15.三角形ABC中，已知tan[(A+B)/2]=sinC,下列四个论断是否正确，说明理由。(1)tanA*cotB=1 (2)0<sinA+sinB≤√2.(3)(sinA)^2+(cosB)^2=1 (4)(cosA)^2+(cosB)^2=(sinC)^2

热心网友

解: ∵A+B+C=180°∴tan[(A+B)/2]=tan[(180°-C)/2]=tan[90°-C/2]=cot(C/2)=sinC/(1-cosC)又由已知tan[(A+B)/2]=sinC∴sinC/(1-cosC)=sinC ∵sinC≠0(C≠0°)∴1-cosC=1,cosC=0∴A+B=C=90°,△ABC是直角三角形(1)tanA*cotB=1 不正确.应该是tanA*tanB=1(2)0

热心网友

都不正确吧

热心网友

解:tan[(A+B)/2]=tan[(180°-C)/2]=tan[90°-C/2]=-cot(C/2)而cot(C/2)=sinC/(1-cosC)∴-cot(C/2)=sinC/(cosC-1)原式=sinC/(cosC-1)=sinC sinCcosC=2sinC ∵sinC≠0(C≠0°)∴cosC=0 C=90°(1)tanA*cotB=1 正确.∵△ABC是直角三角形. ∴tanA*cotB=1成立