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

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### 2 Answers

We are given that sin a + sin b =1 and cos a + cos b = 1/2. We have to determine cos(a - b).

cos(a - b) = (cos a)(cos b) + (sin a)(sin b)

sin a + sin b =1

square both the sides

=> (sin a)^2 + (sin b)^2 + 2(sin a)(sin b) = 1 ...(1)

cos a + cos b =1/2

square both the sides

=> (cos a)^2 + (cos b)^2 + 2(cos a)(cos b) = 1/4 ...(2)

Add (1) and (2)

=> (sin a)^2 + (sin b)^2 + 2(sin a)(sin b) + (cos a)^2 + (cos b)^2 + 2(cos a)(cos b) = 1 + 1/4

=> 1 + 1 + 2(sin a)(sin b) + 2(cos a)(cos b) = 1 + 1/4

=> 2(sin a)(sin b) + 2(cos a)(cos b) = -3/4

=> (sin a)(sin b) + (cos a)(cos b) = -3/8

=> cos (a - b) = -3/8

**The value of cos(a - b) = -3/8**

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

**cos (a - b) = -3/4**