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 +...
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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
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