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

Given

$sinx=3/5$and $cosy=12/13$

since $sin(x+y)=sinx.cosy+cosx.siny.....1$

Now we have to find $siny$ and $cosx$

We know that

$si{n}^{2}x+co{s}^{2}x=1$

$co{s}^{2}x=1-\frac{9}{25}$

$cosx=\frac{4}{5}$

since xis in the $2_{nd}$ quadrant so $cosx$ is negative

Similarly,

$siny=?$

We know that

$si{n}^{2}y+co{s}^{2}y=1$

$si{n}^{2}y=1=-co{s}^{2}y$

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

$cosy=\frac{5}{13}$

Since $y$lie in the $2_{nd}$ quadrant so $siny$ is positive

Put this value in equation 1 then we get

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

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

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