What is the Cartesian form of $r - θ = - 2 {sin}^{2} θ - {cot}^{3} θ$ $r-theta = -2sin^2theta-cot^3theta$ ?

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Answer 1 · 2016-03-20T13:32:11

Set:

$x=rcosθ$

$y=rsinθ$

Answer is:

$sqrt(x^2+y^2)-arccos(x/sqrt(x^2+y^2))=-2x^2/(x^2+y^2)-x^3/y^3$

According to the following picture:

![https://www.mathsisfun.com/]( useruploads.socratic.org )

Set:

$x=rcosθ$

$y=rsinθ$

So we have:

$cosθ=x/r$

$sinθ=y/r$

$θ=arccos(x/r)=arcsin(y/r)$

$r=sqrt(x^2+y^2)$

The equation becomes:

$r-θ=-2sin^2θ-cot^3θ$

$r-θ=-2sin^2θ-cos^3θ/sin^3θ$

$sqrt(x^2+y^2)-arccos(x/r)=-2x^2/r^2-(x^3/r^3)/(y^3/r^3)$

$sqrt(x^2+y^2)-arccos(x/r)=-2x^2/r^2-x^3/y^3$

$sqrt(x^2+y^2)-arccos(x/sqrt(x^2+y^2))=-2x^2/sqrt(x^2+y^2)^2-x^3/y^3$

$sqrt(x^2+y^2)-arccos(x/sqrt(x^2+y^2))=-2x^2/(x^2+y^2)-x^3/y^3$

What is the Cartesian form of r-theta = -2sin^2theta-cot^3theta r−θ=−2sin2θ−cot3θr-theta = -2sin^2theta-cot^3theta ?