利用正弦定理a/sinA=b/sinB=c/sinC
∵cosC/cosB=(2a-c)/b
∴cosC/cosB=(2sinA-sinC)/sinB
∴cosCsinB=2sinAcosB-sinCcosB
∴cosCsinB+sinCcosB=2sinAcosB
∴ sin(B+C)=2sinAcosB
∵ sin(B+C)=sin(180°-A)=sinA
∴ sinA=2sinAcosB
∴ cosB=1/2,
sinB=√3/2
∵ tan(A＋π／4)=7
∴ (tanA+1)/(1-tanA)=7
∴ tanA+1=7-7tanA
∴ tanA=3/4
∴ sinA=3/5,cosA=4/5
∴ cosC
=cos[π-(A+B)]
=-cos(A+B)
=-(cosAcosB-sinAsinB)
=-[(4/5)*(1/2)-(3/5)*(√3/2)]
=(3√3-4)/10