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在三角形ABC中,已知cos[(A-B)/2]=2sin(B/2),证明三边abc成等差数列
人气:474 ℃ 时间:2020-01-28 19:20:40
解答
应该是cos((A-C)/2)=2sin(B/2)2sin(B/2)cos(B/2)=cos(B/2)cos((A-C)/2)sinB=sin(π-B/2)cos((A-C)/2)=sin((A+C)/2)cos((A-C)/2)=(sinA+sinC)/22sinB=sinA+sinCa/sinA=b/sinB=c/sinC∴2b=a+ca,b,c成等差数列
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