1. In $\triangle A B C$, it is known that $\sin A=\frac{3}{5}, \cos B=\frac{5}{13}$, then the complex number $\cos C+\mathrm{i} \sin C=$
1. $\frac{16}{65}+\frac{63}{65} \mathrm{i}$ Detailed explanation: From $\cos B=\frac{5}{13}$, we get $\sin B=\frac{12}{13}>\frac{3}{5}=\sin A$, by the Law of Sines we know $B>A$, thus $A \in$ $\left(0, \frac{\pi}{2}\right)$, so $\cos A=\frac{4}{5}$; then using trigonometric formulas, we calculate $\cos C=-\cos (A+B)=\sin A \sin B-\cos A \cos B=\frac{36}{65}$ $-\frac{20}{65}=\frac{16}{65}, \sin C=\sin (A+B)=\sin A \cos B+\cos A \sin B=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$, hence $\cos C+\mathrm{i} \sin C=\frac{16}{65}+\frac{63}{65} \mathrm{i}$.