Prove that Cos2A+Cos2B-cos2C=1-4sinAsinBsinC
Given that A+B+C=pi
That identity is false.
It's actually $\displaystyle \cos 2a+\cos 2b+\cos 2c=4\sin (a)\sin (b)\sin (c)+1,$ provided $\displaystyle a+b+c=\frac\pi2.$
The solution is quite simple:
$\displaystyle \begin{aligned}
\cos 2a+\cos 2b+\cos 2c&=2\sin (c)\cos (a-b)-2\sin ^{2}c+1 \\
& =2\sin (c)\big(\cos (a-b)-\cos (a+b)\big)+1 \\
& =4\sin (a)\sin (b)\sin (c)+1.
\end{aligned}$
In order to have $\displaystyle \cos a+\cos b+\cos c=4\sin\frac a2\sin\frac b2\sin\frac c2+1,$ we require $\displaystyle a+b+c=\pi.$