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Problem 1202

AMC 12 late, AIME early

Algebra Difficulty 4.9 Prove it Brazilian Mathematical Olympiad · Brazil

Initially, a calculator displays number $1$ . An operation consists in pressing either key $\sin$ or key $\cos$ , which calculates respectively the sine and cosine with the arguments in radians. After performing $2001$ operations, what is the greatest possible value that can be achieved?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Obviously the value does not exceed $1$ . We cannot get close to $1$ with $\sin$ . So the final press must be $\cos$ and we want the previous value to be as close to $0$ as possible. We cannot get close to $0$ with $\cos$ , so the $2000$ th press must be $\sin$ . Thus we want the previous value to be as close to $0$ as possible. So, by a simple induction, all presses except the last must be $\sin$ .

Put $a_1 = \sin 1$ . Put $a_{n+1} = \sin a_n$ for $n = 1, \dots, 1999$ and put $a_{2001} = \cos a_{2000}$ . The best value is $a_{2001}$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.