Answer
$$\sin(A+B)=-\frac{63}{65}$$
Work Step by Step
To calculate $\sin(A+B)$, we again need the values of $\sin A$, $\cos A$, $\sin B$ and $\cos B$.
After the calculations in the first part, their values are as follows:
$$\cos A=\frac{3}{5}\hspace{1cm}\sin B=-\frac{5}{13}\hspace{1cm}\sin A=\frac{4}{5}\hspace{1cm}\cos B=-\frac{12}{13}$$
We can calculate $\sin(A+B)$ now.
According to the identity of sine sum, we have
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
$$\sin(A+B)=\frac{4}{5}\Big(-\frac{12}{13}\Big)+\frac{3}{5}\Big(-\frac{5}{13}\Big)$$
$$\sin(A+B)=-\frac{48}{65}-\frac{15}{65}$$
$$\sin(A+B)=-\frac{63}{65}$$
