Grade 12th passTrigonometry
Cos(A-B)+COS(B-C)+COS(C-A)=-3/2then SIN(A)+SIN(B)+SIN(C)=
1 Answer
Arun
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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