Basic Trigonometry Question

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.

Answer 1

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 $$\cos(2B)=\cos((A+B)-(A-B))=$$ $$=\cos(A+B)\cos(A-B)+\sin(A+B)\sin(A-B)=$$ $$=\frac{5}{13}\frac{3}{5}+\frac{12}{13}\frac{4}{5}=\frac{63}{65}$$

Answer 2

$$\cos(2B)=\cos(A+B+\underbrace{A-B})=\cos(A+B)\cos(A-B)-\sin(A+B)\sin(A-B)$$

$$A-B=\arccos\frac3{15}$$

Using the principal value, $$\implies 0<A-B<\frac\pi2\implies\sin(A-B)>0$$