Limit superior of a sequence of oscillating functions related to Chebyshev polynomials Let $n \in \mathbb N$ and consider the polynomial function $f_n \colon \mathbb R \to \mathbb R$ defined by
$$f_n(x) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} (1 - x^2)^{n-k} x^{2k}$$
for any $x \in \mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)
By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of  $f_3, f_5, f_7$:

As $n \to \infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_{50}$:

Let $f \colon D \to \mathbb R$ be defined by
$$f(x) = \limsup_{n \to \infty} f_n(x)$$
whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.
So far, I have only been able to show the following (see the update below):


*

*$f$ is an even function, since all of the $f_n$'s are even.

*$0 \notin D$. Indeed, $f_n(0) = 2n + 1 \to \infty$ as $n \to \infty$.

*$f(\pm 1) = 1$, because $f_n(\pm 1) = (-1)^n$ for any $n \in \mathbb N$.

*$f \left (\pm \frac {\sqrt 2} 2 \right ) = 1$. This is because:
$$f_n \left ( \pm \frac {\sqrt 2} 2 \right ) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} \left ( \frac 1 2 \right )^n = (-1)^{\left \lfloor \frac n 2 \right \rfloor} 2^n \left ( \frac 1 2 \right )^n = (-1)^{\left \lfloor \frac n 2 \right \rfloor} \le 1$$
In particular, $f_{4m} \left (\pm \frac {\sqrt 2} 2 \right ) = 1$ for any $m \in \mathbb N$, so $\limsup_{n \to \infty} f_n \left ( \pm \frac {\sqrt 2} 2 \right ) = 1$.
By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.

What else can we say about $f$? Is it possible to find a "simple" expression?

Thank you in advance for any reply.

Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that
$$f_n(\sin \alpha) = \frac {\sin ((2n+1) \alpha)}{\sin \alpha}$$
for any $\alpha \in \mathbb R \smallsetminus \pi \mathbb Z$, which immediately provides us with many other values of $f$. For instance,
$$f_n \left (\sin \frac \pi 6 \right ) = \frac{\sin \left ( (2n+1) \frac \pi 6 \right )}{\sin \frac \pi 6} \le \frac 1 {\frac 1 2} = 2$$
In particular,
$$f_{6m+1} \left (\sin \frac \pi 6 \right ) = \frac{\sin \left ( (12 m + 3) \frac \pi 6 \right )}{\sin \frac \pi 6} = \frac{\sin \left ( 2 m \pi + \frac \pi 2 \right )}{\sin \frac \pi 6} = \frac 1 {\frac 1 2} = 2$$
for any $m \in \mathbb N$, and thus $f \left (\pm \frac 1 2 \right ) = 2$.

How can we get a simple expression for $f$ using this information?

 A: This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 \sqrt{x^2 \left(x^2-1\right)} \qquad \text{and}\qquad v=1-2 x^2+2 \sqrt{x^2 \left(x^2-1\right)}$$ a CAS produced
$$f_n(x)=\frac{ \left(u^n+v^n\right)}{2 }+\frac{ \left(u^n-v^n\right)}{2 }\,\frac{\sqrt{x^2 \left(x^2-1\right)}
   }{ x^2}$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_n\left(\sin \left(\frac{\pi  k}{12}\right)\right)$ and obtained the (may be) interesting values
$$\left(
\begin{array}{cc}
 k & f_n\left(\sin \left(\frac{\pi  k}{12}\right)\right) \\
 0 & 2 n+1 \\
 1 & \cos \left(\frac{n \pi }{6}\right)+\left(2+\sqrt{3}\right) \sin \left(\frac{n
   \pi }{6}\right) \\
 2 & \cos \left(\frac{n \pi }{3}\right)+\sqrt{3} \sin \left(\frac{n \pi }{3}\right)
   \\
 3 & \cos \left(\frac{n \pi }{2}\right)+\sin \left(\frac{n \pi }{2}\right) \\
 4 & \cos \left(\frac{2 n \pi }{3}\right)+\frac{1}{\sqrt{3}}\sin \left(\frac{2 n \pi
   }{3}\right) \\
 5 & \cos \left(\frac{5 n \pi }{6}\right)+\left(2-\sqrt{3}\right) \sin
   \left(\frac{5 n \pi }{6}\right) \\
 6 & (-1)^n 
\end{array}
\right)$$
A: Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.
Otherwise, for $\alpha \in (0,\pi/2]$ where $\sin(\alpha)>0$ $$f(\sin(\alpha)) = \frac{\sup_n \sin((2n+1)\alpha)}{\sin(\alpha)}.$$
If $\alpha$ is an irrational multiple of $2\pi$ then $(2n+1)\alpha \pmod{2\pi}$ is dense in $[0, 2\pi]$ and therefore $\sup_n \sin((2n+1)\alpha)=1$. So in this case $$f(\sin(\alpha))=\sin^{-1}(\alpha).$$ If $\alpha = 2\pi \frac{p}{q}$ for some coprime integers $p, q$ then a case by case examination on $q \pmod 8$ shows
$$s(q) = \sup_n \sin((2n+1)\alpha) = \begin{cases}
\cos(\frac{\pi}{2q}) & \textrm{If $q$ is odd}\\
\cos(\frac{\pi}q) & \textrm{If $q \equiv 2, 6 \pmod 8$}\\
\cos(\frac{2\pi}q) & \textrm{If $q \equiv 0 \pmod 8$}\\
1 & \textrm{If $q \equiv 4 \pmod 8$}
\end{cases}$$
so in this case $$f(\sin(\alpha))=\frac{s(q)}{\sin(\alpha)}.$$ Since you already noted that $f$ is even this completely describes $f$. 
