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f(x)=sin^2(2x-π/4)的单调减区间
人气:146 ℃ 时间:2020-02-06 05:33:43
解答
f(x)=sin^2(2x-π/4)= (1-cos(4x-π/2)/2=1/2-(cos(4x-π/2))/2当2kπ-π≤4x-π/2≤2kπ时,即kπ/2-π/8≤x≤kπ/2+π/8时f(x)单调递减, 当2kπ≤4x-π/2≤2kπ+π时,即kπ/2+π/8≤x≤kπ/2+3π/8时f(x)单调递增
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