A curious family of Chebyshev-like polynomials Consider the family $f_n(x)$ of functions of $x$ for $0\leq x\leq1$, each indexed by a variable $n \in \mathbb{N}$, described by the following equation:
$$f_n(x) = \sin^2\left(n \arcsin\left(\sqrt{x}\right)\right)$$
Evaluation in numerical algebra programs yields the following polynomial forms of each function $f_n(x)$ for the first few $n$:

*

*$f_1(x) = x$

*$f_2(x) = 4x - 4x^2$

*$f_3(x) = 16 x^3 - 24 x^2 + 9 x$

*$f_4(x) = -64 x^4 + 128 x^3 - 80 x^2 + 16 x$
and so on.
Is there any way to analytically derive the polynomial form of this family of functions in terms of $n$ and $x$ (i.e., $f_n(x) = g(n,x)$ where $g$ is a polynomial in $x$)? Or perhaps this family of polynomials is related to another "named" family of polynomials under an appropriate transformation?
COMMENT: The "Chebyshev-like" qualifier in the title comes from the fact that the Chebyshev polynomials $T_n(x)$ can be defined by a similar trigonometric identity: $T_n(x) = \cos(n\arccos(x))$.

Chebyshev polynomials of the second kind satisfy $\,U_{n-1}(\cos \theta)\,\sin \theta =\sin(n\theta)\,$.
With $\,\theta=\arcsin\left(\sqrt{x}\right)\,$ it follows that:
$$
\sqrt{x} \; U_{n-1}\left(\sqrt{1-x}\right) = \sin\left(n \arcsin(\sqrt{x})\right) \;\;\iff\;\; f_n(x) = x \cdot U_{n-1}^2\left(\sqrt{1-x}\right)
$$
Since $\,U_n(x)\,$ contains only powers of the same parity, the square root in $\,\sqrt{1-x}\,$ will either vanish or factor out then get squared again, so the end result is in fact a polynomial.
For example:

*

*$n=3:$
$$
\begin{align}
U_2(x) &= 4x^2-1
\\ f_3(x) &= x \cdot U_2^2\left(\sqrt{1-x}\right)
\\ &= x \cdot \left(4\,(1-x)-1\right)^2
\\ &= x\,\left(-4x+3\right)^2
\\ &= 16 x^3 - 24 x^2 + 9 x
\end{align}
$$

*

*$n=4:$
$$
\begin{align}
U_3(x) &= 4x\,(2x^2-1)
\\ f_4(x) &= x \cdot U_3^2\left(\sqrt{1-x}\right)
\\ &= x \cdot 16(1-x) \, \left(2\,(1-x)-1\right)^2
\\ &= 16x(1-x)\,\left(-2x+1\right)^2
\\ &= -64 x^4 + 128 x^3 - 80 x^2 + 16 x
\end{align}
$$
