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在△ABC中,已知ln(sinA+sinB)=lnsinA+ln(sinB-sinA)+lnsinB,且cos(A+B)+cosC=1-cos2C
(1)三角形ABC的形状
(2)(a+b0/b的取值范围
人气:214 ℃ 时间:2019-10-11 17:38:57
解答
1,1-cos2C = cos(A+B) + cosC =0cos2C = 1C =π/2直角三角形2,ln(sinA+sinB)=lnsinA+ln(sinB-sinA)+lnsinBsinA+sinB = sinAsinB(sinB-sinA) (a+b)/b = (sinA+sinB)/sinB = sinA(sinB-sinA)注意sinB-sinA>0所以(a+b)/...
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