陈凌霞：在三角形abc中角a b c的对边分别为abc，已知4sin[(A-B)/2]+4sinAsinB=2+根号2

袁洁的回答：

2absinC ∴1/[(A-B)/2 cos(A+B)=-√2/+42-2×3√2×4×√2/2 cosAcosB+sinAsinB-2sinAsinB=-√2/2 cosAcosB-sinAsinB=-√2/2]+4sinAsinB=2+根号2 2[1-cos(A-B)]+4sinAsinB=2+根号2 2-2cos(A-B)+4sinAsinB=2+根号2 -2cos(A-B)+4sinAsinB=根号2 -cos(A-B)+2sinAsinB=√2/2 A+B=135° ∴C=45° S△ABC=1/4=3 a=3/2 cos(A-B)-2sinAsinB=-√2/+b22)=3√2 ∴c22absinC=6 absin45°=12 asin45°=12/-2abcos45° =(3√2)2(√2/=a2漏了平方：4sin2

张芳瑜的回答：

2absinC ∴1/[(A-B)/2 cos(A+B)=-√2/+42-2×3√2×4×√2/2 cosAcosB+sinAsinB-2sinAsinB=-√2/2 cosAcosB-sinAsinB=-√2/2]+4sinAsinB=2+根号2 2[1-cos(A-B)]+4sinAsinB=2+根号2 2-2cos(A-B)+4sinAsinB=2+根号2 -2cos(A-B)+4sinAsinB=根号2 -cos(A-B)+2sinAsinB=√2/2 A+B=135° ∴C=45° S△ABC=1/4=3 a=3/2 cos(A-B)-2sinAsinB=-√2/+b22)=3√2 ∴c22absinC=6 absin45°=12 asin45°=12/-2abcos45° =(3√2)2(√2/=a2漏了平方：4sin2

蔡红梅的回答：

2absinC ∴1/[(A-B)/2 cos(A+B)=-√2/+42-2×3√2×4×√2/2 cosAcosB+sinAsinB-2sinAsinB=-√2/2 cosAcosB-sinAsinB=-√2/2]+4sinAsinB=2+根号2 2[1-cos(A-B)]+4sinAsinB=2+根号2 2-2cos(A-B)+4sinAsinB=2+根号2 -2cos(A-B)+4sinAsinB=根号2 -cos(A-B)+2sinAsinB=√2/2 A+B=135° ∴C=45° S△ABC=1/4=3 a=3/2 cos(A-B)-2sinAsinB=-√2/+b22)=3√2 ∴c22absinC=6 absin45°=12 asin45°=12/-2abcos45° =(3√2)2(√2/=a2漏了平方：4sin2