Question

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

Answer 1

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

Answer 2

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

追问

好像 你这个 第一步就不对

追答

tan(a+b)=(tana+tanb)/(1-tanatanb)
tan(a-b)=(tana-tanb)/(1+tanatanb)
tan(a+b)tan(a-b)=-1
((tana+tanb)/(1-tanatanb))*((tana-tanb)/(1+tanatanb))=-1
tan^2a-tan^2b=-(1-tan^2atan^2b)

追问

据说... -1*1/2=-1/2 的吧

Answer 3

sin2a = sin[(a+b)+(a-b)] = sin(a+b)cos(a-b) +cos(a+b)sin(a-b)
sin2b= sin[(a+b)-(a-b)] = sin(a+b)cos(a-b) -cos(a+b)sin(a-b)
则，
sin2a/sin2b=[sin(a+b)cos(a-b) +cos(a+b)sin(a-b)]/[ sin(a+b)cos(a-b) -cos(a+b)sin(a-b)]
分子分母同时除以cos(a+b)cos(a-b)，得
=[tan(a+b)+tan(a-b)]/[tan(a+b)-tan(a-b)]
=(-1+1/2)/(-1-1/2)
=3