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If $\cos\theta$ is rational, then is $\cos k\theta$ also rational, where $k$ is a positive integer? I tried a few simple cases, and they worked. Is it true? And if it is then how do I go about rigorously proving this?

I've tried using the identity of $\cos(A+B)$ with induction, but am not able to prove it.

Thanks.

Parcly Taxel
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TheRandomGuy
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    https://math.stackexchange.com/questions/125774/how-to-expand-cos-nx-with-cos-x – Winther Feb 08 '18 at 09:24
  • The expansion of $\cos k\theta$ in terms of $\cos x$ can help us with the idea, but just to prove that $\cos k\theta$ is rational we don't need to expand it. Use $\cos(k+2)\theta+\cos k\theta = 2\cos\theta\cos(k+1)\theta$ and induction. – Jianing Song Nov 15 '23 at 18:17

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We have $$\cos k\theta=T_k(\cos\theta)$$ where $T_k$ is the $k$th Chebyshev polynomial of the first kind. Since all Chebyshev polynomials have integral coefficients and $\cos\theta$ is rational, $\cos k\theta$ must also be rational.

For a lower-level proof, de Moivre's formula gives $\cos k\theta$ as the real part of $(\cos\theta+i\sin\theta)^k$, expanding to $$\cos k\theta=\sum_{l=0}^{\lfloor k/2\rfloor}\binom k{2l}\sin^{2l}\theta\cos^{k-2l}\theta$$ Since $\sin^2x=1-\cos^2x$: $$\cos k\theta=\sum_{l=0}^{\lfloor k/2\rfloor}\binom k{2l}(1-\cos^2\theta)^l\cos^{k-2l}\theta$$ Since binomial coefficients are integers, the RHS is an integer polynomial in $\cos\theta$, so $\cos k\theta$ is rational. In fact, this polynomial is one way to define the $T_k$.

Parcly Taxel
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