在△ABC中若面积S=(a^2+b^2

一道较难的数学题，大家帮帮忙

解:sinA=2√2/3,因为是锐角三角形,所以cosA=1/3 tan^2[(B+C)/2]+sin^2 (A/2)=tan^2(π-A)/2+sin^2(A/2) =cot^2(A/2)+sin^2(A/2)=(cos^2(A/2)/sin^2(A/2)+sin^2(A/2) =[cos^2(A/2)+sin^2(A/2)*sin^2(A/2)]/sin^2(A/2) =[cos^2(A/2)+(1-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2) =[cos^2(A/2)+sin^2(A/2)-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2) =[...全部

  解:sinA=2√2/3,因为是锐角三角形,所以cosA=1/3 tan^2[(B+C)/2]+sin^2 (A/2)=tan^2(π-A)/2+sin^2(A/2) =cot^2(A/2)+sin^2(A/2)=(cos^2(A/2)/sin^2(A/2)+sin^2(A/2) =[cos^2(A/2)+sin^2(A/2)*sin^2(A/2)]/sin^2(A/2) =[cos^2(A/2)+(1-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2) =[cos^2(A/2)+sin^2(A/2)-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2) =[1-(1/4)sin^2A]/[(1-cosA)/2] =[1-(1/4)2√2/3]/[(1-1/3)/2] =(6-√2)/2 (2) S=(1/2)bcsinA=(1/2)bc*2√2/3=√2 所以bc=3 根据余弦定理 a^2-b^2-c^2+2bccosA=0,即4-b^2-c^2+2bc*1/3=0 解得:b=c=√3。
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