Question

How do you convert `r² = costheta` into polar form?

Answer 1

`(sqrt(cos(theta)),theta)`

Explanation:

Standard polar form is of format `(r,theta)`

Where `y=rsin(theta)" and "x=rcos(theta)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that in the question we have `r^2 -> r=sqrt(r^2)=sqrt(cos(theta))`

Implying that standard form of `y=rsin(theta)` translates into `y=sqrt(cos(theta))xxsin(theta)`

Implying that the standard form of `x=rcos(theta)` translates into `x=sqrt(cos(theta))xxcos(theta)`

So in polar form of type `(r,theta)` we have:

`(sqrt(cos(theta)),theta)`