Graph of $r=4\cos(6\theta)$ I'm learning about polar coordinates. What is the best way to understand $r=4\cos(6\theta)$ without a computer?
I already knew what the graph of $r=\cos(2\theta)$ looks like. Is there a way to get an idea about $r=4\cos(6\theta)$ using this "parent graph"?
What about the graph of $r=4\cos(5\theta)$. Does the fact that $5$ is odd while $6$ is even make a big difference as to the number of "leaves" that appear on the "rose"? Thanks
 A: We have that $r=\cos ( \theta)$ makes a loop for any interval $\left[-\frac \pi 2 +k\pi ,\frac \pi 2 +k\pi\right]$ but since it leads to negative values for $k$ odd we have that only the loops with $k$ even can be plotted, which is a circle in this case (note that the same plot is obtained if we allow negative values for $r$ since in this case the plots coincide).
Therefore $r=\cos ( 2\theta)$ makes a loop for any interval $\left[-\frac \pi 4 +k\frac \pi 2 ,\frac \pi 4 +k\frac \pi 2\right]$ which correspond to $2$ loops if we restrict to positive values for $r$ and to $4$ loops if we allow negative values also.
So also for $\cos ( 6\theta)$ we have a loop for any interval $\left[-\frac \pi {12} +k\frac \pi 6 ,\frac \pi {12} +k\frac \pi 6\right]$ which correspond to $6$ loops if we restrict to positive values for $r$ and to $12$ loops if we allow negative values also.
For $r=\cos ( 5\theta)$ we have a loop for any interval $\left[-\frac \pi {10} +k\frac \pi {5} ,\frac \pi {10} +k\frac \pi {5}\right]$ which correspond in any case to $5$ loops since the loops for $k$ even ($r>0$) are the same we obtain for $k$ odd ($r<0$).

Summarizing, the plot for $r= \cos (n\theta)$ has

*

*$n$ loops for $n$ odd

*$n$ loops for $n$ even if negative value for $r$ are not allowed

*$2n$ loops for $n$ even if negative value for $r$ are allowed



A: A particular way to understand how these operations transform the graph is to notice that $f(kx)$ is a horizontal compression of scale factor $\frac{1}{k}$. This idea still applies to polar coordinates, but with the small change that you are now compressing with respect to $\theta$ instead of $x$.
So for the original graph, you could fit 4 pedals from $[0,2\pi]$, but with the factor of 3 in front, you would be now fitting $4 \times 3 = 12$ pedals into $[0,2\pi]$. Formally, this is changing the period of the function from $\pi$ to $\pi/3$.
Similarly we can think of $r=4\cos(5\theta)$ as changing the period to $2\pi/5$, but the difference here is that now the function doubles back on itself from $\pi$ to $2\pi$, so although there should be $4\times\frac{5}{2}=10$ pedals, half of them overlap to create the image of only 5 pedals.
The constant outside of the $\cos$ serves to increase the maximum radius. Originally, the maximum of cosine is 1, but by multiplying the value by 4, all of the r values are increased four times, causing the maximum to increase to 4.
A: An efficient method is to find the $\theta$-values where $r(\theta)=0$ and $r(\theta)=a$, where a is the amplitude of the cosine. Combine this with your knowledge that the graph is a rose-type. Now plot the corresponding points $(r,\theta)$, tracing in order of increasing $\theta$.  This should give you a reliable skeleton on which you can draw the rose.
This strategy is generally a useful one when graphing other types of well-know polar curves.
