tan(a+b)=3,tan(a-b)=1/2,求sin2a/sin2b

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/08 20:28:53

tan(a+b)=3,tan(a-b)=1/2,求sin2a/sin2b
tan(a+b)=3,tan(a-b)=1/2,求sin2a/sin2b

tan(a+b)=3,tan(a-b)=1/2,求sin2a/sin2b
tan(a+b)tan(a-b)=-1
tan^2a-tan^2b=-(1-tan^2atan^2b)
tan^2a+1=tan^2b(1+tan^2a)
(tan^2a+1)(tan^2b-1)=0
tan^2b=1
tanb=1,或tanb=-1
tana+tanb=-2(1-tanatanb)
所以tana=3,或tana=-3
sin2a/sin2b=sinacosa/sinbcosb=(tana/tanb)*(cos^2a/cos^2b)
=3*(1+tan^2b)/(1+tan^2a)=3*2/10=3/5

tan(a+b)=3,tan(a-b)=1/2,
tan2a=tan(a+b+a-b)=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]=(7/2)/(1-3/2)=-7
sin2a=-7/5√2,或7/5√2
同理
tan2b=tan(a+b-a+b)=[tan(a+b)-tan(a-b)]/[1+tan(a+b)tan(a-b)]=(5/2)/(1+3/2)=1
sin2b=√2/2,-√2/2
sin2a/sin2b=7/5或-7/5