Question

已知sin(a+β)=2/3,sin(a-β)=2/5

已知sin(a+β)=2/3,sin(a-β)=2/5，则tana/tanβ的值是？

Answer 1

sin（a+b）=2/3
sinacosb+cosasinb=2/3 (1)
sin（a-b）=2/5
sinacosb-cosasinb=2/5 (2)

(1)+(2),2sinacosb=16/15
(1)-(2),2cosasinb=-4/15

tana/tanb =2sinacosb/2cosasinb=-4

Answer 2

tana/tanβ

= sinacosβ/cosasinβ

sin(a+β)=2/3

sinacosβ+cosasinβ=2/3

sin(a-β)=2/5

sinacosβ-cosasinβ=2/5

2sinacosβ=2/3+2/5, sinacosβ=1/3+1/5=8/15

cosasinβ=1/3-1/5=2/15

tana/tanβ=sinacosβ/cosasinβ=(8/15)/(2/15)=4