Question

已知三角形ABC中，内角A,B,C内角的对边的边长为a,b,c，且bcosC=(2a-c)cosB. 若y=cos^2A+cos^2C求y的最小值

Answer 1

bcosC=(2a-c)cosB
b/sinB=a/sinA=c/sinC
a=bsinA/sinB c=bsinC/sinB
cosC=(2sinA-sinC)cotB
tan(A+C)cosC=sinC-2sinA
(tanA+tanC)/(1-tanAtanC)cosC=sinC-2sinA
(sinAcosC+cosAsinC)cosC=(sinC-2sinA)(cosAcosC-sinAsinC)
sin2A=1/(sinC-cosC)

Answer 2

bcosC=2acosB-ccosB
sinBcosC=2sinAcosB-sinCcosB
sinBcosC+sinCcosB=2sinAcosB
sin(B+C)=2sinAcosB
sinA=2sinAcosB
sinA≠0
∴2cosB=1
cosB=1/2
B=60°
A+C=120°
y=cos2A+cos2C=2cos(A+C)cos(A-C)=2cos120°cos(A-C)=-cos(A-C)
-120°<A-C<120°
∴-1/2<y<=1