$\newcommand\th x$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):= 4 \left(\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)\right)>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{128 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 $t\in(0,t_*)$. So, \eqref{10} immediately follows.