Question

How do you graph `r=cos(theta/n)` for n=3, 5, 7, 9 and how do you predict the shap of the graph for `r=cos(theta/11)`?

Answer 1

See the method and the idiosyncratic graph of `r = cos (theta/11)`
I would edit my answer myself, if necessary, before closure, on 8/15.

Explanation:

Why even n is not included? In my earlier answers, I had discussed all, including `r = cos ((m/n)theta)`.

It appears that there is no identity ( even pseudonym ) for the

source of this question. In this medium, I have had disseminated,

within the limit of what I really know, without fear or favor. So, for

just one more day, I will do it..

Here, I had already solved such problems, that were never solved

before, through Cartesian conversions that care for suppressing

pixels in the graph, for invalid non-negative `r = sqrt( x^2+ y^2 )`

A reference:. .
https://socratic.org/questions/how-do-you-graph-r-cos-theta-n-for-n-2-4-6-and-8-and-how-do-you-predict-the-shap?source=search

I would repeat the method for `r = cos (theta/11)`.

#x/r = cos theta = cos 11(theta/11),

( and upon expanding in the Euler's mode, and using

`a = theta/11, c = cos a and s = sin a`

`x = c^12 - 55( c^10 s^2 - c^3 s^9) + 330 ( c^8 s^4 - c^4 s^7 )`

`-462 c^6 s^6`

Use c = r and s = sqrt ( 1-c^2 ).

The function for Socratic graphics utility is immediate.
graph{x - ((x^2+y^2)^6-55((x^2+y^2)^5(1-x^2-y^2)-(x^2+y^2)^1.5(1-(x^2+y^2))^4.5)+ 330((x^2+y^2)^4(1-x^2-y^2)^2-(x^2+y^2)^2.5(1-(x^2+y^2))^3.5)-462(x^2+y^2)^3(1-(x^2+y^2))^3)=0[-2. 2 -1 1 ] }

Answer 2

I am continuing my answer, for some remarks. I dislike editing my answer now, despite my statement that I would do it myself.. Anyone liking that can do so, after Aug 15.