已知tan[(A+B)/2]=根號下6/2,tanAtanB=13/7求cos(A-B)答案2/3
熱心網(wǎng)友
tan(A+B)=2tan(A+B)/2/[1-tan^2(A+B)/2]=√6/[1-3/2]=-2√6tanA+tanB=tan(A+B)(1-tanAtanB)=12√6/7由根與系數(shù)關(guān)系可算出tanA,tanB,由此可得cos(A-B)
已知tan[(A+B)/2]=根號下6/2,tanAtanB=13/7求cos(A-B)答案2/3
tan(A+B)=2tan(A+B)/2/[1-tan^2(A+B)/2]=√6/[1-3/2]=-2√6tanA+tanB=tan(A+B)(1-tanAtanB)=12√6/7由根與系數(shù)關(guān)系可算出tanA,tanB,由此可得cos(A-B)