Question

Is it possible to convert `r=3cos(3sintheta)` to Cartesian form?

Answer 1

Yes but it requires 4 equations.

Explanation:

The polar graph of `color(red)(r=3cos(3sin(theta)))` is:

Divide both sides of the equation by 3:

`r/3=cos(3sin(theta))`

Take the inverse cosine of both sides:

`cos^-1(r/3) = 3sin(theta)`

Substitute `r = sqrt(x^2+y^2)` and `sin(theta) = y/sqrt(x^2+y^2)`:

`color(blue)(cos^-1(sqrt(x^2+y^2)/3) = 3y/sqrt(x^2+y^2))`

I shall overlay the polar graph on a Cartesian plane with the above equation in blue:

To graph the bottom half of the horizontal loop we need to negate the right side:

`color(blue)(cos^-1(sqrt(x^2+y^2)/3) = -3y/sqrt(x^2+y^2))`

The following is a graph of the two equations (in blue) overlaying the polar equation:

The two vertical loops are overlain when we negate the argument of the inverse cosine:

`color(blue)(cos^-1(-sqrt(x^2+y^2)/3) = 3y/sqrt(x^2+y^2))`

and

`color(blue)(cos^-1(-sqrt(x^2+y^2)/3) = -3y/sqrt(x^2+y^2))`