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)=
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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)=