已知△abc满足(b-c)cos^2A=bcos^2B-ccos^2C,试判断△abc的形状

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(b-c)cos^2A=bcos^2B-ccos^2C
(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C
(b-c)cos^2(B+C)=bcos^2B-ccos^2C
b(cos^2(B+C)-cos^2B)=c(cos^2(B+C)-cos^C)
cos^2(B+C)-cos^2B
=(cos(B+C)+cosB)(cos(B+C)-cosB)
=2cos(2B+C)/2cosC/2*(-sin(2B+C)/2sinC/2)
=-[2sin(2B+C)/2cos(2B+C)/2]*[2sinC/2cosC/2]
=-sin(2B+C)sinC
同样
cos^2(B+C)-cos^C=-sin(B+2C)sinB
所以
bsin(2B+C)sinC=csin(B+2C)sinB
b/sinB*sin(2B+C)=c/sinC*sin(B+2C)
因为:b/sinB=c/sinC
所以,sin(2B+C)=sin(B+2C)
2B+C=B+2C,或,2B+C+(B+2C)=π
B=C,或,B+C=π/3
所以,△abc是等腰三角形,或,A=120度的三角形

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