Question

在三角形ABC中,2bcosB=acosC+ccosA,求B的值和2｛(sinA)的平方｝+cos(A-C)的范围

Answer 1

2bcosB=acosC+ccosA,
根据正激液卜弦定理得：2sinB cosB=sinAcosC+sinCcosA,
2sinB cosB=sin(A+C),
2sinB cosB= sinB,
cosB=1/2,B=60°.
A+C=120°.

2｛(sinA)的平方｝+cos(A-C)=1-cos2A+ cos(A-C)
=1-cos2A+ cos(A-(120°-A))
=1-cos2A+ cos(2A-120°)
=1-cos2A+ cos2A cos120°+sin2A sin120°
=1-cos2A-1/2 cos2A+√3/埋稿2 sin2A
=1-3/2 cos2A+√3/2 sin2A
=1-√3(√3/2 cos2A -1/2 sin2A)
=1-√3 cos(2A+30°)
因明穗为0°<A<120°，30°<2A+30°<270°，
cos(2A+30°)∈[-1, √3/2），
√3 cos(2A+30°) ∈[-√3, 3/2），
所以所求范围是[1-√3, 5/2）.