题目内容：

三角形中ABC中,bcosC=(2a-c)cosB
1求角B大小.2若y=cos^2A+cos^2C,求y取值范围

优质解答

(2a-c)cosB=bcosC
正弦定理得:
(4RsinA-2RsinC)cosB=2RsinBcosC
2sinAcosB=sinBcosC+sinCcosB
2sinAcosB=sin(B+C)
2sinAcosB=sinA
cosB=1/2
得B=60°
y=COS^2A+COS^2C
=(cos2A+cos2C+2)/2
=[2cos(A+C)cos(A-C)+2]/2
=cos(A+C)cos(A-C)+1
=1-cos(A-C)/2
0°