Question

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

Answer 1

Step $1:$
Given data$:\sin x=\frac{3}{5}$ and $\cos y=-\frac{12}{13}$
$\sin (x+y)=\sin x\cos y+\cos x\sin y...\left(i\right)$

We know the value of $\sin x$ and $\cos y$
But we do not the value of $\cos x$ and $\sin y$

Step $2:$ Solve for value of $\cos x$
We know that ${\sin }^{2}x+{\cos }^{2}x=1$
$\Rightarrow {\left(\frac{3}{5}\right)}^2+{\cos }^{2}x=1$
$\Rightarrow {\cos }^{2}x=1-\frac{9}{25}=\frac{16}{25}$
$\Rightarrow \cos x=\pm \sqrt{\frac{16}{25}}\Rightarrow \cos x=\pm \frac{4}{5}$
Since $x$ is in $2^{nd}$ quadrant, $\cos x$ is negative $\Rightarrow \cos x=-\frac{4}{5}$

Step $3:$ Solve for value of $\sin y$
We know that ${\sin }^{2}y+{\cos }^{2}y=1$
$\Rightarrow {\sin }^{2}y+{(-\frac{12}{13})}^2=1$
$\Rightarrow {\sin }^{2}y=1-\frac{144}{169}=\frac{25}{169}$
$\Rightarrow \sin y=\pm \sqrt{\frac{25}{169}}\Rightarrow \sin y=\pm \frac{5}{13}$

Since $y$ is in $2^{nd}$ quadrant, $\sin y$ is positive
$\therefore \sin y=\frac{5}{13}$

Step $4:$
Solve for value of $\sin (x+y)$

Now putting value of $\sin x,\sin y,\cos x,\cos y$ in equation $\left(i\right)$
$\sin (x+y)=\frac{3}{5}\times (-\frac{12}{13})+(-\frac{4}{5})\times \left(\frac{5}{13}\right)=-\frac{56}{65}$

The value of $\sin (x+y)$ is $-\frac{56}{65}$.