30th Swedish 1990 Problem 3 Find all a, b such that sin x + sin a ≥ b cos x for all x.
a = (π/2 + 2nπ), b = 0 Solution Obviously a = (4n+1)π/2, b = 0 works. The question is whether any other a, b work. Suppose sin a  0. So we must have sin a = 1 (and hence a = (4n+1)π/2). Suppose b ≠ 0. Take tan θ = b, and x = 3π/2 + θ. Then sin x = - cos θ, cos x = sin θ. So we require - cos θ + 1 ≥ b sin θ. But cos θ = 1/√(1+b 2 ), sin θ = b/√(1+b 2 ), so we need: -1 + √(1+b 2 ) ≥ b 2 , which is false for b ≠ 0. So we must have b = 0. Thanks to Suat Namli 30th Swedish 1990 © John Scholes jscholes@kalva.demon.co.uk 28 February 2004 Last corrected/updated 28 Feb 04