How to prove the curves $s_1(\theta)=\{\sin\frac\theta2,\cos\frac\theta2\}$ and $s_2(\theta)=\{-\sin\frac\theta2,-\cos\frac\theta2\}$ form a circle

Question

Say I have a set of two surfaces:

$$s_1\left(\theta\right)=\{\sin\frac{\theta}{2},\cos\frac{\theta}{2}\}$$

$$s_2\left(\theta\right)=\{-\sin\frac{\theta}{2},-\cos\frac{\theta}{2}\}$$

where $0\leq\theta\leq 3\pi$. If we plot them, we see that they each plot out $3/4$ of the unit circle. Together, it appears that they cover the entire unit circle (with some overlap).

How can we prove that $s_1$ and $s_2$ cover the unit circle without plotting them?

As in, how can we demonstrate that for any point $\chi=\{x,\pm\sqrt{1-x^2}\}$ (where $|x|\leq 1$) on the unit circle, there exists at least one $\theta_\chi$ such that $s_1\left(\theta_\chi\right)=\chi$ and/or $s_2\left(\theta_\chi\right)=\chi$?

Eventually I plan to generalize something involving this to multiple dimensions so it seems like plotting will quickly go out the window.

Answer 1

Let me suggest following view: for simplicity denote $d_1(\theta) = (\sin \theta, \cos \theta)=s_1(2\theta)$ and $d_2(\phi) = (-\sin \phi, -\cos \phi)=s_2(2\phi)$ and, obviously we can write $d_2(\phi) = (\sin (\pi+\phi), \cos (\pi+\phi))$. And so $d_2(\phi)=d_1(\pi+\phi)$ i.e. $d_1(\theta)=d_2(-\pi+\theta)$.

Same points on surface i.e. covering same part of circle, we will have when $d_2(\phi) = d_1(\theta)$. This means $$(\sin \theta, \cos \theta) = \left(\sin (\pi+\phi), \cos (\pi+\phi)\right)$$ So for part of domain for $\theta$ we have part of domain for $\phi$, that holds $\theta = \pi+\phi$, then we can speak about "covering same curve".

Answer 2

Since $0\leq\theta\leq 3\pi$ we have that $s_1\left(\theta\right)$ covers $3/4$ of a unit circle starting from $(0,1)$ to $(-1,0)$ passing clockwise through $(1,0)$ and $(0,-1)$ while $s_2\left(\theta\right)$ covers $3/4$ of a unit circle starting from $(0,-1)$ to $(1,0)$ passing clockwise through $(-1,0)$ and $(0,1)$.