Question

高一数学题，求助！！只需答案

已知sin(x+5π)=3/5,则sinx

化简：cos(x-π/2)/sin(x+5π/2)·sin(x-2π)·cos(2π-x)=

已知sinα=2根号5/5,α∈(π/2,π)则tanα=

Answer 1

1,sin(x+5π)=-sinx=3/5,则sinx=-3/5
2,cos(x-π/2)/sin(x+5π/2)·sin(x-2π)·cos(2π-x)=sinx/cosx·sinx·cosx=sin²x
3,由α∈(π/2,π),sinα=2√5/5，cosα=-√5/5
tanα=-2
>.<2若为cos(x-π/2)/[sin(x+5π/2)·sin(x-2π)·cos(2π-x)]=sinx/[cosx·sinx·cosx]=1/cos²x

Answer 2

1 sinx=-3/5
2 1/cos²x
3 tanα=-2

Answer 3

-3/5 1/2cos2x -2

Answer 4

-3/5
-(sinx)^2
-2