Given that $\cos(\theta) = -\frac{1}{4}$.

Prove $\cos (n\theta) \neq 1$, for any $n \in \mathbb{Z}^+$.

My attempt: If $\cos(n\theta) = 1$, then $e^{in\theta} = 1$.

$e^{in\theta} = e^{i(n-1)\theta} \cdot e^{i \theta}$

If $e^{in\theta} = 1$, then $e^{i(n-1)\theta} = e^{i(-1)\theta}$.

Then $(-\frac{1}{4} + \frac{2\sqrt{2}}{3}i)^{n-1} = (-\frac{1}{4} - \frac{2\sqrt{2}}{3}i)$.

However, I failed to prove that $(-\frac{1}{4} + \frac{2\sqrt{2}}{3}i)^{n-1} \neq (-\frac{1}{4} - \frac{2\sqrt{2}}{3}i)$, for any $n\in \mathbb{Z}^+$.

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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – Kavi Rama Murthy Feb 26 at 8:22
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    $\begingroup$ Is it $\cos n\theta\neq\frac12$ as in your question title, or $\cos n\theta\neq 1$ as in the second line of your question? $\endgroup$ – user10354138 Feb 27 at 13:17

Suppose by contradiction that $\cos(n \theta)=1$.

Use the identity $$\cos( \theta) = \frac{e^{i \theta} + e^{-i \theta}}{2}$$ Call $z=e^{i \theta}$, then $z$ satisfies the following equation $$\frac{z + z^{-1}}{2}=- \frac{1}{3}$$ which can be rewritten as $$z^{2} + \frac{2}{3}z+1=0 $$

Clearly, the polynomial $x^2+ \frac{2}{3}x+1$ is the minimal polynomial of $z$ over the rational numbers (because it is irreducible over the rationals). This means that $z$ is not an algebraic integer.

On the other hand, $\cos(n \theta)=1$ implies that $z^n=1$. This means that $z$ is a root of unity, and in particular it is an algebraic integer. This is a contradiction.

EDIT: if you don't know what an algebraic integer is, look at the following theorem:

Let $\alpha \in \Bbb C$ be an algebraic number over the rationals. Then TFAE:

  1. the monic minimal polynomial of $\alpha$ has integer coefficients
  2. $\alpha$ is a root of a monic polynomial with integer coefficients

In such a case, $\alpha$ is called an algebraic integer.

  • $\begingroup$ Thanks, but are there some easier ways to prove it. I'm a high school student so I don't know what is a minimal polynomial. And this problem is designed for high school students so I'm pretty sure that there is a more simple method. $\endgroup$ – 李若谷 Feb 26 at 13:14

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