Question

高一数学三角函数问题

已知tanA=1/2,tan(A-B)=-2/3,试求tan(B-2A)
速解，谢谢，在线等，详细点哦

Answer 1

tan(A-B)=-tan(B-A)
-tan(B-A)=-2/3
tan(B-A)=2/3

tan(B-2A)
=tan[(B-A)-A]
=[tan(B-A)-tanA]/[1+tan(B-A)tanA]
=[2/3-1/2]/[1+1/2*2/3]
=[4/6-3/6]/[1+1/3]
=(1/6)/(4/3)
=1/6*3/4
=1/8

Answer 2

tan(B-A)=-tan(A-B)=2/3
tan(B-2A)
=tan[(B-A)-A]
=[tan(B-A)-tanA]/[1-tan(B-A)tanA]
=[2/3-1/2]/[1-2/3*1/2]
=1/4

Answer 3

tan(B-2A)=tan(B-A-A)=[tan(B-A)-tanA]/[1+tan(B-A)*tanA]=1/4

Answer 4

TAN(B-2A)=TAN(2(B-A)+A)

Answer 5

tan(B-2A)=tan(B-A-A)=sin(B-A-A)/cos(B-A-A)
=[sin(B-A)cosB-cos(B-A)sinB]/[cos(B-A)cosB+sin(B-A)sinB]
=[-sin(A-B)cosB-cos(A-B)sinB]/[cos(A-B)cosB-sin(A-B)sinB]
=[-tan(A-B)-tanB]/[1-tan(A-B)tanB]
=[-(-2/3)-1/2]/[1-(-2/3)(1/2)]
=(1/6)/(4/3)
=1/8

Answer 6

tan(B-2A)
=tan[(B-A)-A]
=[tan(B-A)-tanA]/[1+tan(B-A)tanA]
=[2/3-1/2]/[1+1/2*2/3]
=
1/8