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研究高考试题的解法,对高考复习具有重要的意义,本文采取配对的方法,可以获得一些高考题的巧解。下面举例说明配对法在解高考题中的应用。 一、和式配对 例1 sin20°cos70°+sin10°sin50°的值是( ). A.1/4 B.3~(1/2)/2 C.1/2 D.3~(1/2)/4 (1993年全国高考理科试题) 分析:本题原型见高中《代数(必修)》上册P.190,3(3)题。根据该题的特点,可以利用和差角公式sin(α±β)=Sinαcosβ±cosαsinβ和cos(α±β)=cosαcosβ于sinαsinβ配对解之。 解:设a=sin20°cos70°+sin10°sin50°, b=cos20°sin70°+com10°cos50°. 则 a+b=sin90°+cos40°=1+cos40°, ① b-a=sin50°+cos60°=1/2+cos40°. ② 由①一②得 2a=1/2,即a=1/4.故选A.
Studying the solution to the high examination questions is of great significance to the review of the college entrance examination. This article adopts a pairing method to obtain some clever solutions to the college entrance examination questions. The following example illustrates the application of pairing method in solving college entrance examination questions. One, and the pairing example 1 sin20°cos70°+sin10°sin50° is (). A.1/4 B.3~(1/2)/2 C.1/2 D.3~(1/ 2)/4 (1993 national college entrance examination science question) Analysis: The prototype of this question is referred to in the high school “Algebra (Compulsory)” volume P.190, 3 (3). According to the characteristics of the question, the sin(α±β)=Sinαcosβ±cosαsinβ and the cos(α±β)=cosαcosβ can be paired with the sinαsinβ. Solution: Let a=sin20°cos70°+sin10°sin50°, b=cos20°sin70°+com10°cos50°. Then a+b=sin90°+cos40°=1+cos40°, 1 ba=sin50°+cos60 ° = 1 / 2 + cos 40 ° 2 from 1 to 2 2a = 1/2, that is, a = 1/4. So choose A.