已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1 已知函数f(x)=2cos(2x+2π/3)+√3sin2x
\u5df2\u77e5\u51fd\u6570f x=sin(\u5140/2-x)sinx-\u6839\u53f73cos^2x\u4eb2\uff0c\u7f51\u53cb\uff0c\u60a8\u8bf4\u7684\u662f\u4e0d\u662f\u4e0b\u9762\u7684\u95ee\u9898\uff1a
\u5df2\u77e5\u51fd\u6570f x=sin(\u5140/2-x)sinx-\u6839\u53f73cos^2x\uff0c\u6c42\u5468\u671f\u3001\u6700\u503c\u3002
f(x)=1/2 sin2x-\u221a3/2(cos2x+1)
=sin(2x-\u03c0/3)-\u221a3/2
T=2\u03c0/2=\u03c0\u3002
f max=1-\u221a3/2, f min=-1-\u221a3/2.
\u9001\u60a8 2015 \u590f\u797a \u51c9\u5feb
\u89e3\u7531f(x)=2cos(2x+2\u03c0/3)+\u221a3sin2x
=2cos2xcos(2\u03c0/3)-2sin2xsin2\u03c0/3+\u221a3sin2x
=2(-1/2)cos2x-2(\u221a3/2)sin2x+\u221a3sin2x
=-cos2x
\u6545\u51fd\u6570\u7684\u5468\u671fT=2\u03c0/2=\u03c0
\u51fd\u6570\u7684\u6700\u5927\u503c\u4e3a1.
=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，2x+π/4属于［-π/4，3π/4］
所以，√2*sin(2x+π/4)属于［-1，√2］
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