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

How to prove $a\cos\left(\frac{2n\pi}{N}\right) \neq \cos\left(\frac{(n-1)\pi}{N}\right)$ for $n = \{0,1,...,N-1\}$ where $a$ is a scalar and $N \geq 3$.

I proceeded like this

$a\cos\left(\frac{2n\pi}{N}\right) = \cos\left(\frac{(n-1)\pi}{N}\right)$

if $\cos\left(\frac{2n\pi}{N}\right) \neq 0$ then $a = \sin\left(\frac{2\pi}{N}\right) + \tan\left(\frac{2n\pi}{N}\right)\cos\left(\frac{2\pi}{N}\right)$

How to proceed further?

share|cite|improve this question
Your "acos" is the inverse cosine? – J. M. May 1 '11 at 12:52
No, $a$ is a scalar. I admit I don't quite understand the question. Is $a$ supposed to be rational? – Qiaochu Yuan May 1 '11 at 12:57
@Yuan $a$ can be any real number. – Vinod May 1 '11 at 13:00
@Vinod: even $\cos ((n-1) \pi/N) / \cos (2 n \pi / N)$? – Qiaochu Yuan May 1 '11 at 13:02
So you want to prove that $\frac{\cos\left(\frac{(n-1)\pi}{N}\right)}{\cos\left(\frac{2n\pi}{N}\right)}$ cannot be a constant sequence? – J. M. May 1 '11 at 13:03
up vote 2 down vote accepted

Your question isn't clear. I think you're asking for a proof that there is no constant $a$ such that for all those values of $n$ we have your equation. If $n=0$, both cosines are positive, so $a$ would have to be positive. If $n$ is a little bigger than $N/4$, the cosine on the left is negative, the one on the right is positive, so $a$ has to be negative. Contradiction, QED.

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.