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以三角函数的形式出现并且和三角形有关的不等式很多,而且其中有些不等式的证明难度是相当大的,但是如果我们利用一些大家比较熟悉的不等式并采用代换方法就能很容易地解决某些问题。本文的目的就是通过几个事例来说明如何使用代换方法来证明或推导这些不等式。下面的不等式是大家所熟知的。设A′+B′+C′=π,则有sinA′/2sinB′/2sinC′/2≤1/8;(1)cosA′/2cosB′/2cosC′/2≤3(3~(1/3))/8 (2)(1)、(2)两式中等号当且仅当A′=B′=C′=π/3时成立。不等式(1)常出现在数学书刊上,故略去其证明。下面我们借助于(1)来证明(2)。若cosA′/2cosB′/2cosC′/2≤0, 则cosA′/2cosB′/2cosC′/2≤3(3~(1/3))/8显然成立,以下假定cosA′/2cosB′/2cosC′/2>0
There are many inequalities that appear in the form of trigonometric functions and are triangle-related, and some of these inequalities are quite difficult to prove, but if we use some of the more familiar inequalities and use substitution methods, we can easily solve some problems. . The purpose of this article is to illustrate how to use the substitution method to prove or infer these inequalities through several examples. The following inequalities are well known. Let A’ + B’ + C’ = π, then there sinA ’/2sinB’ / 2sinC’/2 ≤ 1/8; (1) cosA ’/2cosB’/2cosC’/2 ≤ 3 (3 ~ (1/ 3))/8 (2) (1), (2) The two equations are established if and only if A’ = B’ = C’ = π/3. Inequality (1) often appears in mathematics books, so its proof is omitted. Below we prove (2) by means of (1). If cosA′/2cosB′/2cosC′/2≤0, then cosA′/2cosB′/2cosC′/2≤3 (3~(1/3))/8 is clearly established. The following assumes cosA′/2cosB′/2cosC. ‘/2>0