# What is sin(x+y) if sinx+cosy=1/4 and cos x+siny=1/2 ?

sin (x + y) = sin x* cos y + cos x * sin y

It is given that sin x + cos y = 1/4 and cos x + sin y = 1/2

sin x + cos y = 1/4

square the two sides

=> (sin x)^2 + (cos...

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sin (x + y) = sin x* cos y + cos x * sin y

It is given that sin x + cos y = 1/4 and cos x + sin y = 1/2

sin x + cos y = 1/4

square the two sides

=> (sin x)^2 + (cos y)^2 + 2*sin x*cos y = 1/16 ...(1)

cos x + sin y = 1/2

square both the sides

=> (sin y)^2 + (cos x)^2 + 2*sin y*cos x = 1/4 ...(1)

(1) + (2)

=> (sin x)^2 + (cos y)^2 + 2*sin x*cos y + (sin y)^2 + (cos x)^2 + 2*sin y*cos x = 1/16 + 1/4

use the property (sin a)^2 + (cos x)^2 = 1

=> 2 + 2*sin x*cos y + 2*sin y*cos x = 5/16

=> 2*sin x*cos y + 2*sin y*cos x = 5/16 - 2

=> sin x*cos y + sin y*cos x = 5/32 - 1

=> sin x*cos y + sin y*cos x = -27/32

The value of sin (x + y) = -27/32

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