已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1(1)求函数的最小正周期 (2)求f(x)在区间{-π/4,π/4}上的最大值和最小值,

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已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1(1)求函数的最小正周期 (2)求f(x)在区间{-π/4,π/4}上的最大值和最小值,
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1
(1)求函数的最小正周期 (2)求f(x)在区间{-π/4,π/4}上的最大值和最小值,

已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1(1)求函数的最小正周期 (2)求f(x)在区间{-π/4,π/4}上的最大值和最小值,
f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1
=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x
=2sin2xcosπ/3+cos2x
=sin2x+cos2x
=√2*(√2/2*sin2x+√2/2*cos2x)
=√2*(sin2xcosπ/4+cos2xsinπ/4)
=√2*sin(2x+π/4)
T=2π/2=π
x∈[-π/4,π/4]
2x∈[-π/2,π/2]
2x+π/4∈[-π/4,3π/4]
-1<=√2*sin(2x+π/4)<=√2
f(x)在区间[-π/4,π/4]上的最大值为:√2
f(x)在区间[-π/4,π/4]上的最小值为:-1

f(x)=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x(sin(α±β)=sinα·cosβ±cosα·sinβ ,cos(2α)==2cos^2α-1)
=sin2x+cos2x
=√2*(sin(2x+π/4))
所以 周期是π
因,x属于[-π/4,π/4]
所以,...

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f(x)=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x(sin(α±β)=sinα·cosβ±cosα·sinβ ,cos(2α)==2cos^2α-1)
=sin2x+cos2x
=√2*(sin(2x+π/4))
所以 周期是π
因,x属于[-π/4,π/4]
所以,2x+π/4属于[-π/4,3π/4]
所以,√2*sin(2x+π/4)属于[-1,√2]

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