在三角形ABC中.cosC／cosB=2a－c／b.若tan(A＋π／4)=7求cosC

1/2 ..........

利用正弦定理，得：
cosC/cosB=(2sinA－sinC)/(sinB)
sinBcosC=2sinAcosB－cosBsinC
sinBcosC＋cosBsinC=2sinAcosB
sin(B＋C)=2sinAcosB
sinA=2sinAcosB
因为：sinA不等于0，则：
cosB=1/2
B=60°
又：tan(A＋π/4)=7
则：(tanA＋1)/(1－tanA)=7
得：tanA=3/4
则：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

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