Polar to Cartesian conversion of $r = \sin(5\theta)$ I was doing some conversions in my free time, and after tackling $r = \sin(3\theta)$ 
I decided to do $r = \sin(5\theta)$. 
So far, I have converted it a simplified form containing only thetas:
$\sin(5\theta) = \sin(\theta)(16\cos^4(\theta)-12\cos^2(\theta)+1)$ 
I am nearly certain the two are equivalent, because their two graphs appear identical (https://www.desmos.com/calculator/jetdy6vngf) and I went through the process of simplifying. 
Anyhow, using $\cos(\theta) = x/r$ and $\sin(\theta) = y/r$, as well as $r^2 = x^2 + y^2$, I obtained $(x^2+y^2)^3 = y(16x^2-12x^2(x^2+y^2)+(x^2+y^2)^2)$, but still was not able to produce an equivalent rectangular equation (at least, not when I entered the equation into a graphing application. Could someone help me out? Thank you!

The Cartesian equation should read
$$(x^2+y^2)^3 = y\left[16x^4-12x^2(x^2+y^2)+(x^2+y^2)^2\right]$$
which graphs as below,

