Are the rationals on a unit circle dense?

Consider a nonvertical line through the point $(1, 0)$ of slope $m$. This line meets the circle at exactly one other point, and it's not hard to show that the coordinates of that point are $$P_m=\left({m^2-1\over m^2+1}, {-2m\over m^2+1}\right).$$ As long as $m$ is rational, $P_m$ has rational coordinates; so now think about the lines through $(1, 0)$ of rational slope . . .

We are interested in angles $\theta$ such that $(\cos \theta, \sin \theta)$ are both rational. Call such an angle a $\mathbb Q-$ angle.

Claim: $\theta$ is a $\mathbb Q-$ angle iff $\tan \frac {\theta}2 \in \mathbb Q$.

Proof:

$\implies$ follows from the half angle formula, $\tan \frac {\theta}2 =\frac {1-\cos \theta}{\sin \theta}$

To see $\impliedby$ it may be easiest to work geometrically. Note that $\frac {\theta}2$ is the angle formed by the line connecting $(-1,0)$ to the point $(\cos \theta, \sin \theta)$. That line has equation $y=h(x+1)$ where $h=\tan \frac {\theta}2$. Solving this and $x^2+y^2=1$ simultaneously we see that we are trying to solve $$x^2+h^2(x^2+2x+1)-1=0\implies x^2+\frac {2h^2}{1+h^2}x+\frac {h^2-1}{1+h^2}=0$$ of course one root is given by $x=-1$ and it follows at once that the other root must also be rational. Thus $\cos \theta \in \mathbb Q$. Now we can use the half-angle formula again to see that this implies that $\sin \theta \in \mathbb Q$.

Note: this is the key point behind the Weierstrass substitution, also known as the $\tan \frac {\theta}2$ substitution.

It is clear that the angles $\theta$ such that $\tan \frac {\theta}2\in \mathbb Q$ are dense so we are done.

If the two given points are in different quadrants, then one of the points $(0, \pm1), (\pm1, 0)$ may fairly be said to be between them. So wlog they are both in the first quadrant. Then for any rational point $P$ on the unit circle, there are $t, u\in \mathbb N$ with $t>u$, where $$P=P(t,u)=\left(\dfrac{2tu}{t^2+u^2},\dfrac{t^2-u^2}{t^2+u^2}\right).$$ If $t+u$ is odd and $t$ and $u$ are coprime, that even yields the fractions in their lowest terms.

So given $X=P(t_1,u_1)$ and $Y=P(t_2,u_2)$, $Z=P(t_1+t_2,u_1+u_2)$ is a rational point between $X$ and $Y$ as required.

For example,

$X=P(2,1)=(\frac45,\frac35)=(.8,.6), Y=P(4,3)=(\frac{24}{25},\frac7{25})=(.96,.28), Z=P(6,4)=(\frac{48}{52},\frac{20}{52})=(\frac{12}{13},\frac5{13})\approx(.92,.38).$