题目

已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)求函数在区间[0,π/2]上的值域
求函数单调区间
打错了...f（x）=cos(2x-π/3)+2sin(x-π/4)sin（x+π/4）

提问时间：2020-11-02

答案

f(x)＝cos(2x-π/3)+2sin(x-π/4)sin（x+π/4）＝cos(2x－π/3)＋2sin(x－π/4)cos[π/2－(x＋π/4)]
＝cos(2x－π/3)＋2sin(x－π/4)cos(x－π/4)＝cos(2x－π/3)＋sin(2x－π/2)
＝cos(2x－π/3)＋cos2x＝2cos(2x－π/6)cosπ/6＝√3cos(2x－π/6)
∵x∈[0,π/2] ∴2x－π/6∈[﹣π/6,5π/6] ∴f(x)∈[﹣3/2,√3]
∴当2x－π/6∈[2kπ,2kπ＋π] 即x∈[kπ＋π/12,kπ＋7π/12] 时,单调递增
当2x－π/6∈[2kπ＋π,2kπ＋2π] 即x∈[kπ＋7π/12,kπ＋13π/12] 时,单调递减