# Converting A Polar Equation to Rectangular Form

## Homework Statement

Convert the polar equation to rectangular form.

r=2sin(3θ)

## The Attempt at a Solution

I can expand this out to

$$r=2(\sin\theta\cos2\theta+\cos\theta\sin2\theta)$$

multiply both sides by r

$$r^2=2r(\sin\theta\cos2\theta+2r\cos\theta\sin2\theta)$$

simplify

$$r^2=2y\cos2\theta+2x\sin2\theta)$$

Then I could expand the 2θ and get

$$r^2=2y\cos^2\theta-2y\sin^2\theta+4x\sin\theta\cos\theta$$

I'm not sure where to go from here.

Plugging it into $$x^2+y^2=r^2$$ from here dosen't seam to help.

The answer is supposed to be: $$(x^2+y^2)^2=6x^2y-2y^3$$

Related Precalculus Mathematics Homework Help News on Phys.org
eumyang
Homework Helper

## The Attempt at a Solution

I can expand this out to

$$r=2(\sin\theta\cos2\theta+\cos\theta\sin2\theta)$$
From here I would go ahead and distribute the 2, and then use the double-angle identities. For cosine, use the cos 2θ = cos2 θ - sin2 θ variant, like you did later on.

After simplifying (you'll be able to combine like terms along the way), multiply both sides by $$r^3$$ instead of $$r$$, so that each trig function on the right side can be "paired" with an r. You will eventually be able to get to the answer you posted.

69

ehild
Homework Helper
Use the expressions x=r cosθ, y=r sinθ.

ehild

From here I would go ahead and distribute the 2, and then use the double-angle identities. For cosine, use the cos 2θ = cos2 θ - sin2 θ variant, like you did later on.

After simplifying (you'll be able to combine like terms along the way), multiply both sides by $$r^3$$ instead of $$r$$, so that each trig function on the right side can be "paired" with an r. You will eventually be able to get to the answer you posted.

69
Ahhh.. yes.

Thank you, that was a big help.

Cant believe I didn't see that before.