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?
 A: 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 $\sin 2t$. 
A: In fact, this property is the tip of an iceberg. Here is a generalization.
There exists a family of polynomials $U_n(x)$ (called Chebyshev polynomials of the second kind (http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html)) allowing to express, for any $n>0$,
$$\sin((n+1)\theta)=\sin\theta \ \times U_n(\cos \theta)$$
Thus, by squaring both sides, for any $n$,
$$\tag{1}\sin^2((n+1)\theta)=(1-\cos^2\theta) \times [U_n(\cos\theta)]^2$$
which is a polynomial is $\cos \theta$.
Here are the first $U_n(x)$:
$$\begin{array}{l}U_1(x)=2x\\U_2(x)=-1 + 4\,x^2\\U_3(x)=-4\,x + 8\,x^3\\U_4(x)=1 - 12\,x^2 + 16\,x^4,\\
  U_5(x)=6x - 32\,x^3 + 32\,x^5,\\U_6(x)=-1 + 24\,x^2 - 80\,x^4 + 64\,x^6 \end{array}$$
etc.
Let us see what formula (1) gives with $n=2$:
$$\sin^2(3\theta)=(1-\cos^2\theta) \times (4\,\cos^2\theta - 1)^2$$
$$=1 - 9\,\cos^2\theta + 24\,\cos^4\theta - 16\,\cos^6\theta$$
as desired.
The $U_n(x)$ constitute an important family of orthogonal polynomials, with many properties, but that's another story...
A: $$\sin^2(3t)=1-\cos^2(3t)=1-(4\cos^3 t-3\cos t)^2=-16(\cos^2 t)^3+24 (\cos^2t)^2-9\cos^2t+1$$
