Question

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

Answer 1

We know that 
sin (x + y) = sin x cos y + cos x sin y ... (1) 
Now cos²x = 1 – sin²x = 1 – 9/25 = 16/25 
Therefore cos x = ± 4/5. 
Since x lies in second quadrant, cos x is negative. 
Hence cos x = −4/5 
Now sin²y = 1 – cos²y = 1 – 144/169 = 25/169 
i.e. sin y = ± 5/13. 
Since y lies in second quadrant, hence sin y is positive. 

Therefore, sin y =5/13. 

Substituting the values of sin x, sin y, cos x and cos y in (1), we get 
sin(x + y) 3/5 × (-12/13) + (−4/5) × 5/13 = (-36/65) –(20/65) = -56/65

Therefore, sin(x+y) = -56/65

Answer 2

We know that  sin(x)² + cos(x)² = 1 

(3/5)² + cos(x)² = 1 
9/25 + cos(x)² = 25/25 
cos(x)² = 16/25 
cos(x) = +/- (4/5) 
Since it is in second quadrant cos(x) = -4/5 
sin(y)² + (-12/13)² = 1 
sin(y)² + 144/169 = 169/169 
sin(y)² = 25/169 
sin(y) = +/- 5/13 
Since it is in second quadrant sin(y) = 5/13 
sin(x) = 3/5 
cos(x) = -4/5 
sin(y) = 5/13 
cos(y) = -12/13 
sin(x + y) => sin(x)cos(y) + sin(y)cos(x) 

=> (3/5) * (-12/13) + (5/13) * (-4/5) 

=> -36 / 65 + -20 / 65

=> -56/65

So, sin(x+y) = -56/65