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.
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 $T_k$.