2011/005) if cos A + cos B + cos C = 0 = sin A + sin B + sin C

prove that

cos 3A + cos 3B + cos 3C = 3 cos( A+ B+ C)

proof:

we have

cos A + cos B + cos C = 0 ...1

sin A + sin B + sin C = 0 ....2

multiply 2nd by i and add

(cos A + i sin A ) + ( cos B + i sin B) + (cos C + i sin C) = 0

or e^(iA) + e^(iB) + e^(iC)= 0 as e^ix = cos x+ i sin x

and as if (x+y+z) = 0 then x^3+y^3 + z^3 = 3xyz

so e^(i3A) + e^(i3B ) + e^(i3C) = 3 e^i(A+B+C)

so ( cos 3A + i sin 3 A) + (cos 3B + i sin 3B ) + ( cos 3C + i sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)

or (cos 3A + cos 3 B + cos 3C) + i( sin 3A + sin 3B + sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)

equating real and imaginary parts on both sides we get
cos 3A + cos 3 B + cos 3C = 3 cos (A+B+C)
sin 3A + sin 3B + sin 3C = 3 sin (A+B+ C)