PUMaC 2013 · 团队赛 · 第 15 题

PUMaC 2013 — Team Round — Problem 15

Prove: | sin a | + | sin a | + | sin a | + ... + | sin a | + | cos ( a + a + a + ... + a ) | ≥ 1 . 1 2 3 n 1 2 3 n ◦

解析

Prove: | sin a | + | sin a | + | sin a | + ... + | sin a | + | cos ( a + a + a + ... + a ) | ≥ 1 . 1 2 3 n 1 2 3 n SOLUTION: Use induction on n . For n = 1, | sin a | + | cos a | ≥ 1 1 1 2 2 ⇔ sin a + cos a + 2 | sin a cos a | ≥ 1 , 1 1 1 2 which is obvious. Assume n = k is true. For n = k + 1, we first have | sin a | + | sin a | + | sin a | + · · · + | sin a | + | cos ( a + a + a + · · · + a ) | ≥ 1 . 1 2 3 k 1 2 3 k So it suffices to prove | sin a | + | cos ( a + a + · · · + a ) | ≥ | cos ( a + a + · · · + a ) | , k +1 1 2 k +1 1 2 k or simply | sin x | + | cos ( x + y ) | ≥ | cos y | , 5 where x = a and y = a + a + · · · + a . Now, k +1 1 2 k | sin x | + | cos ( x + y ) | ≥ | cos y | 2 2 2 ⇔ sin x + (cos x cos y − sin x sin y ) + 2 | sin x cos ( x + y ) | ≥ cos y 2 2 ⇔ sin x sin y + | sin x cos ( x + y ) | ≥ sin x sin y cos x cos y ⇔ | sin x cos ( x + y ) | ≥ sin x sin y cos ( x + y ) , which is obvious. ◦