How do you graph #r=cos(theta/n)# for n=2, 4, 6 and 8 and how do you predict the shap of the graph for #r=cos(theta/10)#?

#r=cos (theta/n) >=0 to theta/n in [-pi/2, pi/2] to theta in [-n/2pi, n/2pi], n = 2, 4, 6, ...#

Using conversion to (x, y), I have inserted n = 2 graph,

for #r = cos (theta/2)#

I have also succeeded in making it, for n = 4, 5, 6, 7, 8, 9, 10, ....

The trend is clear, for the limiting case. As #n to oo, r to cos 0 =1#,

representing a #unit# circle. Also, for n = 1, the graph is a

#1/2unit# circle

Graph for n = 1 Circle.
graph{(x^2+y^2)-x=0[-2 2 -1 1]}
Graph for n = 2 dimpled Cardioid.
graph{x-(x^2+y^2)^0.5(2(x^2+y^2)-1)=0[-2 2 -1 1]}
Graph for #n = 4#, with a large innie.
graph{x-(x^2+y^2)^0.5(1-8(x^2+y^2)(1-x^2-y^2))=0 [-2 2 -1.2 1.2]}

5-in-1 graph, for n = 1, 2, 4, 10 and #oo#:
graph{(x-(x^2+y^2)^0.5((x^2+y^2)^5-(1-x^2-y^2)^5-45(x^2+y^2)(1-x^2-y^2)((x^2+y^2)^3-(1-x^2-y^2)^3)+210(x^2+y^2)^2(1-x^2-y^2)^2(2(x^2+y^2)-1)))(x-(x^2+y^2)^0.5(1-8(x^2+y^2)(1-x^2-y^2)))(x-(x^2+y^2)^0.5(2(x^2+y^2)-1))((x^2+y^2)^0.5-1)(x^2+y^2-x)=0 [-2 2.2 -1.2 1.2]}

The period is #20pi#. For half period #10pi# only #r>=0#.

Observe five #2pi#-rotations of tracing, before returning to the

start at (1, 0) that is fixed for all n. The opposite end moves

towards (-1, 0), for the limit.Slide the graph #rarr darr larr

uarr#, to see all sides of this zoomed graph, on uniform scale.

De Moivre's theorem is used.
graph{x-(x^2+y^2)^0.5((x^2+y^2)^5-(1-x^2-y^2)^5-45(x^2+y^2)(1-x^2-y^2)((x^2+y^2)^3-(1-x^2-y^2)^3)+210(x^2+y^2)^2(1-x^2-y^2)^2(2(x^2+y^2)-1))=0[-1 1 -0.5 0.5]}.

See also the 5-loop neighbor behind in the sequence, #r = cos (theta/9)#. Long pending issues are now settled.

graph{x -(x^2+y^2)^5+36(x^2+y^2)^4(1-x^2-y^2)-126(x^2+y^2)^3(1-x^2-y^2)^2+84(x^2+y^2)^2(1-x^2-y^2)^3-9(x^2+y^2)(1-x^2-y^2)^4=0[-2 2 -1 1]}