已知sin(A B)=2/3,sin(A-B)=3/4,求(tan(a b)-tana-tanb)/(tan^2b*tan(a b))
幫下忙網友廻答：
- sin(A B)=sinAcosB cosAsinB=2/3(1)
sin(A-B)=sinAcosB-cosAsinB=3/4(2)
(1) (2)
2sinAcosB=17/12,sinAcosB=17/24 (3)
(1)-(2)
2cosAsinB=-1/12,cosAsinB=-1/24 (4)
tan(a b)-tana-tanb=tan(a b)-(tana tanb)=tan(a b)-[tan(a b)*(1-tanatanb]=tana*tanb*tan(a b)
(tan(a b)-tana-tanb)/(tan^2b*tan(a b)) =tana*tanb*tan(a b)/(tan^2b*tan(a b)) =tana*tanb/tan^2b
=tana/tanb
=sinacosb/cosasinb
=(3)/(4)
=-17
admin=23,sin(A-B)=34,求(tan(a+b)-tana-tanb)(tan^2b*tan(a+b))幫下忙.jpg "已知sin(A+B)=23,sin(A-B)=34,求(tan(a+b)-tana-tanb)(tan^2b*tan(a+b))幫下忙,第1張")
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