在三角形ABC中

三角形ABC中，三条边a、b、c

∵2b=a+c ∴2sinB=sinA+sinC(利用正弦定理) 将左右两边分别化开，得如下： sinA+sinC=2[sin(A+c)/2]*[cos(A-c)/2]=2cos(B/2)*[cos(A-c)/2] 2sinB=4*sin(B/2)*cos(B/2) 即2cos(B/2)*[cos(A-c)/2]=4*sin(B/2)*cos(B/2) [∵cos(B/2)＞0] 即cos(A-c)/2=2sin(B/2)…………① tan(A/2)tan(C/2) =[sin(A/2)/cos(A/2)]*[sin(C/2)/cos(C/2)] =sin(A/2)*sin(C/2)/...全部

  ∵2b=a+c ∴2sinB=sinA+sinC(利用正弦定理) 将左右两边分别化开，得如下： sinA+sinC=2[sin(A+c)/2]*[cos(A-c)/2]=2cos(B/2)*[cos(A-c)/2] 2sinB=4*sin(B/2)*cos(B/2) 即2cos(B/2)*[cos(A-c)/2]=4*sin(B/2)*cos(B/2) [∵cos(B/2)＞0] 即cos(A-c)/2=2sin(B/2)…………① tan(A/2)tan(C/2) =[sin(A/2)/cos(A/2)]*[sin(C/2)/cos(C/2)] =sin(A/2)*sin(C/2)/[cos(A/2)*cos(C/2)] =-(1/2)[cos(A+c)/2- cos(A-c)/2]/(1/2)[cos(A+c)/2+ cos(A-c)/2] =-[cos(A+c)/2- cos(A-c)/2]/[cos(A+c)/2+ cos(A-c)/2] =-[sin(B/2)-cos(A-c)/2]/[sin(B/2)+cos(A-c)/2] 以①代入 =-[sin(B/2)-2sin(B/2)]/[sin(B/2)+2sin(B/2)] =-(-1/3) =1/3。
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