In $\Delta ABC$ if $\sin A.\sin (B-C)=\sin C.\sin (A-B)$ then (where $A\ne B\ne C$)

$\begin{array}{l}\sin A\sin (B-C)=\sin C.\sin (A-B)\\ \Rightarrow \sin (B+C)\sin (B-C)=\sin (A+B)\sin (A-B)\\ \Rightarrow 2{\sin }^{2}B={\sin }^{2}A+{\sin }^{2}C\\ \Rightarrow 1-\cos 2B=\frac{1}{2}(1-\cos 2A)+\frac{1}{2}(1-\cos 2C)\\ \Rightarrow 2\cos 2B=\cos 2A+\cos 2C\\ \therefore \cos 2A,\cos 2B,\cos 2CareinA.P\end{array}$