Are the rationals on a unit circle dense? It seems evident from infinitely many primitive pythagorean triples $(a,b,c)$ that there are infinitely many rational points $\left(\frac{a}{c}, \frac{b}{c}\right)$ on the unit circle. 
But how would one go about, and show that they are dense, in the sense that for two rational points $x$ and $y$ of angles $α$ and $β$ on the unit circle, if $α<β$ there is a third rational point $z$ of angle $γ$ on the unit circle, such that $α<γ$ and $γ<β$.
Is it to expect that this conjecture holds and that it isn't an unsolved number theory problem?
 A: 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 . . .
A: 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.
A: 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,
$X=P(2,1)=(\frac45,\frac35)=(.8,.6), Y=P(4,3)=(\frac{24}{25},\frac7{25})=(.96,.28), Z=P(6,4)=(\frac{48}{52},\frac{20}{52})=(\frac{12}{13},\frac5{13})\approx(.92,.38).$
A: It's super easy to prove. For any complex number on the unit circle, there exists a sequence of complex numbers with integer parts such that if you replace each term with itself divided by its magnitude, the new sequence is a sequence of points on the unit circle that approaches the square root of the chosen complex number. If you replace each term of the original sequence with its square, then every term in the sequence will have an integral magnitude. If after that, you replace each term of the new sequence with itself divided by its magnitude, the new sequence is a sequence of complex numbers on the unit circle with rational parts that approaches the chosen complex number.
