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Suppose there are parametric equations

$$

x(t) = at – h\sin(t)

$$

$$

y(t) = a – h\cos(t)

$$

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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.

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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.

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.

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