2. In $\triangle A B C$, if $2 \cos B \sin A=\sin C$, then the shape of $\triangle A B C$ must be
A. Isosceles right triangle
B. Right triangle
C. Isosceles triangle
D. Equilateral triangle
2. $C 2 \cos B \sin A=\sin C \Leftrightarrow 2 \cos B \sin A=\sin (A+B)=\sin A \cos B+\cos A \sin B \Leftrightarrow \sin A \cos B-\sin B \cos A=$ $0 \Leftrightarrow \sin (A-B)=0$, so $A=B$