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Given that cos(a-b) + cos(b-c) + cos(c-a) = -3/2. Then find the value of cos(a)+cos(b)+cos(C).

Given that cos(a-b) + cos(b-c) + cos(c-a) = -3/2. Then find the value of cos(a)+cos(b)+cos(C).

Grade:11

1 Answers

Himanshu
103 Points
6 years ago
First of all, multiply 2 and then add 3 on both sides.
Resolve 3 as 1+1+1 on L.H.S and open the identities of cos(x-y).
Replace every 1 as sin2a + cos2a , sin2b + cos2b and sin2c + cos2c respectively.
You will get the two terms (cosa + cosb + cosc)2 + (sina +sinb + sinc)2 = 0
Hence, cosa + cosb + cosc = 0

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