$\newcommand\th x$As in my previous answers [1][1] and [2][2] to your questions, use the substitution 
$$t=\tan\frac\th4,\quad \sin\frac\th2=\frac{2t}{1+t^2},
\quad \cos\frac\th2=\frac{1-t^2}{1+t^2},
\quad \sin\th=\frac{4t(1-t^2)}{(1+t^2)^2},
\quad \th=4\tan^{-1}t$$
in this case to rewrite the inequality in question as 
\begin{equation*}
	f(t):= \frac{t \left(41 t^6-11 t^4-285
   t^2-225\right)}{\left(t^2+1\right)^2 \left(41 t^4-90
   t^2+225\right)}+\tan ^{-1}(t)>0	\tag{10}\label{10}
\end{equation*} 
for all 
\begin{equation*}
	t\in(0,t_*],\quad t_*:=\tan\frac\pi8=\sqrt{2}-1. 
\end{equation*}
One has 
\begin{equation*}
	f'(t)= -\frac{32 t^6 \left(41 t^4-2490
   t^2-2175\right)}{\left(t^2+1\right)^3 \left(41 t^4-90
   t^2+225\right)^2}>0
\end{equation*} 
for $t\in(0,t_*)$. 

Also, $f(0+)=0$. So, \eqref{10} immediately follows. 


  [1]: https://mathoverflow.net/a/464190/36721
  [2]: https://mathoverflow.net/a/475377/36721