Question

if sinx=3/5, cosy=-12/13 where both x and y lie in the second quadrant find value of sin(x+y)

Answer 1

Please find the attached image of solution.

Answer 2

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

Step-by-step explanation:

Given,

$sinx=\frac{3}{5}$ and $cosy=-\frac{12}{13}$ where both $x$ and $y$ lie in the second quadrant.

$cosx=\pm\sqrt{1-sin^{2} x}$

⇒$cosx=\pm\sqrt{1-(\frac{3}{5}) ^{2} }$

⇒$cosx=\pm\sqrt{\frac{25-9}{25}}$

⇒$cosx=\pm\frac{4}{5}$

In second quadrant $cos$ is negative.

So,

⇒$cosx=-\frac{4}{5}$

Similarly,

$siny=\pm\sqrt{1-cos^{2}y }$

⇒$siny=\pm\sqrt{1-(\frac{-12}{13}) ^{2} }$

⇒$siny=\pm\frac{5}{13}$

In second quadrant $sin$ is positive.

⇒$siny=\frac{5}{13}$

∴$sin(x+y)=sinx.cosy+cosx.siny$

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

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

                 $=\frac{-56}{65}$

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