Question

已知0＜a＜90，90＜b＜180，且tan（a/2）=1/2，sin（a+b）=5/13。1.分别求cosa和cosb的值。2.求tan（a-b）

Answer 1

1) tana=2tan(a/2)/[1-(tan(a/2))^2]=1/(1-1/4)=4/3
(sina)^2+(cosa)^2=1
(tana)^2+1=1/(cosa)^2
cosa=3/5
sina=4/5

2) sin(a+b)=5/13
cos(a+b)=-√[1-(sin(a+b))^2]=-12/13
sinb=sin(a+b-a)=sin(a+b)cosa-cos(a+b)sina=5/13*3/5+12/13*4/5=63/65
cosb=-√[1-(sinb)^2]=-16/65
tanb=-63/16

tan(a-b)=(tana-tanb)/(1+tanatanb)=(4/3+63/16)/(1-4/3*63/16)=-253/204

Answer 2

tan（a/2）=1/2，sin（a+b）=5/13
(sina)^2+(cosa)^2=1
(tana)^2+1=1/(cosa)^2
cosa=3/5
sina=4/5
cos(a+b)=-√[1-(sin(a+b))^2]=-12/13
sinb=sin(a+b-a)=sin(a+b)cosa-cos(a+b)sina=5/13*3/5+12/13*4/5=63/65
cosb=-√[1-(sinb)^2]=-16/65
tanb=-63/16
tan(a-b)=-253/204