Question

在銳角三角形中,求證:sinA+sinB+sinC＞cosA+cosB+cosC.

Answer 1

思路解析:此題采用綜合法通過構造角的不等式轉(zhuǎn)化為利用三角函數(shù)的單調(diào)性來證明,此法比常用的和化積形式簡單.

證明:∵銳角三角形中,A+B＞,

∴A＞-B.∴0＜-B＜A＜.

又∵在(0,)內(nèi)正弦函數(shù)是單調(diào)遞增函數(shù),

∴sinA＞sin(-B)=cosB,即sinA＞cosB.                                          ①

同理,sinB＞cosC,                                                             ②

sinC＞cosA.                                                                 ③

由①+②+③,得sinA+sinB+sinC＞cosA+cosB+cosC.