作业帮已知函数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）
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![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）](/uploads/image/z/3938895-63-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x-%CF%80%2F3%29%2B2sin%28x-%CF%80%2F4%29%E6%B1%82%E5%87%BD%E6%95%B0%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C%CF%80%2F2%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F%E6%B1%82%E5%87%BD%E6%95%B0%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E6%89%93%E9%94%99%E4%BA%86...f%EF%BC%88x%EF%BC%89%3Dcos%282x-%CF%80%2F3%29%2B2sin%28x-%CF%80%2F4%29sin%EF%BC%88x%2B%CF%80%2F4%EF%BC%89)
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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）
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）
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）
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