3i will use A , B and C.
we define three complex numbers as these -
cosA + isinA, cosB + isinB , cosC + isinC.
here i is √-1.
on squaring and adding these we get,
(cos2A + cos2B + cos2C) + i(sin2A + sin2B + sin2C) = 0.
so, Σcos2A = Σsin2A = 0.
cos2A = 2cos2A -1 = 0
cos2B = 2cos2B - 1 = 0
cos2C = 2cos2c -1 = 0
adding all these we get,
Σcos2A = 3/2 = Σsin2A.
