大师作文网求详解,一步步来,对于我这种数学渣渣
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/12 15:12:54
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
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
-1
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
-1