$A\in Q_3,\tan A=\frac{\sqrt{7}}{3} then$$18-16{\sin }^2\frac{A}{2}-32\sin \frac{A}{2}\sin \frac{5A}{2}=$

$A\in Q_3$

$From triangle \cos A=\frac{-3}{4}$
$\begin{array}{l}18-8(1-\cos A)-16\left[2\sin \frac{5A}{2}\sin \frac{A}{2}\right]\\ 18-8\left(1-\left(\frac{-3}{4}\right)\right)-16\left[\cos \left(\frac{5A}{2}-\frac{A}{2}\right)-\cos \left(\frac{5A}{2}+\frac{A}{2}\right)\right]\\ 18-8\left(\frac{4+3}{4}\right)-16[\cos 2A-\cos 3A]\\ 18-14-16\left[\frac{1}{8}-\frac{9}{16}\right]=11\end{array}$