Polar to cartesian form of r=sin(4θ)?

Question

The Polar to cartesian form of $ r = \sin(2\theta)$ is fairly simple. What is the Cartesian form of the polar equation r=sin(4θ)?

[edit] $$r=4sin(θ)cos(θ)(cos(θ)^2-sin(θ)^2)$$, so $$r^5=4rsin(θ)rcos(θ)(r^2cos(θ)^2-r^2sin(θ)^2)$$,so $$r^5=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5/2}=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5}=(4xy(x^2-y^2))^{2}$$, so $$(x^2+y^2)^{5}=16x^2y^2(x^2-y^2)^{2}$$

Answer 1

$$\rho=\sin 4\theta=2\sin2\theta\cos2\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=\frac{4xy(x^2-y^2)}{\rho^4},$$ hence

$$(x^2+y^2)^5=(4xy(x^2-y^2))^2,$$

$$x^{10}+10x^4y^6+32x^4y^4+5x^2y^8-16x^2y^6+y^{10}+10y^4x^6+5y^2x^8-16y^2x^6=0.$$

Answer 2

Hint.

$$\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$$

Then again you may use

$$\sin(2\theta) = 2\sin\theta\ \cos\theta$$

and

$$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$

Can you go on?

Answer 3

Hint: use that $$\cos(4\theta)=\sin (\theta )+\cos ^4(\theta )-6 \sin ^2(\theta ) \cos ^2(\theta )$$