已知sinx+cosx=1/3,则sinxcosx=已知tanx=2,则sinx+cosx/sinx-cosx= sin^2x-cos^2x+sinxcosx=化简2sin^2x-1/1-2cos^2x=若sin(π/6-x)=a,则cos(2π/3-x)=已知sinα=3/5,α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),求cos(α-β)=

已知sinx+cosx=1/3,则sinxcosx=已知tanx=2,则sinx+cosx/sinx-cosx= sin^2x-cos^2x+sinxcosx=化简2sin^2x-1/1-2cos^2x=若sin(π/6-x)=a,则cos(2π/3-x)=已知sinα=3/5,α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),求cos(α-β)=
已知sinx+cosx=1/3,则sinxcosx=
已知tanx=2,则sinx+cosx/sinx-cosx= sin^2x-cos^2x+sinxcosx=
化简2sin^2x-1/1-2cos^2x=
若sin(π/6-x)=a,则cos(2π/3-x)=
已知sinα=3/5,α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),求cos(α-β)=

已知sinx+cosx=1/3,则sinxcosx=已知tanx=2,则sinx+cosx/sinx-cosx= sin^2x-cos^2x+sinxcosx=化简2sin^2x-1/1-2cos^2x=若sin(π/6-x)=a,则cos(2π/3-x)=已知sinα=3/5,α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),求cos(α-β)=
已知sinx+cosx=1/3,则(sinx+cosx)²=1/9 得1+2sinxcosx=1/9
sinxcosx=-4/9
已知tanx=2,把所求式子的分子父母同除cosx 则sinx+cosx/sinx-cosx= tanx+1/tanx-1=3
sin^2x-cos^2x+sinxcosx=（sin^2x-cos^2x+sinxcosx）/1
=(sin^2x-cos^2x+sinxcosx)/（sin²x+cos²x）分子分母同除cos²x
=（tan²x-1+tanx）/tan²x+1=（4-1+2）/(4+1)=1
化简2sin^2x-1/1-2cos^2x=（sin²x-cos²x）/（sin²x-cos²x）=-1
若sin(π/6-x)=a,则cos(2π/3-x)=cos（π/2+(π/6-x)）=-sin(π/6-x)=-a
已知sinα=3/5,α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),
α在第二象限,cosα=-4/5; β在第四象限,所以sinβ=-12/13
求cos(α-β)=cosαcosβ-sinαsinβ=-4/5 ×5/13-（3/5×-12/13）=-20/65+36/65=16/65

已知sinα=3/5，α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),求cos(α-β)=