How do you convert #r^2 = cos4θ# into rectangular form?

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Answer 1 · 2016-05-16T05:30:53

#(x^2+y^2)^3+8x^2y^2-(x^2+y^2)^2=0#

#r^2=cos 4theta#

#=1-2sin^2(2theta)#

#=1-2((2sin theta cos theta)^2#

#=1-8sin^2thetacos^2theta#

Use #sin theta=x/r, cos theta=y/r and r^2=x^2+y^2#

#x^2+y^2=1-8(x^2y^2)/(x^2+y^2)^2#

Rearranging to a polynomial form,

#(x^2+y^2)^3+8x^2y^2-(x^2+y^2)^2=0#

graph{(x^2+y^2)^3+8x^2y^2-(x^2+y^2)^2=0[-4 4 -2 2]}

Note that, on par with #r = cos ntheta, r^2 = cos ntheta# indeed

creates rose curves, with wider petals. I made them much wider by

using #r^2 = 4 cos 4theta# instead.

Combined graph, for n = 4:

graph{(0.25(x^2+y^2)^3+8x^2y^2-(x^2+y^2)^2)((x^2+y^2)^2.5+8x^2y^2-(x^2+y^2)^2)(x^2+y^2-.01)=0[-4 4 -2 2]}