已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ,对于x∈R都有f(x)≥0成立,θ∈(0,3π/2),且tan2θ=-3/4,求cosθ的值
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![已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ,对于x∈R都有f(x)≥0成立,θ∈(0,3π/2),且tan2θ=-3/4,求cosθ的值](/uploads/image/z/14336699-59-9.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3Dsin%28%CE%B8%2Bx%29%2Bsin%28%CE%B8-x%29-2sin%CE%B8%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3Dsin%28%CE%B8%2Bx%29%2Bsin%28%CE%B8-x%29-2sin%CE%B8%2C%E5%AF%B9%E4%BA%8Ex%E2%88%88R%E9%83%BD%E6%9C%89f%EF%BC%88x%EF%BC%89%E2%89%A50%E6%88%90%E7%AB%8B%2C%CE%B8%E2%88%88%280%2C3%CF%80%2F2%EF%BC%89%2C%E4%B8%94tan2%CE%B8%3D-3%2F4%2C%E6%B1%82cos%CE%B8%E7%9A%84%E5%80%BC)
已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ,对于x∈R都有f(x)≥0成立,θ∈(0,3π/2),且tan2θ=-3/4,求cosθ的值
已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ
已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ,对于x∈R都有f(x)≥0成立,θ∈(0,3π/2),且tan2θ=-3/4,求cosθ的值
已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ已知f(x)=sin(θ+x)+sin(θ-x)-2sinθ,对于x∈R都有f(x)≥0成立,θ∈(0,3π/2),且tan2θ=-3/4,求cosθ的值
f(x)=sin(θ+x)+sin(θ-x)-2sinθ
=2sinθcosx-2sinθ
=2sinθ(cosx-1)>=0
cosx-1>=0
sinθ>=0
0<θ<3π/2
0<θ<=π
tan2θ=2tanθ/(1-tan^2 θ)=-3/4
tanθ=-1/3 or tanθ=3
1+tan^2 θ=1/cos^2 θ
cosθ=±3√10/10 or cosθ=±√10/10
f(x)=sin(θ+x)+sin(θ-x)-2sinθ
=2sinθcosx-2sinθ
=2sinθ(cosx-1)
∵cosx-1∈[-2,0] ∴sinθ θ∈(0,3π/2) 即θ∈(π,3π\2)
θ∈(π,3π/2),且tan2θ=-3/4
2θ∈(2π,3π) sin2θ\cos2θ=-3\4 sin2θ=3\5 cos2...
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f(x)=sin(θ+x)+sin(θ-x)-2sinθ
=2sinθcosx-2sinθ
=2sinθ(cosx-1)
∵cosx-1∈[-2,0] ∴sinθ θ∈(0,3π/2) 即θ∈(π,3π\2)
θ∈(π,3π/2),且tan2θ=-3/4
2θ∈(2π,3π) sin2θ\cos2θ=-3\4 sin2θ=3\5 cos2θ=-4\5
2tanθ\(1-tanθ^2)=-3\4 tanθ=3 or tanθ=-1\3(舍去 因为θ在第三象限)
tanθ=3 cosθ=-√10\10
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