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.

enter image description here

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?


1 Answer 1


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 $\pi.$

  • $\begingroup$ How would I know to stop at $\pi$ before drawing the cosine wave? Is there a way to see the "polar period" is $\pi$ from the original formula instead of guessing and checking? $\endgroup$ Commented Jan 23, 2017 at 6:40
  • $\begingroup$ Usually you would just solve for the first place that $(r(\theta), \theta)=(r(\theta+x), \theta+x)$ where x is the period. The only two situations where this can happen is if the coordinates are the same or if the coordinates are negations of their counterparts. $\endgroup$ Commented Jan 23, 2017 at 14:27
  • $\begingroup$ Edit: Oops, I meant that $(r, \theta)=(-r, \theta+\pi).$ $\endgroup$ Commented Jan 23, 2017 at 16:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .