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