# Tracing out Polar Graph for $r=\cos(3\theta)$ Exactly Once.

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?

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 $\pi.$
• 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? Commented Jan 23, 2017 at 6:40
• 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. Commented Jan 23, 2017 at 14:27
• Edit: Oops, I meant that $(r, \theta)=(-r, \theta+\pi).$ Commented Jan 23, 2017 at 16:58