作业帮1.已知f(x)=sin(x+θ)+sin（θ-x）-2sinθ对于x∈R都有f（x）≥0成立,θ∈（0,3π/2],且tanθ=3/4,求cosθ/2的值
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![1.已知f(x)=sin(x+θ)+sin（θ-x）-2sinθ对于x∈R都有f（x）≥0成立,θ∈（0,3π/2],且tanθ=3/4,求cosθ/2的值](/uploads/image/z/2666612-20-2.jpg?t=1.%E5%B7%B2%E7%9F%A5f%28x%29%3Dsin%28x%2B%CE%B8%29%2Bsin%EF%BC%88%CE%B8-x%EF%BC%89-2sin%CE%B8%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%EF%BC%880%2C3%CF%80%2F2%5D%2C%E4%B8%94tan%CE%B8%3D3%2F4%2C%E6%B1%82cos%CE%B8%2F2%E7%9A%84%E5%80%BC)
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1.已知f(x)=sin(x+θ)+sin（θ-x）-2sinθ对于x∈R都有f（x）≥0成立,θ∈（0,3π/2],且tanθ=3/4,求cosθ/2的值
1.已知f(x)=sin(x+θ)+sin（θ-x）-2sinθ对于x∈R都有f（x）≥0成立,θ∈（0,3π/2],且tanθ=3/4
,求cosθ/2的值
1.已知f(x)=sin(x+θ)+sin（θ-x）-2sinθ对于x∈R都有f（x）≥0成立,θ∈（0,3π/2],且tanθ=3/4,求cosθ/2的值
f(x)=sinxcosθ+cosxsinθ+sinθcosx-cosθsinx-2sinθ
=2sinθcosx-2sinθ
=2sinθ(1-cosx)
>=0
因为1-cosx>=0
所以sinθ>=0,又
tanθ=3/4>0
θ（0,3π/2]所以θ在（0,π/2）θ/2在（0,π/4)
cosθ=4/5
又cos²θ/2=1/2(1+cosθ)=1/2(1+4/5)=9/10
所以
cosθ/2=3v10/10