笨脑袋 死活不懂为什么sinx-cosx=根号下(1-sin2x),则x的取值范围是2sinx-cosx = √(1-sin2x) = √(sin^2x+cos^2x-2sinxcosx) = √(sinx-cosx)^2 = |sinx-cosx|∴sinx-cosx ≥ 0∴√2 (sinxcosπ/4-cosxsinπ/4) ≥ 0∴√2 sin(x-π/4)

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笨脑袋 死活不懂为什么sinx-cosx=根号下(1-sin2x),则x的取值范围是2sinx-cosx = √(1-sin2x) = √(sin^2x+cos^2x-2sinxcosx) = √(sinx-cosx)^2 = |sinx-cosx|∴sinx-cosx ≥ 0∴√2 (sinxcosπ/4-cosxsinπ/4) ≥ 0∴√2 sin(x-π/4)
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笨脑袋 死活不懂为什么sinx-cosx=根号下(1-sin2x),则x的取值范围是2sinx-cosx = √(1-sin2x) = √(sin^2x+cos^2x-2sinxcosx) = √(sinx-cosx)^2 = |sinx-cosx|∴sinx-cosx ≥ 0∴√2 (sinxcosπ/4-cosxsinπ/4) ≥ 0∴√2 sin(x-π/4)
笨脑袋 死活不懂为什么
sinx-cosx=根号下(1-sin2x),则x的取值范围是
2sinx-cosx = √(1-sin2x) = √(sin^2x+cos^2x-2sinxcosx) = √(sinx-cosx)^2 = |sinx-cosx|
∴sinx-cosx ≥ 0
∴√2 (sinxcosπ/4-cosxsinπ/4) ≥ 0
∴√2 sin(x-π/4) ≥ 0
∴ sin(x-π/4) ≥ 0
∴x-π/4 ∈【2kπ,2kπ+π】,其中k∈Z
∴x∈【2kπ+π/4,2kπ+5π/4】,其中k∈Z
完全看不懂 什么四分之派

笨脑袋 死活不懂为什么sinx-cosx=根号下(1-sin2x),则x的取值范围是2sinx-cosx = √(1-sin2x) = √(sin^2x+cos^2x-2sinxcosx) = √(sinx-cosx)^2 = |sinx-cosx|∴sinx-cosx ≥ 0∴√2 (sinxcosπ/4-cosxsinπ/4) ≥ 0∴√2 sin(x-π/4)
cosπ/4和sinπ/4是相等的,因为sinx-cosx ≥ 0,所以第三行就成立了