在△ABC中,满足sin2A-sin2C+sin2B=sinAsinB。那么C的大小为??
1个回答
展开全部
(sinA+sinB)^2-sin^2C=3sinAsinB
sin²A+2sinAsinB+sin²稿姿纤B-3sinAsinB=sin²C
sin²A-sinAsinB+sin²B=sin²(A+B)
sin²A(sin²册尺B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)=(sinAcosB+sinBcosA)²
展开得
sin²A(sin²B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)
=(sinAcosB+sinBcosA)²
=sin²Acos²B+2sinAsinBcosAcosB+sin²Bcos²A
两边减去相同项得
sin²键仿Asin²B-sinAsinB+sin²Bsin²A=2sinAsinBcosAcosB
2sin²Asin²B-sinAsinB-2sinAsinBcosAcosB=0
sinAsinB(2sinAsinB-1-2cosAcosB)=0
2sinAsinB-1-2cosAcosB=0
sinAsinB-cosAcosB=1/2
-cos(A+B)=1/2
cos(A+B)=-1/2
A+B=120°
则角C=60度
sin²A+2sinAsinB+sin²稿姿纤B-3sinAsinB=sin²C
sin²A-sinAsinB+sin²B=sin²(A+B)
sin²A(sin²册尺B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)=(sinAcosB+sinBcosA)²
展开得
sin²A(sin²B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)
=(sinAcosB+sinBcosA)²
=sin²Acos²B+2sinAsinBcosAcosB+sin²Bcos²A
两边减去相同项得
sin²键仿Asin²B-sinAsinB+sin²Bsin²A=2sinAsinBcosAcosB
2sin²Asin²B-sinAsinB-2sinAsinBcosAcosB=0
sinAsinB(2sinAsinB-1-2cosAcosB)=0
2sinAsinB-1-2cosAcosB=0
sinAsinB-cosAcosB=1/2
-cos(A+B)=1/2
cos(A+B)=-1/2
A+B=120°
则角C=60度
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