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三角形ABC中,已知sinA*sinB=[cos(c/2)]^2,求此三角形的形状?

热心网友

sinA*sinB=[cos(c/2)]^22sin(A/2)cos(A/2)*2sin(B/2)cos(B/2)=[cos(180-A-B)/2]^2={cos[90-(A+B)/2]}^2={sin[A/2+B/2]}^2={(sinA/2)*cos(B/2)+cos(A/2)*sin(B/2)}^2=[sin(A/2)*cos(B/2)]^2+2[sin(A/2)]*[cos(B/2)]*[cos(A/2)]*[sin(B/2)]++[cos(A/2)*sin(B/2)]^2==0==[sin(A/2)*cos(B/2)]^2-2[sin(A/2)]*[cos(B/2)]*[cos(A/2)]*[sin(B/2)]++[cos(A/2)*sin(B/2)]^2={(sinA/2)*cos(B/2)-cos(A/2)*sin(B/2)}^2==(sinA/2)*cos(B/2)=cos(A/2)*sin(B/2)==tan(A/2)*cot(B/2)=1==A=B△ABC是等腰三角形。

热心网友

sina*sinb=[cos(c/2)]^2=[cos(c/2)]^2-1/2+1/2=0.5(cosc+1)=0.5(-cos(a+b)+1)=0.5(1-cosacosb+sinasinb)移项0.5(sinasinb+cosacosb)=0.5cos(a-b)=1a-b=0a=b等腰三角形

热心网友

解:因为2sinAsinB=2[cos(c/2)]^2=1+cosC=1+cos(180度-(A+B))=1-cos(A+B)=1-cosAcosB+sinAsinB所以2sinAsinB=1-cosAcosB+sinAsinB即 cosAcosB-sinAsinB=1 cos(A-B)=1 A-B=0 A=B 所以此三角形的形状是等腰三角形.

热心网友

等腰三角形sinBsinC=cos(A/2)^2的平方,则三角形＿＿＿＿？左边＝-(1/2)[cos(B+C)-cos(B-C)]=-(1/2)[-cosA-cos(B-C)]=(1/2)[cosA+cos(B-C)]右边＝cos(A/2)^2=(1+cosA)/2左右边相等则cosA+cos(B-C)＝1+cosA则cos(B-C)＝1B＝C为等腰三角形

热心网友

等边三角形