# How to prove $a\cos(\frac{2n\pi}{N}) \neq \cos(\frac{(n-1)\pi}{N}))$

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?

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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

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.