在△ABC中,若a+c=2b,则cosA+cosA-cosAcosC+1/3sinAsinC=

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在△ABC中,若a+c=2b,则cosA+cosA-cosAcosC+1/3sinAsinC=
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在△ABC中,若a+c=2b,则cosA+cosA-cosAcosC+1/3sinAsinC=
在△ABC中,若a+c=2b,则cosA+cosA-cosAcosC+1/3sinAsinC=

在△ABC中,若a+c=2b,则cosA+cosA-cosAcosC+1/3sinAsinC=
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)
=1