21st USAMO 1992

Let k = 1^o. Show that S₀⁸⁸ 1/(cos nk cos(n+1)k ) = cos k/sin²k.

Solution

tan(n+1)k - tan nk = ( sin(n+1)k cos nk - sin nk cos(n+1)k )/( cos nk cos(n+1)k ) = sin k /( cos nk cos(n+1)k ). Using this expression the sum telescopes and we get ∑₀⁸⁸ 1/(cos nk cos(n+1)k ) = (tan 89k - tan 0)/sin k. But tan 0 = 0 and tan 89k = cot(π/2 - 89k) = cot k.

21st USAMO 1992

© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002