64th Putnam 2003 Problem A3 Find the minimum of |sin x + cos x + tan x + cot x + sec x + cosec x| for real x.
2√2-1 Solution Put x = y - 3π/4. Then sin x = -(cos y + sin y)/√2, cos x = -(cos y - sin y)/√2, so sin x + cos x = - √2 cos y. After some manipulation, tan x + cot x = 2/(cos 2 y - sin 2 y), sec x + cosec x = -2√2 cos y/(cos 2 y - sin 2 y). After some further manipulation sin x + cos x + tan x + cot x + sec x + cosec x = -c - 2/(c+1), where c = √2 cos y. By AM/GM (c+1) + 2/(c+1) has minimum 2√2 for c+1 positive (and ≤ -2√2 for c+1 negative), so expr has minimum 2√2-1. 64th Putnam 2003 © John Scholes jscholes@kalva.demon.co.uk 8 Dec 2003 Last corrected/updated 8 Dec 03