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

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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

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