$\newcommand\th x$As in my previous answers, 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{4 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{-5043 t^{12}+6806 t^{10}+278307 t^8+262740 t^6+11475 t^4-182250 t^2-151875}{\left(t^2+1\right)^3 \left(41 t^4-90 t^2+225\right)^2}<0 \end{equation*} for $t\in(0,t_*)$. So, in fact we have the inequality opposite to \eqref{10}: \begin{equation*} f(t)<0 \end{equation*} for $t\in(0,t_*]$.