## Trigonometry (11th Edition) Clone

$$\sin(A+B)=-\frac{63}{65}$$
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 $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ $$\sin(A+B)=\frac{4}{5}\Big(-\frac{12}{13}\Big)+\frac{3}{5}\Big(-\frac{5}{13}\Big)$$ $$\sin(A+B)=-\frac{48}{65}-\frac{15}{65}$$ $$\sin(A+B)=-\frac{63}{65}$$