Question

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

Answer 1

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

Explanation:

`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]}