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

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Himanshu · Answer

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 sin 2 a + cos 2 a , sin 2 b + cos 2 b and sin 2 c + cos 2 c 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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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).

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