Question

在三角形ABC中,角A,B,C,所对应的边a,b,c若tanC/tanA+tanC/tanB=1则cosC最小值为

Answer 1

tanC/tanA+tanC/tanB=1,

∴sinCcosA/(sinAcosC)+sinCcosB/(sinBcosC)=1,

去分母得sinC(cosAsinB+sinAcosB)=sinAsinBcosC,

cosAsinB+sinAcosB=sin(A+B)=sinC,

sinAsinB=(1/2)[cos(A-B)-cos(A+B)]=(1/2)[cos(A-B)+cosC],

∴(sinC)^2=(1/2)[cos(A-B)+cosC]cosC,

设cosC=x,m=cos(A-B),则-1