已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1

已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1

题目
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,
答案
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
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