Prove that 1/sin 2x + 1/sin 4x + ... + 1/sin 2nx = cot x - cot 2nx for any natural number n and any real x (with sin 2nx non-zero).
Solution
cot y - cot 2y = cos y/sin y - (2 cos2y - 1)/(2 sin y cos y) = 1/(2 sin y cos y) = 1/sin 2y. The result is now easy. Use induction. True for n = 1 (just take y = x). Suppose true for n, then taking y = 2nx, we have 1/sin 2n+1x = cot 2nx - cot 2n+1x and result follows for n + 1.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
29 Sep 1998
Last corrected/updated 26 Sep 2003