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在一次考试中,我出了这样一道题:求证:(1-cosα+sina)/(1+cosα+sinα)=tga/2(用两种方法证明)。这个等式的构造是由半角公式tgα/2=(1-cosα)/sinα=sinα/(1+cosα)并再由等比定理直接推得: tgα/2=(1-cosα+sinα)/(1+cosα+sinα) ①由①的构造过程我们可得到一种简单方法。证一:右边=(1-cosα)/sina=sinα/(1+cosα)==(1-cosα+sinα)/(1+cosα+sinα)由于大部分学生不会用等比定理,该方法虽然简单,但问鼎者仅两人。大部分学生采取了下面的证法。证二:左边=(1-(1-tg~2(α/2))/(1+tg~2(α/2))+2tg(α/2)/(1+tg~2(α/2)))/(1+(1-tg~2(α/2))/(1+tg~2(α/2)+2tg(α/2)/(1+tg~2(α/2))=(1+tg~2(α/2)-1+tg~2(α/2)+2tg(α/2))/(1+tg~2(α/2)+1-tg~2(α/2)+2tg(α/2))=tgα/2证三:左边=(2sin~2(α/2)+2sin~2(α/2)cos(α/2))/(2cos~2(α/2)+2sin(α/2)cos(α/2))
In an examination, I made this question: Proof: (1-cosα+sina)/(1+cosα+sinα)=tga/2 (proved in two ways). The construction of this equation is derived from the half-angle formula tgα/2=(1-cosα)/sinα=sinα/(1+cosα) and is directly derived from the equal ratio theorem: tgα/2=(1-cosα+sinα)/ (1+cosα+sinα) We can get a simple method from 1’s construction process. Certificate 1: Right = (1-cosα)/sina=sinα/(1+cosα)==(1-cosα+sinα)/(1+cosα+sinα) Since most students do not use the equal ratio theorem, this method Although simple, there are only two people who aspire. Most students take the following proofs. Card 2: Left = (1-(1-tg~2(α/2))/(1+tg~2(α/2))+2tg(α/2)/(1+tg~2(α/ 2)))/(1+(1-tg~2(α/2))/(1+tg~2(α/2)+2tg(α/2)/(1+tg~2(α/2 )) = (1+tg~2(α/2)-1+tg~2(α/2)+2tg(α/2))/(1+tg~2(α/2)+1-tg~ 2(α/2)+2tg(α/2))=tgα/2 Certificate Three: Left=(2sin~2(α/2)+2sin~2(α/2)cos(α/2))/( 2cos~2(α/2)+2sin(α/2)cos(α/2))