已知在△abc中，内角a、b、c满足：（1/2）sin^2（a-b）=1-2sin [（π/4)-(c/2）]

显示全部楼层 · 2012-3-28 11:09:31

1(1/2)sin(A-B)=1-2(sin(π/4-C/2))^2(1/2)sin(A-B)=cos(π/2-C)(1/2)sin(A-B)=sinCsin(A-B)=2sinCsinAcosB-cosAsinB=2sinAcosB+2cosAsinBsinAcosB+3cosAsinB=0sinA/cosA+3sinB/cosB=0tanA+3tanB=02C=π/6(1/2)sin(A-B)=sinC=1/2A-B=π/2A+B=π-π/6A=2π/3B=π/6...

千问 · 2012-3-28 11:09:31

已知在△abc中，内角a、b、c满足：（1/2）sin^2（a-b）=1-2sin [（π/4)-(c/2）]（1）求证：tanA+3tanB=0这一问是不可能成立的请重新检查题目（估计题目输入有误）！理由如下若tanA+3tanB=0恒成立不妨取A=B 则∵（1/2）sin^2（a-b）=1-2sin [（π/4)-(c/2）]∴1-2si...