当 θ∈(0,π/2)时,求证sinθ +cosθ <π/2
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当 θ∈(0,π/2)时,求证sinθ +cosθ <π/2
如题
如题
![当 θ∈(0,π/2)时,求证sinθ +cosθ <π/2](/uploads/image/z/18510037-61-7.jpg?t=%E5%BD%93+%CE%B8%E2%88%88%EF%BC%880%2C%CF%80%2F2%29%E6%97%B6%2C%E6%B1%82%E8%AF%81sin%CE%B8+%2Bcos%CE%B8+%EF%BC%9C%CF%80%2F2)
sinθ+cosθ
=√2(√2/2*sinθ+√2/2cosθ)
=√2(sinθcosπ/4+cosθsinπ/4)
=√2sin(θ+π/4)
θ∈(0,π/2)
π/4
=√2(√2/2*sinθ+√2/2cosθ)
=√2(sinθcosπ/4+cosθsinπ/4)
=√2sin(θ+π/4)
θ∈(0,π/2)
π/4
求证-sin(θ-π/2)=cosθ
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
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