Let point $P=(cos(\alpha), sin (\alpha))$ and point $Q = (cos(\alpha+ \pi), sin (\alpha+\pi))$ be two points moving on a circle ( of center $(0,0)$ and of radius $1$).
The straignt line passing through $P$ and $Q$ is the set of all points satisfying :
$ y = \frac {sin(\alpha)-sin(\alpha+\pi)}{cos(\alpha)- cos(\alpha+\pi)} ( x- cos(\alpha)) + sin(\alpha)$.
Which conditions should be imposed on $\large x$ in order to reduce this straight line to the line segment joining $P$ and $Q$?
My only guess is
$cos(\alpha+\pi)\leq x \leq cos(\alpha)$.
But this "method" does not work apparently ( because the ordering relation between the two limiting values is not constant) :
https://www.desmos.com/calculator/k43zn8gvzj
Clearly the equation $cos(\alpha+\pi)\leq x \leq cos(\alpha)$ is only good when $cos(\alpha+\pi) < cos(\alpha),$ which is about half the time. As $\alpha$ changes the two cosines will swap places and then eventually swap back again.
When $cos(\alpha+\pi)> cos(\alpha)$ you want $cos(\alpha)\leq x \leq cos(\alpha+\pi)$.
A relatively foolproof way to fix this is to write $$ \min(cos(\alpha),cos(\alpha+\pi))\leq x \leq \max(cos(\alpha),cos(\alpha+\pi)). $$ That way, whichever of the two $x$ endpoints is less will be on the left where you want it, and whichever is greater will be on the right where you want it.
This works for any two angles, for example $\alpha$ and $\alpha + \frac\pi2,$ not just for this particular example.
But in this particular example, where the difference between the angles is always exactly $\pi,$ the two points are exactly opposite each other on the circle, and therefore $$ cos(\alpha+\pi) = - cos(\alpha). $$ So really the two extreme $x$ values are just $\cos(\alpha)$ and $-\cos(\alpha).$ Again which one is less than $x$ and which is greater will depend on whether $\cos(\alpha)$ is positive or negative. But you can make sure you get the negative one on the left and the positive on the right like this:
$$ -\lvert \cos(\alpha) \rvert \leq x \leq \lvert \cos(\alpha) \rvert. $$