α,β∈(0,π\2),cos(α+β)=12\13,cos(2α+β)=3\5,则cosβ=
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![α,β∈(0,π\2),cos(α+β)=12\13,cos(2α+β)=3\5,则cosβ=](/uploads/image/z/14770445-5-5.jpg?t=%CE%B1%2C%CE%B2%E2%88%88%EF%BC%880%2C%CF%80%5C2%EF%BC%89%2Ccos%28%CE%B1%2B%CE%B2%EF%BC%89%3D12%5C13%2Ccos%282%CE%B1%2B%CE%B2%29%3D3%5C5%2C%E5%88%99cos%CE%B2%3D)
α,β∈(0,π\2),cos(α+β)=12\13,cos(2α+β)=3\5,则cosβ=
α,β∈(0,π\2),cos(α+β)=12\13,cos(2α+β)=3\5,则cosβ=
α,β∈(0,π\2),cos(α+β)=12\13,cos(2α+β)=3\5,则cosβ=
sin(a+b)=5/13
sin(2a+b)=4/5
cosa
=cos[(2a+b)-(a+b)]
=cos(2a+b)cos(a+b)+sin(2a+b)sin(a+b)
=12/13*3/5+5/13*4/5
=56/65