若sin(π/4+x)=5/13且x∈(π/4,3π/4),则(1-tanx)/(1+tanx)=
问题描述：
若sin(π/4+x)=5/13且x∈(π/4,3π/4),则(1-tanx)/(1+tanx)=
答
(1-tanx)/(1+tanx)=(cosx-sinx)/(cosx+sinx)=(cosx^2-sinx^2)/sin2x=cos2x/sin2x=cot2x
sin(π/4+x)=5/13
根号2/2(sinx+cosx)=5/13
sinx+cosx=5根号2/13
sin2x=-119/169
x∈(π/4,3π/4)
2x∈(π/2,3π/2)
cos2x=-根号(1-119^2/169^2)=120/169
原式=cot2x=120/119
答
sin(π/4+x)=5/13
cos(π/4+x)=-12/13
sinx=sin(π/4+x-π/4)
=sin(π/4+x)cos(π/4)-cos(π/4+x)sin(π/4)
=5/13*√2/2+12/13*√2/2
=17√2/26
cosx=-7√2/26
tanx=-17/7
(1-tanx)/(1+tanx)=-12/5