MotylaNogaTomkaMazura has found a very good parametrization $x=\cos(2t)$. (but it seems that his/her answer has an error) This answer uses this parametrization.
If you parametrize $x=\cos(2t)$ with $t \in [0,\pi/4]$, the original function can be written as
$$ g_k(t)=\frac{\cos ^k(t)-\sin ^k(t)}{\cos (t)-\sin (t)},$$
with $k=2n+1$ and omitting a factor $2^{1 + n}$.
True. Let $c:=\cos(t),s:=\sin(t)$. For $k=2n+1\ge 7$,
$$\begin{align}\frac{d}{dt}\left(\frac{c^k-s^k}{c-s}\right)&=\frac{(kc^{k-1}(-s)-ks^{k-1}c)(c-s)-(c^k-s^k)(-s-c)}{(c-s)^2}\\&=\frac{-k(c^{k-1}s+s^{k-1}c)(c-s)+(c^k-s^k)(s+c)}{(c-s)^2}\end{align}$$
Here, let
$$F(t):=-k(c^{k-1}s+s^{k-1}c)(c-s)+(c^k-s^k)(s+c)$$
Then,
$$F'(t)=-k((k-1)c^{k-2}(-s)s+c^{k-1}c+(k-1)s^{k-2}c^2+s^{k-1}(-s))(c-s)$$$$-k(c^{k-1}s+s^{k-1}c)(-s-c)+(kc^{k-1}(-s)-ks^{k-1}c)(s+c)+(c^k-s^k)(c-s)$$
$$\begin{align}&=-k(-(k-1)c^{k-2}s^2+c^{k}+(k-1)s^{k-2}c^2-s^{k})(c-s)+(c^k-s^k)(c-s)\\&=(c-s)(k(k-1)c^{k-2}s^2-k(k-1)s^{k-2}c^2+c^k-kc^{k}-s^k+ks^{k})\\&=(k-1)(c-s)(kc^{k-2}s^2-ks^{k-2}c^2+s^k-c^k)\\&=c^k(k-1)(c-s)(k\tan^2(t)-k\tan^{k-2}(t)+\tan^k(t)-1)\end{align}$$
Here, let
$$G(u):=ku^2-ku^{k-2}+u^k-1$$
Then,
$$G'(u)=2ku-k(k-2)u^{k-3}+ku^{k-1}$$
$$G''(u)=2k-k(k-2)(k-3)u^{k-4}+k(k-1)u^{k-2}$$
$$\begin{align}G'''(u)&=-k(k-2)(k-3)(k-4)u^{k-5}+k(k-1)(k-2)u^{k-3}\\&=k(k-2)u^{k-5}((k-1)u^2-(k-3)(k-4))\end{align}$$
This is negative, so $G''(u)$ is decreasing with $G''(0)=2k\gt 0$ and $G''(1)=-k(k-1)(k-5)\lt 0$.
So, there exists only one real $0\lt\alpha\lt 1$ such that $G''(\alpha)=0$, and $G'(u)$ is increasing for $0\lt u\lt\alpha$ and is decreasing for $\alpha\lt u\lt 1$ with $G'(0)=0$ and $G'(1)=-k(k-5)\lt 0$.
So, there exists only one real $0\lt \beta\lt 1$ such that $G'(\beta)=0$, and $G(u)$ is increasing for $0\lt u\lt\beta$ and is decreasing for $\beta\lt u\lt 1$ with $G(0)=-1\lt 0$ and $G(1)=0$.
So, there exists only one real $0\lt \gamma\lt 1$ such that $G(\gamma)=0$.
It follows from this that there exists only one real $0\lt\arctan\gamma\lt\pi/4$ such that $F'(\arctan\gamma)=0$. So, $F(t)$ is decreasing for $0\lt t\lt\arctan\gamma$ and is increasing for $\arctan\gamma\lt t\lt \pi/4$ with $F(0)=1\gt 0$ and $F(\pi/4)=0$.
It follows from this that $f_n(x)$ where $x\in ]0,1[$ has only one critical point for every value of $n \in \{3,4,5,\dots\}$. $\blacksquare$
Proof:
We need to prove that $f(x) \ge f(0) = 2\csc^2\frac{\pi}{2n}$ for all $0 \le x < \frac{\pi}{2}$.
If $n = 3$, we have
$$f(x) = - 16\left(\cos^2 \frac{x}{3} - \frac34\right)^2 + 9 \ge 8 = 2\csc^2 \frac{\pi}{6}$$
where we have used $\frac{3}{4} \le \cos^2 \frac{x}{3} \le 1$.
In the following, assume that $n \ge 4$.
We have
$$f(x) = \cos^2 x \cdot
\frac{2\cos\frac{\pi}{n}\sin^2 \frac{x}{n} + 2\sin^2 \frac{\pi}{2n}}{(\sin^2 \frac{\pi}{2n} - \sin^2 \frac{x}{n})^2}.$$
Fact 1: $\sin \frac{x}{n} \ge \frac{2}{\pi} x \sin \frac{\pi}{2n}$.
(Hint: Take derivative.)
Fact 2: $\cos^2 x \cdot
\frac{4x^2\pi^2\cos \frac{\pi}{4} + \pi^4}{\left(\pi^2 - 4x^2 \right)^2} \ge 1$.
(The proof is given at the end.)
By using Fact 1, we have
\begin{align*}
f(x) &\ge \cos^2 x \cdot
\frac{2\cos\frac{\pi}{n}\cdot (\frac{2}{\pi} x \sin \frac{\pi}{2n})^2 + 2\sin^2 \frac{\pi}{2n}}{\left[\sin^2 \frac{\pi}{2n} - (\frac{2}{\pi} x \sin \frac{\pi}{2n} )^2\right]^2}\\[5pt]
&= 2\csc^2\frac{\pi}{2n} \cdot \cos^2 x \cdot
\frac{\frac{4x^2}{\pi^2}\cos\frac{\pi}{n} + 1}{\left(1 - \frac{4x^2}{\pi^2} \right)^2}\\[5pt]
&\ge 2\csc^2\frac{\pi}{2n} \cdot \cos^2 x \cdot
\frac{\frac{4x^2}{\pi^2}\cos \frac{\pi}{4} + 1}{\left(1 - \frac{4x^2}{\pi^2} \right)^2}\\[5pt]
&\ge 2\csc^2\frac{\pi}{2n} \tag{1}
\end{align*}
where we have used Fact 2 to obtain (1).
We are done.
Proof of Fact 2:
Fact 3: $\cos x \ge \frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1 \ge 0$ for all $0 \le x < \pi/2$.
(The proof is given at the end.)
By using Fact 3, it suffices to prove that
$$\left(\frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1\right)^2\cdot \frac{4x^2\pi^2\cos \frac{\pi}{4} + \pi^4}{\left(\pi^2 - 4x^2 \right)^2} \ge 1$$
or
$$\frac{x^2}{4\pi^6}(Ax^4 + Bx^2 + C) \ge 0$$
where
\begin{align*}
A &= 2\,\sqrt{2}\, {\pi }^{4}-32\,\sqrt {2}\,{\pi }^{2}+128\,\sqrt {2}, \\
B &= {\pi }^{6}-8\,\sqrt{2}\,{\pi }^{4}-16\,{\pi }^{4}+64\,\sqrt{2}\,{\pi }^{
2}+64\,{\pi }^{2}
,\\
C &= -4\,{\pi }^{6}+8\,\sqrt {2}\,{\pi }^{4}+32\,{\pi }^{4}.
\end{align*}
It is easy to prove that $Ax^4 + Bx^2 + C \ge 0$ for all $0 \le x < \pi/2$.
We are done.
Proof of Fact 3:
We split into two cases:
- $0 \le x \le \pi/3$:
It is easy to prove that
$$\cos x \ge 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 - \frac{1}{720}x^6.$$
It suffices to prove that
$$1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 - \frac{1}{720}x^6 \ge \frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1$$
which is true (easy).
- $\frac{\pi}{3} < x < \pi/2$:
It is easy to prove that
$$\cos x \ge - \left(x - \frac{\pi}{2}\right) + \frac{1}{6}\left(x - \frac{\pi}{2}\right)^3.$$
It suffices to prove that
$$- \left(x - \frac{\pi}{2}\right) + \frac{1}{6}\left(x - \frac{\pi}{2}\right)^3
\ge \frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1$$
which is true (easy).
We are done.
Best Answer
You have a more than decent approximation of the function if you use the $[2n,2n]$ Padé approximant $P_n$ of it.
For example $$P_1=\frac 16\,\,\frac {\left(4-n^2\right)+\frac{\left(3 n^6-98 n^4-98 n^2+1768\right) }{105 n^2 \left(n^2+14\right)}x^2 } {1-\frac{2 \left(n^4+11 n^2+93\right) }{21 n^2 \left(n^2+14\right)}x^2 }$$
To give an idea, consider the infinite norm $$\Phi_n=\int_0^{\frac \pi 2} \Big(f(x)-P_1\Big)^2\,dx$$
A few values $$\left( \begin{array}{cc} n & \Phi_n \\ 3 & 5.6139\times 10^{-6} \\ 4 & 3.3417\times 10^{-7} \\ 5 & 6.6111\times 10^{-7} \\ 6 & 3.8840\times 10^{-6} \\ 7 & 9.1023\times 10^{-6} \\ 8 & 1.6175\times 10^{-5} \\ \end{array} \right)$$
This should help since the numerator of the derivative of $P_1$ with respect to $x$ is $$-(n-2) n^2 (n+2) \left(n^2+14\right)^3 x$$