Question

已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1)求证:tanA=2tanB(2)tan

Answer 1

（1）sin(A+B)=3/5,sin(A-B)=1/5

sin(a+b)=sinAcosB+sinBcosA=3/5

sin(a-b)=sinAcosB-sinBcosA=1/5

两式相加相减后可得：

sinAcosB=2/5

sinBcosA=1/5

将两式相除,可得tanA=2tanB

（2）tan(B)=sinB/cosB=sinBcosA/cosAcosB

cos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(根号6-2)/5

tanB=1/(根号6-2)=(根号6+2)/2