Basic Trigonometry Question If $\cos{(A-B)}=\frac{3}{5}$ and $\sin{(A+B)}=\frac{12}{13}$, then find $\cos{(2B)}$.
Correct answer = 63/65.
 I tried all identities I know but I have no idea how to proceed. 

From
$$\cos{(A-B)}=\frac{3}{5}\Rightarrow \sin(A-B)=\sqrt{1-(3/5)^2}=\frac{4}{5}$$ 
and from 
$$\sin{(A+B)}=\frac{12}{13}\Rightarrow \cos(A+B)=\sqrt{1-(12/13)^2}=\frac{5}{13}$$
then
$$\cos(2B)=\cos((A+B)-(A-B))=$$
$$=\cos(A+B)\cos(A-B)+\sin(A+B)\sin(A-B)=$$
$$=\frac{5}{13}\frac{3}{5}+\frac{12}{13}\frac{4}{5}=\frac{63}{65}$$
