If A + B + C = 3 π 2 , then cos 2 A + cos 2 B + cos 2 C =

$\cos 2A+\cos 2B+\cos 2C$

$=2\cos \left(A+B\right)\cos \left(A-B\right)+\cos 2C$ $\left[\because \cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\right]$

$=2\cos \left(\frac{3\pi }{2}-C\right)\cos \left(A-B\right)+\left(1-2{\sin }^{2}C\right)$ $\left[\because A+B+C=\frac{3\pi }{2} \& \cos 2\theta =1-2{\sin }^2\theta \right]$

$=1-2\sin C\cos \left(A-B\right)-2{\sin }^{2}C$

$=1-2\sin C\left[\cos \left(A-B\right)+\sin \left\{\frac{3\pi }{2}-\left(A+B\right)\right\}\right]$ $\left[\because A+B+C=\frac{3\pi }{2}\right]$

$=1-2\sin C\left[\cos \left(A-B\right)-\cos \left(A+B\right)\right]$

$=1-4\sin A\sin B\sin C$ $\left[\because \cos C-\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{D-C}{2}\right)\right]$