If $\mathrm{A}$ is in the third quadrant and $\tan \mathrm{A}=\frac{\sqrt{7}}{3},$ then $18-16{\sin }^2\frac{\mathrm{A}}{2}-32\sin \frac{\mathrm{A}}{2}\sin \frac{5\mathrm{A}}{2}=$

$$\begin{gathered} \tan \mathrm{A}=\frac{\sqrt{7}}{3}, A\in Q_3 \\ \cos A=-\frac{3}{4} \\ 18-16{\sin }^2\frac{\mathrm{A}}{2}-32\sin \frac{\mathrm{A}}{2}\sin \frac{5\mathrm{A}}{2} \\ =18-8(1-\cos A)-16(\cos 2A-\cos 3A) \\ =10+8\cos A-16(2{\cos }^{2}A-1)+16(4{\cos }^{3}A-3\cos A) \\ =10-6-18+16-27+36 \\ =11 \end{gathered}$$