A + B + C is an integral multiple of π. x, y, z are real numbers. If x sin A + y sin B + z sin C = x2 sin 2A + y2 sin 2B + z2 sin 2C = 0, show that xn sin nA + yn sin nB + zn sin nC = 0 for any positive integer n.
Solution
The juxtaposition of x2 and sin 2A strongly suggests considering cos A + i sin A. So put u = x(cos A + i sin A), v = y(cos B + i sin B), w = z(cos C + i sin C). Put an = un + vn + wn. So a1 and a2 are real. Hence also uv + vw + wu = (a12 - a2)/2 is real. Also uvw = xyz exp( i(A+B+C) ) = ± xyz, since A+B+C is an integral multiple of π. Thus u, v, w are roots of some cubic p3 + ap2 + bp + c = 0 with real coefficients. Putting p = u, v, w and adding, we get a3 + a a2 + b a1 + 3c = 0, so a3 is real. Also multiplying through by pn, then putting p = u, v, w and adding, we get: an+3 + a an+2 + b an+1 + c an = 0. So by a trivial induction an is real for all positive n. Hence result.
© John Scholes
jscholes@kalva.demon.co.uk
20 Aug 2002