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Dec 2, 2014 at 2:39 comment added C. E. You should read Aron's answer more carefully, he shows you how you can write a better code which is even simpler. Another simple one-liner: b = PolarPlot[Cos[2 [Theta]], {\[Theta], $MachineEpsilon, #}, PlotRange -> 1] & /@ Range[0, 2 Pi, 0.02 Pi];
Dec 1, 2014 at 14:41 comment added Nimbigli Look at both of them. you should find out difference of them. My answer is just answer of the principal question.
Dec 1, 2014 at 14:28 history answered Nimbigli CC BY-SA 3.0