Given that sina + sinb =1 and cosa + cosb = 1/2 calculate cos(a-b)

We'll note the given relations
as:

sina + sinb = 1 (1) 

cosa
+ cosb = 1/2 (2)

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

 (sina + sinb)^2 =
1^2

We'll expand the
square:

(sin a)^2 + 2sina*sinb + (sin b)^2 = 1
(3)

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

(cosa + cosb)^2 =
(1/2)^2

We'll expand the
square:

(cos a)^2 + 2cos a*cos b + (cos b)^2 = 1
(4)

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

(sin
a)^2 + 2sina*sinb + (sin b)^2 + (cos a)^2 + 2cos a*cos b + (cos b)^2 = 1 +
1/4

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

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

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

1
+ 1 + 2(cos a*cos b + sina*sinb) = 5/4

We'll subtract 2
both sides:

2(cos a*cos b + sina*sinb) = 5/4 -
2

2(cos a*cos b + sina*sinb) =
-3/2

We'll divide by 2:

cos
a*cos b + sina*sinb = -3/4

We'll recognize the formula in
the sum from the left side, the formula of cos (a -
b):

cos (a - b) =
-3/4