If $\sin x=\frac{3}{5}, \cos y=-\frac{12}{13},$ where $x$ and $y$ both lie in second quadrant, find the value of $\sin (x+y)$.

We know that

$\sin (x+y)=\sin x \cos y+\cos x \sin y$.......$(1)$

Now $\cos ^{2} x=1-\sin ^{2} x=1-\frac{9}{25}=\frac{16}{25}$

Therefore $\cos x=\pm \frac{4}{5}$

since $x$ lies in second quadrant, cos $x$ is negative.

Hence $\cos x=-\frac{4}{5}$

Now $\sin ^{2} y=1-\cos ^{2} y=1-\frac{144}{169}=\frac{25}{169}$

i.e. $\sin y=\pm \frac{5}{13}$

since $y$ lies in second quadrant, hence sin $y$ is positive. Therefore, $\sin y=\frac{5}{13} .$ Substituting the values of $\sin x, \sin y, \cos x$ and $\cos y$ in $(1),$ we get

$\sin (x+y)=\frac{3}{5} \times\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right) \times \frac{5}{13}$

$\frac{36}{65}-\frac{20}{65}=-\frac{56}{65}$