论文部分内容阅读
1.一道试题题目已知α、β都是锐角,且3sin2α+2sin2β=1①3sin2α-2sin2β=0②求证:α+2β=(π/2) 这是1978年全国统一高考中的一道试题,已被收录在许多复习资料中,本文以几何出发,给出两种新的解法. 2.几何解法I——利用正弦定理与射影先看条件②式
1. The subject of a test is known to be both acute and acute, and 3sin2α+2sin2β=113sin2α-2sin2β=02 Proof: α+2β=(π/2) This is a test item in the National Unified College Entrance Examination in 1978 and has been included. In many review materials, this article starts with geometry and gives two new solutions. 2. Geometric solution I—using sine theorem and projection to look at condition 2