Question

已知函数f（x）=4cossin（x+π/6）-1

已知函数f（x）=4cossin（x+π/6）-1
1）求f（x）的最小正周期
2）求f（x）在区间上【-π/6,π/4】的最大值和最小值

Answer 1

f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=4×√3/2sinxcosx+4×1/2×cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
由2kπ+π/2≤2x+π/6≤2kπ+3π/2
得 kπ+π/6≤x≤kπ+2π/3，k∈Z
∴f(x)单调递减区间为 [kπ+π/6,kπ+2π/3]，k∈Z

Answer 2

不知