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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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