Polar to cartesian form of r=sin(4θ)? 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 $$(x^2+y^2)^{5}=16x^2y^2(x^2-y^2)^{2}$$

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?
