HMMT 二月 2016 · 冲刺赛 · 第 15 题

HMMT February 2016 — Guts Round — Problem 15

解析

[ 9 ] Compute tan tan tan . 7 7 7 Proposed by: Alexander Katz √ Answer: 7 7 ix Consider the polynomial P ( z ) = z − 1. Let z = e = cos x + i sin x . Then ( ( ) ( ) ( ) ) 7 7 7 2 4 6 7 7 5 3 z − 1 = cos x − cos x sin x + cos x sin x − cos x sin x − 1 2 4 6 ( ( ) ( ) ( ) ) 7 7 7 7 5 2 3 4

i − sin x + sin x cos x − sin x cos x + sin x cos 6 x 2 4 6 7 Consider the real part of this equation. We may simplify it to 64 cos x − . . . − 1, where the middle ( ) ∏ 7 2 π 4 π 2 πk 1 terms are irrelevant. The roots of P are x = , , . . . , so cos = . But k =1 7 7 7 64 ( ) 2 ( ) ( ) 7 3 ∏ ∏ 2 πk kπ cos = cos 7 7 k =1 k =1 ( ) ∏ 3 kπ 1 so cos = . k =1 7 8 11 Now consider the imaginary part of this equation. We may simplify it to − 64 sin x + . . . + 7 sin x , 10 where again the middle terms are irrelevant. We can factor out sin x to get − 64 sin x + . . . + 7, and ( ) ∏ 6 2 π 12 π 2 πk 7 this polynomial has roots x = , . . . , (but not 0). Hence sin = − . But, like before, k =1 7 7 7 64 we have ( ) 2 ( ) ( ) 6 3 ∏ ∏ 2 πk 2 πk sin = − sin 7 7 k =1 k =1 √ √ √ 7 ( ) ∏ 3 kπ 7 8 hence sin = . As a result, our final answer is = 7 . 1 k =1 7 8 8 n