已知α∈(π/2,π),β∈(π/2,3π/2),sinα=2√2/3,sinβ=7/9,则cos(α-β)=
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![已知α∈(π/2,π),β∈(π/2,3π/2),sinα=2√2/3,sinβ=7/9,则cos(α-β)=](/uploads/image/z/14410187-35-7.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E2%88%88%28%CF%80%2F2%2C%CF%80%29%2C%CE%B2%E2%88%88%28%CF%80%2F2%2C3%CF%80%2F2%29%2Csin%CE%B1%3D2%E2%88%9A2%2F3%2Csin%CE%B2%3D7%2F9%2C%E5%88%99cos%28%CE%B1-%CE%B2%29%3D)
已知α∈(π/2,π),β∈(π/2,3π/2),sinα=2√2/3,sinβ=7/9,则cos(α-β)=
已知α∈(π/2,π),β∈(π/2,3π/2),sinα=2√2/3,sinβ=7/9,则cos(α-β)=
已知α∈(π/2,π),β∈(π/2,3π/2),sinα=2√2/3,sinβ=7/9,则cos(α-β)=
因为α∈(π/2,π),sinα=2√2/3,所以cosα=-1/3
因为),β∈(π/2,3π/2),sinβ=7/9所以cosβ=-4√2/9
cos(α-β)= cosαcosβ+sinαsinβ =2√2/3