Why does the tracing of this polar curve stop at the origin?

Question

[ The red graph is -2sinx; The blue is the polar equation $r =-2sin \theta$.From $0≤\theta≤\pi$, r≤0. I get that. But r >0 when $\theta$ passes π, so shouldn't the blue curve continue above the x-axis? Thanks in advance.

Answer 1

As $\theta$ passes through any multiple of $\pi$ you would cross the horizontal axis provided that $r$ maintains its sign. But instead, $r$ changes sign so that your intended crossing the horizontal axis is superposed on an inversion through the origin. You end up back on the same side of the horizontal axis, crossing the vertical axis instead.

Answer 2

Note that for small positive $\epsilon$, $$r(\pi +\epsilon) = -2\sin(\pi+\epsilon) = -2(\sin\pi\cos\epsilon+\cos\pi\sin\epsilon)$$ $$=-2(0\cos\epsilon + (-1)\sin\epsilon)= 2\sin\epsilon > 0$$ So the point is in the 3rd quadrant.