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

We'll note the given relations
as:

sin x + cos y = 1/4
(1)

cos x + sin y = 1/2
(2)

We'll raise to square (1), both
sides:

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

We'll expand the
square:

(sin x)^2 + 2sinx*cosy + (cos y)^2 = 1/16
(3)

We'll raise to square (2), both
sides:

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

We'll expand the
square:

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

We'll add (3) + (4):

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

But, from the fundamental formula of trigonometry,
we'll get:

(sin ax)^2 + (cos x)^2 =
1

(sin y)^2 + (cos y)^2 = 1

1
+ 1 + 2(sinx*cosy + cos x*sin y) = 5/16

We'll subtract 2
both sides:

2(sinx*cosy + cos x*sin y) = 5/16 -
2

2(sinx*cosy + cos x*sin y) =
-27/16

We'll divide by
2:

sinx*cosy + cos x*sin y =
-27/32

But the sum from the left side represents the
expanding of sin(x+y):

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

The value of the sine of the sum of angles
x and y is: sin (x+y) =
-27/32