How is $r(\theta) = \sin \frac\theta2$ symmetric about the x-axis?

Question

I understand how it is symmetric about the $y$-axis.

because $r(-\theta) = \sin \left(-\frac\theta2\right)=-\sin \left(\frac\theta2\right)=-r(\theta)$

But how is it symmetric about $x$-axis?

Answer 1

Hint

$$r(\pi+\theta)=\sin\left(\frac\pi2+\frac\theta2\right)=\cos\frac\theta2$$ $$r(\pi-\theta)=\sin\left(\frac\pi2-\frac\theta2\right)=\cos\frac\theta2$$

Alternatively, $r(-\theta)=-r(\theta)$, hence your $y$-axis symmetry, but you have also $$r(2\pi-\theta)=\sin\left(\pi-\frac\theta2\right)=\sin\frac\theta2=r(\theta)$$ $$r(2\pi+\theta)=\sin\left(\pi+\frac\theta2\right)=-\sin\frac\theta2=-r(\theta)$$

Hence the points associated to angles $\theta$, $-\theta$, $2\pi-\theta$, $2\pi+\theta$ form a rectangle.