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Cos(A-B)+COS(B-C)+COS(C-A)=-3/2then SIN(A)+SIN(B)+SIN(C)=

Cos(A-B)+COS(B-C)+COS(C-A)=-3/2then SIN(A)+SIN(B)+SIN(C)=

Grade:12th pass

1 Answers

Arun
25758 Points
4 years ago
Dear student
 
 
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, sina + sinb + sinc = 0
 
 
Regards
Arun (askIITians forum expert)

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