Why does my parametric trigonometric function appears to be a polynomial? I was fooling around with $(\cos^2(t),\sin^2(at))$ with varying values of $a$, and found that if $a=3$ then $(\cos^2(t),\sin^2(3t))$ gives the graph of $y=-16x^3+24x^2-9x+1$ on the domain $[0,1]$

The calculator won't do parametrics but it just looks like the graph from 0 to 1.
Eliminating the parameter gave me $\sin ^2\left(3\cos ^{-1}\left(\sqrt{x}\right)\right)$
Why is this true? And, how can I solve $\sin ^2\left(3\cos ^{-1}\left(\sqrt{x}\right)\right)=-16x^3+24x^2-9x+1$ to prove that the Cartesian form is equal to the polynomial?

We have $$\begin{align*}y=\sin^2 3t &= (3\sin t - 4\sin^3 t)^2 \\ &= \sin^2 t(3-4\sin^2 t)^2 \\& = (1-\cos^2 t)(3-4(1-\cos^2 t))^2 \\ & = (1-x)(4x-1)^2 \\ & =-16x^3+24x^2-9x+1\end{align*}$$
where we derived $\sin 3t = 3\sin t - 4\sin^3 t$ either by expanding $e^{3it}$ and taking imaginary parts or by expanding $\sin 3t = \sin (2t + t)$ and then expanding $\sin 2t$. 
