# Polar graph $r = cos(5\theta)$ in xy-plane

I see that there is a more or less similar question in the list, but my one is not a duplicate. I will be grateful if you take a look at how I try to approach the task, and help me to find my mistakes - I am still struggling to get to grips with sketching polar graphs on xy-plane. Important: the goal is to do that "by hand", namely, first sketch the fundamental cycle of the polar graph on $r\theta$ plane, and then manually (and logically - here is one of my main problems) transfer it to the xy-plane.

Please, tell me which of the two approaches I shall follow, as I am a bit confused.

(1)

(1.1) Do I correctly choose values for $\theta$ and do I then correctly compute values of the fundamental cycle of the $r = cos(5\theta)$?

(please, see the image below)

(1.2) Then, based on those values, do I correctly compute x and y coordinates to plot the graph on xy-plane?

But this looks not right to me. Eventually I need to get this:

(2) On the other hand, according to graphing rules, if (x) is multiplied by a certain value, then the its value is "shrieked" by that value, namely we need to divide all original x-s by that value; in this case by 5. But how it is supposed to work in this case when I manually need to first find values of r, and then using them to find x and y values?

Thank you very much!

• Well, the pattern will repeat after $2\pi/5$ so I would concentrate on the range $[0, 2\pi/5]$. Calculate more samples in that range to get a better idea of the shape. Better still, work on the range $[-\pi/5, \pi/5]$ as it will be zero at the two extremes and you should get one of the loops. – badjohn May 31 '17 at 8:07
• @badjohn Thank you! Do I understand correctly that I have to follow the pattern I showed in the table, but only choose a smaller range? – Vitale May 31 '17 at 8:37
• Because $\cos$ has period $2\pi$. After that it will repeat. Due to the $5$ in your formula, it will repeat after $2pi/5$. So calculating outside my suggested range is just repeating work. So, you effort is better expended in a smaller range within which it does not repeat. You can save a bit more calculation by exploiting the symmetry of $\cos$ about $0$. So, you could only calculate in the range $[0, \pi/5]$ and just mirror it in the other part of my suggested range. – badjohn May 31 '17 at 8:51

I have not checked your arithmetic but the approach seems valid. Plot those $(x, y)$ pairs. However, you might not have enough detail to spot the shape.
You can get more detail at no cost by remembering the behaviour of $\cos$. It has period $2\pi$ so calculating values of $\cos$ over $2\pi$ is a waste, it will just repeat. Due to the $5$ in your formula, it will repeat after $\theta = 2\pi/5$. So, concentrate on a range just that wide.
Now remember that $\cos$ is symmetric about $0$ so if you use the range $[-\pi/5, \pi/5]$ then you can get the negative half for free by just reflecting the positive half.
Also, think of the behaviour of $cos$ after $\pi/2$. So, you don't actually need detailed calculations for that range $(\pi/10, \pi/5]$ either.
So, in summary, calculate and plot multiple points in the range $[0, \pi/10]$ and you should be able to predict the rest by symmetry.