在△ABC中,2b=a+c,求证tan(A/2)•tan(C/2)=1/3

(1)由条件有2sinB=sinA+sinC
因此4sin((A+C)/2)cos((A+C)/2)=2sin(A+C)=2sinB=sinA+sinC=2sin((A+C)/2)cos((A-C)/2)
约去2sin((A+C)/2)由2cos((A+C)/2)=cos((A-C)/2)
即2(cos(A/2)cos(C/2)-sin(A/2)sin(C/2))=cos(A/2)cos(C/2)+sin(A/2)sin(C/2)
即cos(A/2)cos(C/2)=3sin(A/2)sin(C/2)即tan(A/2)tan(C/2)=1/3
(2)设tan(A/2)=x tan(C/2)=y
则cosA=(1-x^2)/(1+x^2) cosC=(1-y^2)/(1+y^2)
则5cosA-4cosAcosC+5cosC
=5(1-x^2)/(1+x^2)-4(1-x^2)(1-y^2)/((1+x^2)(1+y^2))+5(1-y^2)/(1+y^2)
=1/((1+x^2)(1+y^2))*[5(1-x^2)(1+y^2)+5(1+x^2)(1-y^2)-4(1-x^2)(1-y^2)]
=1/((1+x^2)(1+y^2))*[6+4x^2+4y^2-14(xy)^2]
=(40/9+4x^2+4y^2)/(10/9+x^2+y^2)=4 (xy=1/3代入)
(3)则cosA=-sinC 代入(2)有4=-5sinC+4sinCcosC+5cosC
设t=cosC-sinC 则2sinCcosC=1-t^2
故4=5t+2(1-t^2) 解得t=1/2(舍去t=2) 于是sin2C=3/4 由t>0知C