在△ABC中,若a+c=2b,则cosA+cosC-cosAcosC+sinAsinC/3=?

问题描述:

在△ABC中,若a+c=2b,则cosA+cosC-cosAcosC+sinAsinC/3=?

因为tanA/2tanC/2=1/3所以cosA+cosC-cosAcosC+(1/3)sinAsinC=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...