已知cos(3π+α/2)=-2/3,则cos2α已知cos(3π+α/2)=-2/3,则cos2α=?
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![已知cos(3π+α/2)=-2/3,则cos2α已知cos(3π+α/2)=-2/3,则cos2α=?](/uploads/image/z/15064215-15-5.jpg?t=%E5%B7%B2%E7%9F%A5cos%283%CF%80%2B%CE%B1%2F2%29%3D-2%2F3%2C%E5%88%99cos2%CE%B1%E5%B7%B2%E7%9F%A5cos%EF%BC%883%CF%80%2B%CE%B1%2F2%EF%BC%89%3D-2%2F3%2C%E5%88%99cos2%CE%B1%3D%3F)
已知cos(3π+α/2)=-2/3,则cos2α已知cos(3π+α/2)=-2/3,则cos2α=?
已知cos(3π+α/2)=-2/3,则cos2α
已知cos(3π+α/2)=-2/3,则cos2α=?
已知cos(3π+α/2)=-2/3,则cos2α已知cos(3π+α/2)=-2/3,则cos2α=?
cos(α+β)=4/5,cos(α-β)=-4/5,α+β∈(7/4π,2π),α-β∈(3/4π,π) sin(α+β)=-3/5,sin(α-β)=3/5, cos2α=cos[(α+β)+(α-β)] =cos(α+β)cos(α-β)-sin(α+β)sin(α-β) =4/5*(-4/5)-(-3/5)*(3/5) =-7/25
cos(3π+α/2)=-2/3
cos(π+α/2)=-2/3
-cosα/2=-2/3
cosα/2=2/3
cosα=2cos²α/2-1=2×4/9-1=-1/9
所以
cos2α=2cos²α-1=2×1/81-1=-79/81
cos(3π+α/2)=cos(π+α/2)=-cosα/2=-2/3
cosα/2=2/3
cosα=2cosα/2*cosα/2-1=2*2/3*2/3-1=-1/9
cos2α=2cosα*cosα-1=2*(-1/9)*(-1/9)-1=-79/81