设cos(α-β/2)=-1/9 sin(α/2-β)=2/3 α∈(π/2,π)β∈(0,π/2)求cos(α+β)

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设cos(α-β/2)=-1/9 sin(α/2-β)=2/3 α∈(π/2,π)β∈(0,π/2)求cos(α+β)
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设cos(α-β/2)=-1/9 sin(α/2-β)=2/3 α∈(π/2,π)β∈(0,π/2)求cos(α+β)
设cos(α-β/2)=-1/9 sin(α/2-β)=2/3 α∈(π/2,π)β∈(0,π/2)求cos(α+β)

设cos(α-β/2)=-1/9 sin(α/2-β)=2/3 α∈(π/2,π)β∈(0,π/2)求cos(α+β)
提示:把α+β / 2 看做α-β/2与α/2-β的差,用两角差余弦公式求
设cos(x-y/2)=-1/9,sin(x/2-y)=2/3,且0sin(2x-y)=8√5/81
sin(x/2-y)=2/3---->cos(x-2y)=1/9------>sin(x-2y)=4√5/9
cos(x+y)=cos[(2x-y)-(x-2y)]=(-79/81)(1/9)+(8√5/81)(4√5/9)=1/9