y=cos(π/4-2x)=cos(2x-π/4)
∵y=cosx在[2kπ,(2k+1)π]上为减函数
∴2kπ≤2x-π/4≤(2k+1)π
∴π/8+kπ≤x≤5π/8+kπ
∴y=cos(π/4-2x)的单调递减区间为[π/8+kπ,5π/8+kπ],k∈Z第一步不理解,能解释下,根据cos(-α)=cos(α)即两角互为相反数时,对应的余弦值相等,π/4-2x+2x-π/4=0∴cos(π/4-2x)=cos(2x-π/4)也可以根据y=cosx为偶函数得出cos(π/4-2x)=cos(2x-π/4
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