Question

Let $A$ and $B$ be acute angles. If $\sin (A+B)=\frac{12}{13}$ and $\cos (A-B)=\frac{3}{5}$, then $\sin \left(2A\right)$ is equal to

• A. $\frac{49}{65}$
• B. $\frac{56}{65}$
• C. $\frac{12}{25}$
• D. $\frac{13}{25}$

Answer 1

The correct option is B $\frac{56}{65}$

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