I am given the following polar function: $r(\theta)=2\,\cos{2\theta}$

And I am asked to find the domain of $f=r(\theta)$ if $r>0$

I understand that, in this context, finding the domain means to find the allowed values of $\theta$ when $r>0$, am I right?

I plotted the function (which, by the way, is the ultimate objective of this problem):


So what I did was to impose: $$r(\theta)>0 \Leftrightarrow 2\,\cos{2\theta}>0\Leftrightarrow 2\theta \in \left(-\dfrac \pi 2,\, \dfrac \pi 2 \right)\Leftrightarrow \boxed{\theta \in \left(-\dfrac \pi 4,\, \dfrac \pi 4 \right)}$$

But I don't know what this really means (I think it's the rightmost petal in the plot?). The function is well defined everywhere.

My book's solution is: $$\theta \in \left[0,\, \dfrac \pi 4 \right] \cup\left[\dfrac {3\pi}{4},\, \dfrac{5\pi}{4} \right] \cup\left[ \dfrac{7\pi}{4}, 2\pi\right]$$

Maybe there is something I'm missing. How exactly was this solution obtained? How does it relate to mine? Is it possible to use negative angles in polar coordinates? If the function is well-defined everywhere, what does "domain" mean in this context? I'm still learning, apologies if this question was a bit too elementary. Thank you!

  • $\begingroup$ It is usual to take $[0,2\pi]$ as the range for $\theta$, but you could take $[-\pi,\pi]$. Try plotting $\cos2\theta$. You can easily see that the book's range gives $\cos2\theta\ge0$. Your plot is wrong if you are trying to plot only the part with $r\ge0$ - you only want the horizontal lobes. The vertical ones have $r<0$. $\endgroup$
    – almagest
    Feb 12, 2018 at 16:45
  • $\begingroup$ When you learn about "deceptive Points", you will understand why this question actually is quite important. $\endgroup$
    – imranfat
    Feb 12, 2018 at 16:46

2 Answers 2


The book seems to be assuming the domain is a subset of $[0,2\pi]$, and is simply removing the values of $\theta$ for which $\cos2\theta$ is negative. When you require $r(\theta)\ge0$, you only get the right- and left-pointing lobes in your graph. The upper and lower lobes come by allowing $r(\theta)\lt0$: the lower lobe is swept out as $\theta$ runs from $\pi/4$ to $3\pi/4$, and the upper lobe is swept out as $\theta$ runs from $5\pi/4$ to $7\pi/4$.

  • $\begingroup$ Thanks for the answer. I think I'm having trouble to find when $\cos(2\theta)>0$ (how to find the intervals) Would you help me with that? $\endgroup$ Feb 12, 2018 at 18:00
  • 1
    $\begingroup$ @JoseLopezGarcia, I find it helps to picture which quadrant you land in when you double an angle. For angles up to $\pi/4$ (i.e., $45$ degrees), you stay in the first quadrant, where the $x$ coordinate is positive, so that $\cos2\theta$ is positive. For angles from $\pi/4$ to $\pi/2$, you land in the second quadrant, where the $x$ coordinate is negative, so that $\cos2\theta$ is negative. For angles from $\pi/2$ to $3\pi/4$, you land in the third quadrant, where the $x$ coordinate is still negative, etc. I hope all this makes sense. $\endgroup$ Feb 12, 2018 at 18:11
  • $\begingroup$ It makes a lot of sense, but what if we had a more general angle, like $\cos(k\theta)>0$ for any $k>0$. Is there a more systematic way of finding those intervals, as a function of $k$? It would be helpful if you could expand your answer or maybe give me a link to another resource, that answers this question. I appreciate all your help @BarryCipra $\endgroup$ Feb 12, 2018 at 18:36
  • $\begingroup$ @JoseLopezGarcia, you might do well to ask that as its own, stand-alone question. (You could link to your question here as a way to provide context.) $\endgroup$ Feb 12, 2018 at 19:09
  • $\begingroup$ Nevermind, I think I just have to divide the interval $[0, \frac \pi 2]\cup [\frac {3\pi}{2}, 2\pi]$ by the constant $k$ right? Thank you for your time $\endgroup$ Feb 12, 2018 at 19:53

I may not be qualified to answer your question, and I am not trying to. From what I understand you have already figured it out, but i am posting this to help those like me coming to this topic for the first time. I want to say thanks for all the comments and the answers above they explained it well and helped me understand it but I just want to put my answer in hopes that it will make it easier for beginners like me to understand it.

So the first thing to keep in mind is the domain of a polar equation which is usually [$0$,$2\pi$] (and this is because of the fact that $\theta$ would be repeating it self after that point), but it could also some times be [-$\pi$,$\pi$] . it would usually be given or you might have to figure it out depending on the problem. Then what you have to do is find the domain of your equation that is the subset of this domain. Lets take the above example.


First lets find the zeros of this equation. $r(\theta) = 0$ whenever $\cos{2 \theta} = 0$, and the first zero is at $\frac{\pi}{4}$ and because $cos(2 \theta)$ has a period of $\frac{2\pi}{2}$ the other zero's are $(...,-\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4},...)$ and so on. Because it is easy to graph the $cos$ function we can identify on which intervals $\cos(2 \theta)>0$ hence we will know on which intervals $r >0$, but for the more harder equation you may need to use first derivative test and the like to find turning points and get a nice sketch of the graph to determine on which intervals it is greater than zero. In our case $\cos(2 \theta)>0$ for $(...,[\frac{-5\pi}{4},\frac{-3\pi}{4}],[\frac{-\pi}{4},\frac{\pi}{4}],[\frac{3\pi}{4},\frac{5\pi}{4}][\frac{7\pi}{4},\frac{9\pi}{4}],...)$. Now from this set of intervals, let choose the once that satisfy our first rule. ($\theta \in \left(0 , 2\pi \right)$

when we do that we can see that $\theta \in \left[0,\, \dfrac \pi 4 \right] \cup\left[\dfrac {3\pi}{4},\, \dfrac{5\pi}{4} \right] \cup\left[ \dfrac{7\pi}{4}, 2\pi\right]$ so that would be our solution. The graph would then look like $\infty$. The graph if you are interested


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