Question

If sin x = 3/5 , cos y = − 12/13 , where x and y both lie in second quadrant, find the value of sin (x + y).

Answer 1

Given,

sin x = \(\frac 35\) and  cos y = \(\frac {12}{13}\)

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

 Now we have to find siny and cosx

 We know that

sin²x + cos²x = 1

cos²x = 1 − \(\frac 9{25}\)

cosx = \(\frac 45\)

Since x lie in the 2^nd​ quadrant so cosx is negative

Similarly,

 siny = ?

 We know that

sin²y + cos²y = 1

sin²y = 1 = −cos²y

= 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

\(\sin(x + y) = \frac 35 \times \frac{-12}{13} + \frac{-4}5 \times \frac 5{13}\)

\(= \frac{-36}{65} - \frac{20}{65}\)

\(= \frac{-56}{65}\)

Answer 2

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