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

$sinx=\frac{3}{5}$

$cosx=\sqrt{1-si{n}^{2}x}$

$cosx=\sqrt{1-\frac{9}{25}}$

$cosx=\sqrt{\frac{16}{25}}=\pm \frac{4}{5}$

$x$ lie in second quadrant, so $cosx=-\frac{4}{5}$

$cosy=-\frac{12}{13}$

$siny=\sqrt{1-co{s}^{2}y}$

$siny=\sqrt{1-\frac{144}{169}}$

$siny=\sqrt{\frac{25}{169}}=\pm \frac{5}{13}$

$y$ lie in second quadrant, so $siny=\frac{5}{13}$

Now,

$\sin (x+y)=\sin x\cos y+\cos x\sin y$ ......... $\sin (A+B)=\sin A\cos B+\sin B\cos A$

$=\frac{3}{5}(-\frac{12}{13})+(-\frac{4}{5})\frac{5}{13}$

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