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∵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
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