Question

If $\sin A=\frac{12}{13},\cos B=-\frac{3}{5},0<A<\frac{\pi }{2},$ $\pi <B<\frac{3\pi }{2}$, then the value of $\sin (A+B)$ is

• A. $\frac{33}{65}$
• B. $-\frac{1}{63}$
• C. $-\frac{56}{65}$
• D. $-\frac{63}{65}$

Answer 1

The correct option is C $-\frac{56}{65}$

$\sin A=\frac{12}{13},0<A<\frac{\pi }{2}\Rightarrow \cos A=\frac{5}{13}\cos B=-\frac{3}{5},\pi <B<\frac{3\pi }{2}\Rightarrow \sin B=-\frac{4}{5}$
Now,
$\sin (A+B)=\sin A\cos B+\cos A\sin B\Rightarrow \sin (A+B)=-\frac{12}{13}\times \frac{3}{5}-\frac{5}{13}\times \frac{4}{5}\therefore \sin (A+B)=-\frac{56}{65}$