已知f(x)=4cosxsin(x+π/6)-1（1）求f（x）的最小正周期（2）求f（x）在区间[-π/6，π/4]上的最大值和最
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1、
f(x)=4cosxsin(x+π/6)-1
=2[sin(x+x+π/6)-sin(x-x-π/6)]-1
=2sin(2x+π/6)+1-1
=2sin(2x+π/6)
最小正周期为：T=2π/2=π
2、当-π/6≤x≤π/4 时有：
-π/6≤2x+π/6≤2π/3
所以可得：x=π/6 时有最大值为：f(π/6)=2sin(2π/6+π/6)=2sin(π/2)=2
当x=-π/6时有最小值为：f(-π/6)=2sin[2x(-π/6)+π/6]=2sin(-π/6)=-1
f(x)=4cosxsin(x+π/6)-1
=2[sin(x+x+π/6)-sin(x-x-π/6)]-1
=2sin(2x+π/6)+1-1
=2sin(2x+π/6)
最小正周期为：T=2π/2=π
2、当-π/6≤x≤π/4 时有：
-π/6≤2x+π/6≤2π/3
所以可得：x=π/6 时有最大值为：f(π/6)=2sin(2π/6+π/6)=2sin(π/2)=2
当x=-π/6时有最小值为：f(-π/6)=2sin[2x(-π/6)+π/6]=2sin(-π/6)=-1
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