Parametric equations where sin(t) and cos(t) must be rational

Question

Suppose there are parametric equations

$$ x(t) = at - h\sin(t) $$ $$ y(t) = a - h\cos(t) $$

and it is required that both $\sin(t)$ and $\cos(t)$ should be rational.

What the values of $t$ should be in that case?

Thanks.

Answer 1

Your question (at least how you wrote it) is really,

For what values of $t$ are both $\cos t$ and $\sin t$ rational?

Note that if $x$ and $y$ are real numbers with $x^2 + y^2 = 1$, then there exists a real number $t$ such that $x = \cos t$, $y = \sin t$. Therefore, we want solutions to $x^2 + y^2 = 1$ for $x, y \in \mathbb{Q}$. For this we let $x = \frac{a}{c}$, $y = \frac{b}{d}$, $a,b,c,d \in \mathbb{Z}$, with the fractions in reduced form. Then we get $$ a^2 d^2 + b^2 c^2 = c^2 d^2 \tag{1} $$ Let $c = c'k$, $d = d'k$ with $c', d'$ relatively prime. Then the above gives $$ a^2 d'^2 + b^2 c'^2 = k^2 c'^2 d'^2 $$ Mod $d'$, we find $b^2 \equiv 0$. Since $\frac{b}{d}$ was in reduced form, $d' = 1$. Similarly $c' = 1$. So we get $c = d = k$. Then we have, coming back to (1), $$ a^2 + b^2 = c^2 $$ so $(a,b,c)$ are a Pythagorean triple. In particular they are a primitive Pythagorean triple, because $(a,c) = (b,c) = 1$. That means all possible $a, b, c$ (up to sign and switching $a$ and $b$) are given by $$ (a,b,c) = (2mn, m^2 - n^2, m^2 + n^2) $$ where $m,n$ are relatively prime and not both odd. Therefore for any $a,b,c$ the solution $t$ is given by $$ t = \pm \cos^{-1}\left(\frac{a}{c}\right) + 2\pi n = \pm \cos^{-1} \left(\pm \frac{2mn}{m^2 + n^2} \right) \text{ or } \pm \cos^{-1} \left(\pm \frac{m^2 - n^2}{m^2 + n^2} \right) $$ for some $m,n$. Unfortunately we cannot in general simplify the $\cos^{-1}$ out of the formula as far as I know.

Answer 2

Take any two integers $m$ and $n$, and then choose $t$ so that $\tan \tfrac12 t= r = m/n$.

Then, using well-known trig identities $$ \cos t = \frac{1-r^2}{1+r^2} = \frac{m^2-n^2}{m^2+n^2} \quad ; \quad \sin t = \frac{2r}{1+r^2} = \frac{2m}{m^2+n^2} $$ So $\cos t$ and $\sin t$ are both rational.