题目

b/(a+c)=tan（B/2）
判断三角形ABC的形状.

提问时间：2020-12-06

答案

正玄定理 sinB/(sinA + sinC) = tan(B/2)
万能公式sinB = 2tan(B/2) / (1+tan^2 (B/2));
和差化积 sin A+sinC=2sin[(A+C)/2]·cos[(A-C)/2] = 2 cos(B/2)*cos[(A-C)/2]
带入消掉 tan(B/2) != 0
2 = (1+tan^2 (B/2)) * 2 cos(B/2)*cos[(A-C)/2]
1 = cos[(A-C)/2] / cos (B/2)
cos [(A-C)/2] = cos(B/2)
再用和差化积 cos α-cos β=-2sin[(α+β)/2]·sin[(α-β)/2]
sin(((A-C)/2 + B/2)/2) = 0 或者是 sin(((A-C)/2 - B/2)/2) = 0
A+B-C = 180 - 2C,于是有sin ((90-C)/2) = 0 或者是 sin ((A-90)/2) = 0
考查范围,有C=90或者A=90 直角三角形.