Question

已知锐角△ABC中,sin(A+B)=3/5,sin(A-B)=1/5

1求证tanA=2tanB
2求tanA的值

Answer 1

sin(A+B)=3/5,
sin(A-B)=1/5
则：sin(A+B)=3sin(A-B)
sinAcosB+cosAsinB=3sinAcosB-3cosAsinB
2sinAcosB=4cosAsinB
sinA/cosA=2sinB/cosB
tanA=2tanB

sin(A+B)=3/5
sinC=sin[180°-(A+B)]=sin(A+B)=3/5
则：cosC=根号(1-sin²C)=4/5
则：tanC=sinC/cosC=3/4
tan(A+B)=tan(180°-C)=-tanC=-3/4
(tanA+tanB)/(1-tanAtanB)=-3/4
3tanB/(1-2tan²B)=-3/4
2tan²B-1=4tanB
2tan²B-4tanB-1=0
tanB=(4+2√6)/4=(2+√6)/2
tanA=2tanB=2+√6