If $\mathrm{A}+\mathrm{B}+\mathrm{C}=3\mathrm{\pi }/2$, then cos 2A +cos 2B +cos 2C is equal to

$\begin{array}{c}\cos 2\mathrm{A}+\cos 2\mathrm{B}+\cos 2\mathrm{C}=2\cos (\mathrm{A}+\mathrm{B})\cos (\mathrm{A}-\mathrm{B})+\cos 2\mathrm{C}\\ =2\cos \left(\frac{3\mathrm{\pi }}{2}-\mathrm{C}\right)\cos (\mathrm{A}-\mathrm{B})+\cos 2\mathrm{C}\\ =-2\sin \mathrm{Ccos}(\mathrm{A}-\mathrm{B})+1-2{\sin }^2\mathrm{C}\\ =1-2\sin \mathrm{C}(\cos (\mathrm{A}-\mathrm{B})+\sin \mathrm{C})\\ =1-2\sin \mathrm{C}\{\cos (\mathrm{A}-\mathrm{B})+\sin [3\mathrm{\pi }/2-(\mathrm{A}+\mathrm{B})\left]\right\}\\ =1-2\sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B}\left)\right]\\ =1-2\sin C[2\sin A \sin B]\\ =1-4\sin A \sin B \sin C\end{array}$