Question

三角形中ABC中,bcosC=(2a-c)cosB

Answer 1

(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°