Is it possible to convert $r = 3 cos (3 sin θ)$ $r=3cos(3sintheta)$ to Cartesian form?

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Answer 1 · 2018-03-14T15:08:24

Yes but it requires 4 equations.

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

enter image source here

Divide both sides of the equation by 3:

$\frac{r}{3} = cos (3 sin (θ))$ $r/3=cos(3sin(theta))$

Take the inverse cosine of both sides:

${cos}^{- 1} (\frac{r}{3}) = 3 sin (θ)$ $cos^-1(r/3) = 3sin(theta)$

Substitute $r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2+y^2)$ and $sin (θ) = \frac{y}{\sqrt{x^{2} + y^{2}}}$ $sin(theta) = y/sqrt(x^2+y^2)$ :

${cos}^{- 1} (\frac{\sqrt{x^{2} + y^{2}}}{3}) = 3 \frac{y}{\sqrt{x^{2} + y^{2}}}$ $color(blue)(cos^-1(sqrt(x^2+y^2)/3) = 3y/sqrt(x^2+y^2))$

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

![www.desmos.com/calculator]( useruploads.socratic.org )

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

${cos}^{- 1} (\frac{\sqrt{x^{2} + y^{2}}}{3}) = - 3 \frac{y}{\sqrt{x^{2} + y^{2}}}$ $color(blue)(cos^-1(sqrt(x^2+y^2)/3) = -3y/sqrt(x^2+y^2))$

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

![www.desmos.com/calculator]( useruploads.socratic.org )

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

${cos}^{- 1} (- \frac{\sqrt{x^{2} + y^{2}}}{3}) = 3 \frac{y}{\sqrt{x^{2} + y^{2}}}$ $color(blue)(cos^-1(-sqrt(x^2+y^2)/3) = 3y/sqrt(x^2+y^2))$

and

${cos}^{- 1} (- \frac{\sqrt{x^{2} + y^{2}}}{3}) = - 3 \frac{y}{\sqrt{x^{2} + y^{2}}}$ $color(blue)(cos^-1(-sqrt(x^2+y^2)/3) = -3y/sqrt(x^2+y^2))$

![www.desmos.com/calculator]( useruploads.socratic.org )

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