△ABC中,若sinA·sinB=cos^2(C/2),则△ABC的形状是?

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△ABC中,若sinA·sinB=cos^2(C/2),则△ABC的形状是?
△ABC中,若sinA·sinB=cos^2(C/2),则△ABC的形状是?

△ABC中,若sinA·sinB=cos^2(C/2),则△ABC的形状是?
sinAsinB=cos²(C/2)=[1+cosC]/2.
===>2sinAsinB=1+cosC=1+cos[180-(A+B)]=1-cos(A+B)=1-cosAcosB+sinAsinB.
===>cosAcosB+sinAsinB=cos(A-B)=1.
===>A-B=0.
===>A=B.
===>等腰三角形.

cos^2(C/2)=(cosC+1)/2
2sinA·sinB=cosC+1 cosC=-cos(A+B)=-cosAcosB+sinAsinB
2sinA·sinB=-cosAcosB+sinAsinB+1
cosAcosB+sinAsinB=1
cos(A-B)=1
A-B=0
A=B
△ABC的形状是等腰三角形