Question

已知sin(a+b)等于2/3，sin(a-b0 =1/5,求tana:tanb

Answer 1

已知sin(a+b)等于2/3，sin(a-b0 =1/5
sin(a+b)/sin(a-b)
=(sina*cosb+cosa*sinb)/(sina*cosb-cosa*sinb)
分式上下同除以cosa*cosb
=(tana+tanb)/(tana-tanb)
=(tana/tanb+1)/(tana/tanb-1)
=(2/3)/(1/5)
所以tana/tanb=7/13

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Answer 2

sin(a+b)=sinacosb+cosasinb=2/3
sin(a-b)=sinacosb-cosasinb=1/5
两者相加：2sinacosb=13/15 sinacosb=13/30
两者相减：2cosasinb=7/15 cosasinb=7/30
tana:tanb=sinacosb:cosasinb=13/30:7/30=13:7