Can we show that


by considering the seventh roots of unity? If so how could we do it?

Also I have observed that


as well, so just out of curiosity, is it true that $$\sum_{k=1}^{n-1} \cos\frac{k\pi}{n} = 0$$

for all $n$ odd?

  • $\begingroup$ Use complex numbers, something like this $\endgroup$ – rtybase May 21 at 21:05
  • 4
    $\begingroup$ You could use that $\cos (\pi-\theta)=-\cos \theta$ instead, and just pair each cosine with its negative. $\endgroup$ – Mark Bennet May 21 at 21:10
  • $\begingroup$ Also, it works for all $n$'s, not just odd ones. $\endgroup$ – rtybase May 21 at 21:19
  • $\begingroup$ because the unpaired one is $\cos (\pi /2)$ $\endgroup$ – G Cab May 21 at 21:41
  • 1
    $\begingroup$ @Acccumulation Even for $n=1$, one has $\sum_{k=1}^{n-1}\cos(k\pi/n)=0$. $\endgroup$ – Lord Shark the Unknown May 22 at 5:39

Note that $$\cos(\pi - \alpha)= - \cos(\alpha)$$ Therefore $$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{5\pi}{7})+\cos(\frac{6\pi}{7})=$$


The same goes for other natural numbers $n$ instead of $7$.

  • 1
    $\begingroup$ and even number also $\endgroup$ – G Cab May 21 at 21:41

I think you can use Euler's Formula.

The Nth roots of unity = $e^{2\pi k i/N}$ for values of k between $0$ and $N-1$ inclusive.

There sum from k to $N-1$ is a geometric series.

$S= \sum_{k=0}^{N-1} \ e^{2\pi i k/N}=\frac{1\cdot e^{(2\pi i /N)N}-1}{e^{2\pi i /N}-1}$

The numerator is zero for any N.

But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.

  • $\begingroup$ <pedant>This works only for integer N greater than 1</pedant> $\endgroup$ – Acccumulation May 22 at 5:16

Pointing at the link I left in the comments

$$1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$$

Then for $\forall n\in\mathbb{N}, n>0$ $$\cos\frac{\pi}{n+1}+ \cos\frac{2\pi}{n+1} +... + \cos \frac{n\pi}{n+1} = \frac{\sin\left[(n+\frac{1}{2})\frac{\pi}{n+1}\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=\\ \frac{\sin\left[\frac{2n+1}{2(n+1)}\pi\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}= \frac{\sin\left[\pi-\frac{\pi}{2(n+1)}\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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