Question

已知sin(π/4+x)=3/5,5π/12<x<3π/4,求cos2x+tanx的值 在线等！！！完整过程！！！

Answer 1

,5π/12<x<3π/4所以2π/3<x+π/4<π，又sin(π/4+x)=3/5，所以 cos(π/4+x)=-4/5，
cos2x=sin(π/2+2x)=2sin(π/4+x)xos(π/4+x)=2*3/5*(-4/5)=-24/25
又2π/3<x+π/4<π，sin(π/4+x)=3/5<√2/2
所以3π/4<x+π/4<π，从而　π/2<x<3π/4
所以 π<2x<3π/2
所以 sin2x=-√[-1-(-24/25)^2]=-7/25
所以 tanx=sinxcosx/(cosx)^2=sin2x/(1+cos2x)=(-7/25)/(1-24/25)=-7
故　cos2x+tanx=-24/25-7=-199/25

Answer 2

sin(π/4+x)=sinx+cosx=3/5
(sinx)^2+(cosx)^2=1
连理方程，解得sinx=4/5 ,cosx=-1/5 (5π/12<x<3π/)
cos2x+tanx=(cosx)^2-(sinx)^2+tanx=-23/5