Hi,
@pasmith. Did you make slight typo? You said you where going to use ##R_n = 2n \pi##, but you seem to have plugged ##n \pi## into your formula instead.
Assuming you meant to plug in ## R_n = 2n \pi##, I think I have found a bound of the form you postulated...
Starting with
$$
f_n (\theta) = \dfrac{2 n \pi e^{-(1-a) 2n \pi cos \theta}}{\sqrt{e^{-4n \pi cos \theta} + 2 e^{-2n \pi cos \theta} \cos \left( 2 n \pi \sin \theta \right) + 1}}
$$
and using
\begin{align*}
\sin (\frac{\pi}{2} + \phi) = \sin \frac{\pi}{2} \cos \phi + \cos \frac{\pi}{2} \sin \phi = \cos \phi
\nonumber \\
\cos (\frac{\pi}{2} + \phi) = \cos \frac{\pi}{2} \cos \phi - \sin \frac{\pi}{2} \sin \phi = - \sin \phi
\end{align*}
we have
$$
\frac{f_n (\frac{\pi}{2} + \phi)}{n \pi} = \dfrac{2 e^{(1-a) 2n \pi \sin \phi}}{\sqrt{e^{4n \pi \sin \phi} + 2 e^{2n \pi \sin \phi} \cos \left( 2 n \pi \cos \phi \right) + 1}} \qquad (*)
$$
First assume ##\phi \geq 0##. For ##0 \leq \phi \leq \frac{\pi}{2}## we have ##\sin \phi \leq \phi## and ##\cos \phi \geq 1 - \frac{2}{\pi} \phi##. Note for ##\phi## close enough to ##0##,
\begin{align*}
\cos \left( 2 n \pi \cos \phi \right) & \geq \cos \left( 2 n \pi \left(1 - \frac{2}{\pi} \phi \right) \right)
\nonumber \\
& = \cos \left( 2 n \pi - 4 n \phi \right)
\nonumber \\
& = \cos \left( 4 n \phi \right)
\end{align*}
This inequality obviously holds on the interval ##[0, \frac{\pi}{8n}]## (at ##\phi = \frac{\pi}{8n}##, ##\cos \left( 4 n \phi \right)## becomes zero).
Using these inequalities in ##(*)##, we have
\begin{align*}
\frac{f_n (\frac{\pi}{2} + \phi)}{n \pi} & \leq \dfrac{2 e^{(1-a) 2n \pi \phi}}{\sqrt{1 + 2 \cos \left( 2 n \pi (1 - \frac{2}{\pi} \phi) \right) + 1}}
\nonumber \\
& = \dfrac{2 e^{(1-a) 2n \pi \phi}}{\sqrt{2 + 2 \cos \left( 4n \phi \right)}}
\end{align*}
while ##0 \leq \phi \leq \phi_{max} = \frac{\pi}{8n}##, with the RHS of this inequality monotonically increasing on ##[0,\frac{\pi}{8n}]##. And so
$$
\frac{f_n (\frac{\pi}{2} + \phi)}{n \pi} \leq \sqrt{2} e^{(1-a) \pi^2 / 4}
$$
while ##0 \leq \phi \leq \phi_{max} = \frac{\pi}{8n}##.
Now assume ##\phi \leq 0##. Write
\begin{align*}
\frac{f_n (\frac{\pi}{2} + \phi)}{n \pi} & = \dfrac{2 e^{-a 2n \pi \sin \phi}}{\sqrt{e^{-4n \pi \sin \phi} + 2 e^{-2n \pi \sin \phi} \cos \left( 2 n \pi \cos \phi \right) + 1}}
\nonumber \\
& = \dfrac{2 e^{a 2n \pi |\sin \phi|}}{\sqrt{e^{4n \pi |\sin \phi|} + 2 e^{2n \pi |\sin \phi|} \cos \left( 2 n \pi \cos \phi \right) + 1}}
\end{align*}
By similar reasoning, we have
$$
\frac{f_n (\frac{\pi}{2} + \phi)}{n \pi} \leq \sqrt{2} e^{a \pi^2 / 4}
$$
while ##- \frac{\pi}{8n} = - \phi_{max} \leq \phi \leq 0##.
Put
$$
M = max \left( \sqrt{2} e^{(1-a) \pi^2 / 4} , \sqrt{2} e^{a \pi^2 / 4} \right)
$$
we have for ##\frac{1}{2} \epsilon_n \leq \phi_{max} = \frac{\pi}{8n}##,
$$
f_n (\theta) \leq \left\{
\begin{matrix}
m_n & 0 \leq \theta < \frac{1}{2} \left( \pi - \epsilon_n \right) \\
n \pi M & \frac{1}{2} \left( \pi - \epsilon_n \right) \leq \theta \leq \frac{1}{2} \left( \pi + \epsilon_n \right) \\
m_n' & \frac{1}{2} \left( \pi + \epsilon_n \right) < \theta \leq \pi
\end{matrix}
\right.
$$
We choose:
$$
\epsilon_n = \frac{\pi}{4 n^2}
$$
so that
$$
n \pi M \epsilon_n \rightarrow 0 .
$$