Complex numbers help !

Question

anonymous · Answer

Please help me in this question..

anonymous · Answer

@AkashdeepDeb

akashdeepdeb · Answer

Can you tell me the role of the Summation signs then maybe I can help you out with this!

anonymous · Answer

$$\sum_{}^{} \cos2A = \cos2A + \cos2B + \cos2C$$
$$\sum_{}^{}\cos(A+B) = \cos(A+B) + \cos(B+C) + \cos(A+C)$$

akashdeepdeb · Answer

Oh okay then just hold on for a sec! :)

akashdeepdeb · Answer

When you add B and D, the answer comes out to be zero! 
I really have not got much clue about this. O.O

anonymous · Answer

Looks trivial at first site but has a 'bite' to it. The only logical approach I can think of is to start expanding each of the responses in order to find any of the given two equations. I've tried the first one, so SIGMA(cos2A) and end up with something that definitively does NOT go down to zero. Strike out, therefore. 

Same thing for the second option, so now trying the third one. Got the same result as AkashdeepDeb, with B and D. Both answers compatible but difficult to relate to initial problem. Sorry :-(

anonymous · Answer

@mathslover

sirm3d · Answer

$$0=(\cos A +\cos B+ \cos C)(\sin A + \sin B + \sin C)\\quad=(1/2)\left[\sin(2A)+\sin(2B)+\sin(2C)\right]+\sin(A+B)+\sin(B+C)+\sin(A+C)$$
if (b) is true then (d) is also true

sirm3d · Answer

that leaves (a) or (c) as the correct answer, or probably no answer at all.

anonymous · Answer

The answer given is that ALL options are correct.. How is this possible?