I would like to know how to anticipate how large of a domain of $\theta$ I need to consider in order to draw graphs of the form $r=f(\theta)$ in polar coordinates if $f$ is a sine or cosine (where the graph is traced out exactly once).
I can graph $r=\cos(3\theta)$ by considering the graph in rectangular coordinates, but it turns out that if I start at $\theta=0$, then I need to stop at $\theta =\pi$ in order to graph the entire polar graph. I understand why $[0,2\pi/3]$ doesn't suffice.
Where is the $\pi$ coming from? Is there a way to anticipate the upper bound $a$ in $0\le \theta <a$ such that the polar graph traces out exactly once?
Best Answer
Yes it's technically pi, but it's also an illusion in a way.
When $\theta=\pi$ we have $r=-1$ so the circle starts on the negative x-axis and since the angle is $\pi$ the result is the same point as with $\theta=0.$
But you can prove that $r(\theta)=r(\theta+\pi)$ - We have $r(\theta+\pi)=\cos(3\theta+3\pi)=-\cos(3\theta)=-r(\theta)$ and with a simple circle argument you can prove that the "period" is $\pi.$