Question

已知sin阿法=3/5，阿法属于(派/2，派)，tan (派-被打)=1/2，求tan (阿法-2贝打)的值

Answer 1

因sin阿法=3/5，阿法属于(派/2，派)，所以cosa=-4/5,所以tana=-3/4
tana=2tan(a/2)/(1-tan²a)=-3/4,阿法属于(派/2，派)，所以a/2∈（π/4,π/2)，解得tan(a/2)=3
tan (派-被打)=1/2,所以tanb=-1/2,所以tan(a/2 -b)=-7,故tan(a-2b)=tan2(a/2-b)=7/24

Answer 2

sin阿法=3/5，阿法属于(派/2，派)， cosa=-4/5
tana=sina/cosa=-3/4
tan (派-被打)=1/2， tanβ=-1/2
tan2β=2tanβ/(1-tan^2β)=-1/(1-1/4)=-4/3
tan (阿法-2贝打)=(tana-tan2β)/(1+tana*tan2β)
=(-3/4+4/3)/(1+1)
=7/24

Answer 3

α在第二象限
所以cosα<0
sin²α+cos²α=1
所以cosα=-4/5
tanα=sinα/cosα=-3/4

tan（π-β）=-tanβ=1/2
tanβ=-1/2
tan2β=2tanβ/(1-tan²β)=-4/3

所以原式=(tanα-tan2β)/(1+tanαtan2β)=7/24

Answer 4

tan(a-2b)=tan[2(a-b)-a]={tan[2(a-b)]-tan(a)}/{1+tan[2(a-b)]*tan(a)},sina=3/5=>cosa=-4/5,tana=-3/4,tan2(a-b)=2tan(a-b)/[1+tan(a-b)^2]=4/5,所以tan(a-2b)={4/5+3/4}/{1-3/5}=1/32

Answer 5

sina=3/5
cosa=-4/5
tana=-3/4
tan(π-b)=-tanb=1/2
tanb=-1/2
tan2b=2tanb/(1-tan^2b)=-1/3
tan(a-2b)=(tana-tan2b)/(1+tanatan2b)=(-3/4+1/3)/(1+1/4)
=-1/3