已知函数f(x)=2cosxsin(x+兀/6)+cos^4x-sin^4x.求f(x)的最小正周期,若x属于[-兀/1
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已知函数f(x)=2cosxsin(x+兀/6)+cos^4x-sin^4x.求f(x)的最小正周期,若x属于[-兀/12,兀/6]求f(x)的最大值最小值及相应的x的值.
![已知函数f(x)=2cosxsin(x+兀/6)+cos^4x-sin^4x.求f(x)的最小正周期,若x属于[-兀/1](/uploads/image/z/14979907-19-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2cosxsin%EF%BC%88x%2B%E5%85%80%2F6%EF%BC%89%2Bcos%5E4x-sin%5E4x.%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%2C%E8%8B%A5x%E5%B1%9E%E4%BA%8E%5B-%E5%85%80%2F1)
(1)f(x)=2cosx(sinx*√3/2+cosx*1/2)+(cos²x+sin²x)(cos²x-sin²x)
=√3cosxsinx+cos²x+cos²x-sin²x
=√3/2*sin2x+(1+cos2x)/2+cos²2x
=√3/2*sin2x+3/2*cos2x+1/2
=√3(sin2x*1/2+cos2x*√3/2)+1/2
=√3(sin2x*cosπ/3+cos2xsinπ/3)+1/2
=√3sin(2x+π/3)+1/2
T=2π/2=π
(2)∵x∈[-π/12,π/6]
∴2x+π/3∈[π/6,2π/3]
∴sin(2x+π/3)∈[1/2,1]
∴f(x)max=√3+1/2,此时x=π/12;
f(x)min=√3/2+1/2,此时x=-π/12
=√3cosxsinx+cos²x+cos²x-sin²x
=√3/2*sin2x+(1+cos2x)/2+cos²2x
=√3/2*sin2x+3/2*cos2x+1/2
=√3(sin2x*1/2+cos2x*√3/2)+1/2
=√3(sin2x*cosπ/3+cos2xsinπ/3)+1/2
=√3sin(2x+π/3)+1/2
T=2π/2=π
(2)∵x∈[-π/12,π/6]
∴2x+π/3∈[π/6,2π/3]
∴sin(2x+π/3)∈[1/2,1]
∴f(x)max=√3+1/2,此时x=π/12;
f(x)min=√3/2+1/2,此时x=-π/12
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