Question

已知tan(a-b)=1/2,tanb=-1/7,且a、b属于(0,派). 求2a-b的值？

Answer 1

因为A＝（A－B）＋B
tana=tan(a-b+b)=[tan(a-b)+tanb]/[1-tan(a-b)*tanb]
=(1/2-1/7)/(1+1/2*1/7)
=1/3
tan(2a-b)=tan(a+a-b)
=[tana+tan(a-b)]/[1-tana*tan(a-b)]
=(1/3+1/2)/(1-1/3*1/2)
=1
0＜tana＜1　，所以　0 ＜a＜ π/4
tanb=-1/7 ＞－1　,且0＜b＜π　
所以　3π/4 ＜b＜π　，-π＜(2a-b)＜-π/4
所以　2A-B = -3π/4

Answer 2

tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b))=1/2
(tan(a)+1/7)/(1+1/7*tan(a))=1/2
2tan(a)+2/7=1+1/7tan(a)
13/7tan(a)=5/7
tan(a)=5/13
tan(2a-b)=tan(a+a-b)=tan(a)tan(a-b)/(1-tan(a)tan(a-b))=5/13*1/2/(1-5/13*1/2)=5/21
2a-b=arctg(5/21)