1)$ by $y^{3}$ and get the answer.

Another method to look at this:

$7 \theta = \text{an odd multiple of} \: \pi$, therefore $\cos{4\theta} = -\cos{3\theta}$. Take $c=\cos{\theta}$. From this you have \begin{align*} 2\cdot (2c^{2}-1)^{2} -1 &= -(4c^{2}-3c) \\\ 8c^{4} + 4c^{2}-8c^{2} -3c + 1 &=0 \\\ (c+1) \cdot (8c^{3} - 4c^{2} -4c+1) &=0 \end{align*}

The following page also may be helpful:

share|cite|improve this answer

As a variation of my previous answer, it can be shown (see e.g. this paper) that the eigenvalues of the tridiagonal matrix


are $2\cos\dfrac{\pi}{7}$, $2\cos\dfrac{3\pi}{7}$, and $2\cos\dfrac{5\pi}{7}$. Since the trace of a matrix is the same as the sum of its eigenvalues,


and the identity shown in the OP easily follows.

share|cite|improve this answer

To compute $s=\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7)$, one can continue the sum and consider $$t=\sum_{k=1}^7\cos\left(\frac{2k-1}7\pi\right). $$ Since $\cos(2\pi-\alpha)=\cos(\alpha)$ for every $\alpha$ and $\cos(\pi)=-1$, $t=2s-1$. Now, $t$ is the real part of $$ u=\mathrm e^{\mathrm i\pi/7} + \mathrm e^{3\mathrm i\pi/7} + \mathrm e^{5\mathrm i\pi/7} + \mathrm e^{7\mathrm i\pi/7} + \mathrm e^{9\mathrm i\pi/7} + \mathrm e^{11\mathrm i\pi/7} + \mathrm e^{13\mathrm i\pi/7}, $$ and $u=z+z^3+z^5+z^7+z^9+z^{11}+z^{13}$ with $z=\mathrm e^{\mathrm i\pi/7}$. This is a geometric sum with ratio $z^2\ne1$ hence $u=z(1-z^{14})/(1-z^2)$.

Finally, $z^{14}=\mathrm e^{14\mathrm i\pi/7}=1$ hence $u=0$. Thus, $t=0$, and $s=1/2$.

This is the $n=3$ case of the identity, valid for every $n\geqslant1$, $$ \color{red}{\sum\limits_{k=1}^n\cos\left(\frac{2k-1}{2n+1}\pi\right)=\frac12}. $$

share|cite|improve this answer

Here's yet another approach, which has very great overlap with many of the previous answers. Let $\zeta=\cos(\pi/7)+i\sin(\pi/7)$. It’s a primitive fourteenth root of unity, thus satisfies the polynomial relation (A): $\zeta^6-\zeta^5+\zeta^4-\zeta^3+\zeta^2-\zeta+1=0$. Since $\zeta+\zeta^{-1}=2\cos(\pi/7)$, and more generally $\zeta^n+\zeta^{-n}=2\cos(n\pi/7)$ for any $n$, you want to prove that $\zeta^5+\zeta^3+\zeta+\zeta^{-1}+\zeta^{-3}+\zeta^{-5}=1$. All that remains is to note that since $\zeta^7=-1$, $\zeta^5+\zeta^{-5}=-\zeta^2-\zeta^{-2}$, the desired fact falls out from relation (A).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.