已知f(x)=cos^(x+θ)-2cosθcosxcos(x+θ)+cos^θ求f(x)的最大值、最小值和最小正周期!
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![已知f(x)=cos^(x+θ)-2cosθcosxcos(x+θ)+cos^θ求f(x)的最大值、最小值和最小正周期!](/uploads/image/z/14341835-11-5.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dcos%5E%28x%2B%CE%B8%29-2cos%CE%B8cosxcos%28x%2B%CE%B8%29%2Bcos%5E%CE%B8%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E3%80%81%E6%9C%80%E5%B0%8F%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%21)
已知f(x)=cos^(x+θ)-2cosθcosxcos(x+θ)+cos^θ求f(x)的最大值、最小值和最小正周期!
已知f(x)=cos^(x+θ)-2cosθcosxcos(x+θ)+cos^θ求f(x)的最大值、最小值和最小正周期!
已知f(x)=cos^(x+θ)-2cosθcosxcos(x+θ)+cos^θ求f(x)的最大值、最小值和最小正周期!
f(x)=cos(x+θ)[cos(x+θ)-2cosθcosx]+cos^2θ
=-cos(x+θ)[sinxsinθ+cosxcosθ]+cos^2θ
=-cos(x+θ)cos(x-θ)+cos^2θ
=-(cosx)/2-(cosθ)/2+cos^2θ
f(x)的最大值1+cos^2θ-(cosθ)/2、最小值-1+cos^2θ-(cosθ)/2,最小正周期2π
cos²x+cos²(x+θ)-2cosxcosθcos(x+θ)
=cos²x+cos(x+θ)[cos(x+θ)-2cosxcosθ]
=cos²x+cos(x+θ)(-cosxcosθ-sinxsinθ)
=cos²x-(cosxcosθ-sinxsinθ)(cosxcosθ+sinxsinθ)
=cos...
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cos²x+cos²(x+θ)-2cosxcosθcos(x+θ)
=cos²x+cos(x+θ)[cos(x+θ)-2cosxcosθ]
=cos²x+cos(x+θ)(-cosxcosθ-sinxsinθ)
=cos²x-(cosxcosθ-sinxsinθ)(cosxcosθ+sinxsinθ)
=cos²x-cos²xcos²θ+sin²xsin²θ
=cos²x(1-cos²θ)+sin²xsin²θ
=cos²xsin²θ+sin²xsin²θ
=sin²θ
fx=sinx的平方,之后就好解了
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