已知函数f(X)=sin(x+π\3)-根号3cos(2π\3-x)(1)求函数最小正周期.(2)求函数在区间【-π\2
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f(x)=sin(x+π/3)-√3cos(2π/3-x)=sin(x+π/3)+√3cos[π-(2π/3-x)]

=sin(x+π/3)+√3cos(x+π/3)=2[1/2×sin(x+π/3)+√3/2×cos(x+π/3)]

=2sin[(x+π/3)+π/3]=2sin[(x+2π/3)

(1)T=2π

(2)∵x∈[﹣π/2,π/2] ∴x+2π/3∈[π/6,7π/6] ∴sin[(x+2π/3)∈[﹣1/2,1]

∴f(x)∈[﹣1,2]

∴最大值为2,最小值为﹣1

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