因为tanA/2tanC/2=1/3
所以cosA+cosC-cosAcosC+1/3sinAsinC
=cosA+cosC-cosAcosC+(tanA/2tanC/2)sinAsinC
=cosA+cosC-cosAcosC+(1-cosA)/sinA*(1-cosC)/sinC*sinAsinC
=cosA+cosC-cosAcosC+(1-cosA)(1-cosC)
=cosA+cosC-cosAcosC+(1-cosA-cosC+cosAcosC)
=1tanA/2tanC/2=1/3怎么得到∵a＋c＝2b∴sinA＋sinc＝2sinB即sinA＋sinC＝2sin(A＋C)由和差化积、二倍角公式得：2sin[(A+C)/2]×cos[(A-C)/2]＝4sin[(A+C)/2]×cos[(A+C)/2]∵sin[(A+C)/2]≠0∴cos[(A-C)/2]＝2cos[(A+C)/2]　cos(A/2)cos(C/2)＋sin(A/2)sin(C/2)＝2cos(A/2)cos(C/2)－2sin(A/2)sin(C/2)即3sin(A/2)sin(C/2)＝cos(A/2)cos(C/2)∴tan(A/2)×tan(C/2)＝1/3 　∴[(1－cosA)/sinA]×[(1－cosC)/sinC]＝1/3∴cosA＋cosC－cosAcosC＋（1/3)sinAsinC＝1