求详解,一步步来,对于我这种数学渣渣1.已知α∈(0,π/2),β∈(π/2,π),sin(α+β)=-3/5,cosβ=-5/13,则sinα的值等于________2.函数f(x)=cos2x-6cosx+1的值域是____________3.已知π
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![求详解,一步步来,对于我这种数学渣渣1.已知α∈(0,π/2),β∈(π/2,π),sin(α+β)=-3/5,cosβ=-5/13,则sinα的值等于________2.函数f(x)=cos2x-6cosx+1的值域是____________3.已知π](/uploads/image/z/4995930-66-0.jpg?t=%E6%B1%82%E8%AF%A6%E8%A7%A3%2C%E4%B8%80%E6%AD%A5%E6%AD%A5%E6%9D%A5%2C%E5%AF%B9%E4%BA%8E%E6%88%91%E8%BF%99%E7%A7%8D%E6%95%B0%E5%AD%A6%E6%B8%A3%E6%B8%A31.%E5%B7%B2%E7%9F%A5%CE%B1%E2%88%88%EF%BC%880%2C%CF%80%2F2%EF%BC%89%2C%CE%B2%E2%88%88%EF%BC%88%CF%80%2F2%2C%CF%80%EF%BC%89%2Csin%EF%BC%88%CE%B1%2B%CE%B2%EF%BC%89%3D-3%2F5%2Ccos%CE%B2%3D-5%2F13%2C%E5%88%99sin%CE%B1%E7%9A%84%E5%80%BC%E7%AD%89%E4%BA%8E________2.%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dcos2x-6cosx%2B1%E7%9A%84%E5%80%BC%E5%9F%9F%E6%98%AF____________3.%E5%B7%B2%E7%9F%A5%CF%80)
求详解,一步步来,对于我这种数学渣渣1.已知α∈(0,π/2),β∈(π/2,π),sin(α+β)=-3/5,cosβ=-5/13,则sinα的值等于________2.函数f(x)=cos2x-6cosx+1的值域是____________3.已知π
求详解,一步步来,对于我这种数学渣渣
1.已知α∈(0,π/2),β∈(π/2,π),sin(α+β)=-3/5,cosβ=-5/13,则sinα的值等于________
2.函数f(x)=cos2x-6cosx+1的值域是____________
3.已知π
求详解,一步步来,对于我这种数学渣渣1.已知α∈(0,π/2),β∈(π/2,π),sin(α+β)=-3/5,cosβ=-5/13,则sinα的值等于________2.函数f(x)=cos2x-6cosx+1的值域是____________3.已知π
由sinβ=-12/13且β∈(-π/2,0),得cosβ = 5/13
由α∈(π/2,π),β∈(-π/2,0),得2α-β∈(π,5π/2)
又sin(2α-β) = 3/5 > 0,所以2α-β∈(2π,5π/2)
sin(2α-β) = sin(2α)cosβ - cos(2α)sinβ = 5/13sin(2α) + 12/13cos(2α) = 3/5
cos(2α-β) = cos(2α)cosβ + sin(2α)sinβ = 5/13cos(2α) - 12/13sin(2α) = 4/5
解得cos(2α) = 56/65
sin²α = (1/2)(1 - cos(2α)) = 9 / 130
sinα = 3/√130
f(x)=(2cos²x-1)-6cosx+1
=2cos²x-6cosx
=2(cosx-3/2)²-9/2
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