If sinx + siny = 1/4 and cosx + cosy = 1/3, then prove that cos[(x-y)/2]= +/- 5/24

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

cos(x) + cos(y) = 1/3
cos(x)^2 + 2cos(x)cos(y) + cos(y)^2 = 1/9

sin(x)^2 + cos(x)^2 + 2sin(x)sin(y) + 2cos(x)cos(y) + sin(y)^2 + cos(y)^2 = 1/16 + 1/9
1 + 1 + 2 * (sin(x)sin(y) + cos(x)cos(y)) = (9 + 16) / 144
2 + 2 * cos(x - y) = 25 / 144
2 * cos(x - y) = 25/144 - 288/144
cos(x - y) = -263/288

cos((x - y) / 2) =>
sqrt((1/2) * (1 + cos(x - y))) =>
sqrt((1/2) * (1 - 263/288)) =>
sqrt((1/2) * 25/288) =>
sqrt(25 / 576) =>
+/- 5/24

sinx + siny = 1/4
cosx + cosy = 1/3

sin (x+y/2)cos(x-y/2) = 1/8 (1)
cos(x+y/2)cos(x-y/2)=1/6 (2)

cos(x-y/2) = 1/6cos(x+y/2) = 1/8sin(x+y/2)

==> 8sin(x+y/2) = 6cos(x+y/2) (3)

Let a = sin(x+y)/2

(3)==> 8a = 6√(1-a^2)
==> 16a^2 = 9-9a^2 ==> 25a^2 = 9 ==> a = 3/5 or -3/5

(1)==> (+/-3/5) cos(x-y/2) = 1/8
==> cos(x-y/2) = (1/8)*(+/- 5/3) = +/- 5/24