已知函数f(x)=4sinx/2cos(x/2-π/6)求单调递减区间及对称中心以及在区间[π/2,4π/3]上的值域分别是

Question

已知函数f(x)=4sinx/2cos(x/2-π /6)求单调递减区间及对称中心以及在区间[π/2,4π/3]上的值域分别是

Answer 1

sinα*sinβ=[cos(α-β)-cos(α+β)]/2
cos(x/2-π/6)=sin(x/2-π/6+π/2)=sin(x/2+π/3)
f(x)=4sinx/2sin(x/2+π/3)
=4*[cosπ/3-cos(x+π/3)]/2
=2*[1/2-cos(x+π/3)]
=-2cos(x+π/3)+1
=2cos(x-2π/3)+1
x-2π/3∈[2kπ,π+2kπ]单调减,即减区间为[2π/3+2kπ,5π/3+2kπ]
(x-2π/3∈[-π+2kπ,2kπ]单调增,即增区间为[-π/3+2kπ,2π/3+2kπ])
对称中心x-2π/3=π/2+kπ 即x=π/6+kπ k∈z
区间[π/2,4π/3],x-2π/3∈[-π/6,2π/3] cos(x-2π/3)∈[-1/2,1] 2cos(x-2π/3)∈[-1,2]
在区间[π/2,4π/3],值域为[0,3]