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. 

$$\cos(2B)=\cos(A+B+\underbrace{A-B})=\cos(A+B)\cos(A-B)-\sin(A+B)\sin(A-B)$$
$$A-B=\arccos\frac3{15}$$
Using the principal value, $$\implies 0<A-B<\frac\pi2\implies\sin(A-B)>0$$
