Just saw this problem in some Coaching Institute material.
If cos A + cos B + cos C = sin A + sin B + sin C = 0, prove that cos (A-B) + cos (B-C) + cos (C-A) = -3/2
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4 Answers
3sir i hope dat u maynot like this one.
cosA+cosB+cosC=0 (1)
sinA+sinB+sinC=0 (2)
squaring 1 and 2 and then adding will yields us the required result.
341Yeah thats correct.
There is a way to do it without using trigonometry. Can anyone try it?
62\\e^{ia}+e^{ib}+e^{ic}=0 \\e^{-ia}+e^{-ib}+e^{-ic}=0 \\e^{ia}+e^{ib}+e^{ic}=0 \\e^{-ia}+e^{-ib}+e^{-ic}=0 3+\sum{e^{i(a-b)}}=0
Hence proved :)
62\\e^{ia}+e^{ib}=-e^{ic} \\e^{-ia}+e^{-ib}=-e^{-ic} \\2+e^{i(a-b)}+e^{i(b-a)}=1 \\e^{i(a-b)}+e^{i(b-a)}=-1
Hence we have proved that \cos(a-b)=-1/2
same with b-c and c-a
adding them we get the result....
Somehow for a strange reason this result does not go well in my stomach..
can someone point out my mistake ??
Or is it not a mistake at all!