PUMaC 2010 · 代数（A 组） · 第 7 题

PUMaC 2010 — Algebra (Division A) — Problem 7

The expression sin 2 sin 4 sin 6 · · · sin 90 is equal to p 5 / 2 , where p is an integer. Find p .

解析

The expression sin 2 sin 4 sin 6 · · · sin 90 is equal to p 5 / 2 , where p is an integer. Find p . 2 πi/ 90 Solution: 192. Let ω be the root of unity e , so we have 45 45 n ∏ ∑ ω − 1 ◦ sin(2 n ) = . n/ 2 2 iω n =1 n =1 ◦ By the symmetry of the sine (and the fact that sin(90 ) = 1), 45 89 ∏ ∏ ◦ ◦ sin(2 n ) = sin(2 n ) , n =1 n =46 so ∣ ∣ 2 45 89 ∣ ∣ n ∏ ∑ | ω − 1 | 90 ∣ ∣ ◦ sin(2 n ) = = , ∣ ∣ 89 ∣ ∣ 2 2 n =1 n =1 where we’ve used the usual geometric series sum for roots of unity. The product is clearly positive and real, so it is equal to √ √ 45 3 5 = , 44 44 2 2 6 implying that p = 3 · 2 = 192. 2