If: sinx = 3/5, cosy = −12/13, Where:x and y both lie in the second quadrant, Find the value of sin(x + y).

sin(x + y) = sin(x) × cos(y) + cos(x) × sin(y)                  …....(1)

Now, cos²x = 1 − sin²(x) = 1 − 9/25 = 16/25

Therefore  cosx = ±4/5

Since x lies in the second quadrant, cosx is negative.

Hence, cosx = −4/5

now, sin²y = 1 − cos²y = 1 − 144/169 = 25/169

i.e., siny = ±5/13

Since y lies in the second quadrant, siny is positive. Therefore, siny = 5/13. Substituting the values of sinx, siny, cosx and cosy in (1), we get

sin(x + y) = (3/5)×(−12/13) + (−4/5)×(5/13) = −36/65 − 20/65 = = −56/65