asinAsinB+b（1-2sinA^2）= 根号2乘以a
sinA(asinB-2bsinA)=根号2乘以a-b
根据正弦定理asinB=bsinA得
bcosA^2=根号（2)a
b/a=根号（2)/cosA^2 这样做就误解了题意
原始方程左右同时乘2r
sinAsinAsinB+sinB-sinBsinAsinA=根号2乘以sinA
会写了吧?*——*
b/a=sinB/sinA=根号2
△ABC的三个内角A,B,C所对的边分别为a,b,c,asinAsinB+bcos2A= 根号2乘以a ,则 b/a=
△ABC的三个内角A,B,C所对的边分别为a,b,c,asinAsinB+bcos2A= 根号2乘以a ,则 b/a=
数学人气：113 ℃时间：2019-09-03 08:37:36
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