本刊1992。2《优化解题过程的方法》给出的一些解题方法,确实多富精巧解法,给人以启迪。该文例3 已知:α、β∈(-π/2,π/2), 求证:|(sinα+sinβ)/(1+sinαsinβ)|<1。夏季刊文:如果f(x)=1g(1-x)/(1+x), 则f(a)+f (b)=f((a+b)/(1+ab))成立。 (*) 构造函数,由f(sinα)+f(sinβ) =1g (1-sinα)/(1+sinα)+1g(1-sinβ)/(1+sinβ)=… =f(sinα+sinβ)/(1+sinαsinβ)确定|(sinα+sinβ)/(1+sinαsinβ)|<1。巧则巧矣,但“联想到”(高中《代数》第一册P_(70))的“习题”*,这条“蹊径”未免有点偏狭。笔者下面提供一种解法,方法比原解法更 The 1992. 2 “Methods for Optimizing the Problem Solving Process” gives some problem-solving methods that are indeed rich and delicate solutions to give people enlightenment. This example 3 is known to: α, β∈ (-π/2, π/2), Prove: |(sinα+sinβ)/(1+sinαsinβ)|<1. Summer essay: If f(x)=1g(1-x)/(1+x), then f(a)+f(b)=f((a+b)/(1+ab)) holds. (*) Constructor from f(sinα)+f(sinβ) =1g (1-sinα)/(1+sinα)+1g(1-sinβ)/(1+sinβ)=... =f(sinα+sinβ )/(1+sinαsinβ) determines |(sinα+sinβ)/(1+sinαsinβ)|<1. Qiao is clever, but “associated with” (high school “Algebra”, the first book P_ (70)) “exercise” *, this “path” is a bit too narrow. The author provides a solution below, the method is more than the original solution