3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $T_k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.