tan(a+b)=-1,tan(a-b)=1/2,则sin2a/sin2b=多少

tan(a+b)=-1,tan(a-b)=1/2,则sin2a/sin2b=多少

解：sin2a+sin2b=2sin(a+b)cos(a-b)
sin2a-sin2b=2cos(a+b)sin(a-b)
(sin2a+sin2b)/(sin2a-sin2b)
=[2sin(a+b)cos(a-b)]/[2cos(a+b)sin(a-b)]
=tan(a+b)/tan(a-b)
=-2
sin2a+sin2b=-2sin2a+2sin2b
3sin2a=sin2b
sin2a/sin2b=1/3

sin2a/sin2b

=sin[(a+b)+(a-b)]/sin[(a+b)-(a-b)]

=[sin(a+b)cos(a-b)+cos(a+b)sin(a-b)]/[sin(a+b)cos(a-b)-cos(a+b)sin(a-b)]

=[tan(a+b)+tan(a-b)]/[tan(a+b)-tan(a-b)]

=(3+1/2)/(3-1/2)

=7/5

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

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