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亲,是偶数,数。首先,根据题意可得:f(x+π/2) = f(-x)将x替换为(x-π/4),得到:f(x-π/4+π/2) = f(π/4-x)即:f(x+π/4) = f(π/4-x)这表明函数f(x)是一个奇函数。因此,f(π/4-x) = -f(x-π/4)。将x替换为(π/4-x),得到:f(π/4-(π/4-x)) = f(x)即:f(x) = f(x/2+π/8)将x替换为(π/4+x),得到:f(π/4+x) = f(-x/2+π/8)将上面的等式代入f(x) = Asin(x+∮)中:Asin(x+π/4+∮) = Asin(-x/2+π/8+∮)根据正弦函数的周期性质,可得:x+π/4+∮ = -x/2+π/8+∮+2kπ 或 x+π/4+∮ = π-x/2+π/8+∮+2kπ整理得到:x = 2kπ-π/4-∮ 或 x = 2kπ-5π/4-∮因此,函数f(x)的周期为π/2,且∮=π/8。将x替换为(π/4-x),得到:f(π/4-x) = Asin(π/4+2x+π/8)
咨询记录 · 回答于2023-03-29
已知函数f(x)=Asin(x+∮)有f(x+兀/2)=f(一x)则函数f(兀/4一x)是奇函数或偶函
亲,是偶数,数。首先,根据题意可得:f(x+π/2) = f(-x)将x替换为(x-π/4),得到:f(x-π/4+π/2) = f(π/4-x)即:f(x+π/4) = f(π/4-x)这表明函数f(x)是一个奇函数。因此,f(π/4-x) = -f(x-π/4)。将x替换为(π/4-x),得到:f(π/4-(π/4-x)) = f(x)即:f(x) = f(x/2+π/8)将x替换为(π/4+x),得到:f(π/4+x) = f(-x/2+π/8)将上面的等式代入f(x) = Asin(x+∮)中:Asin(x+π/4+∮) = Asin(-x/2+π/8+∮)根据正弦函数的周期性质,可得:x+π/4+∮ = -x/2+π/8+∮+2kπ 或 x+π/4+∮ = π-x/2+π/8+∮+2kπ整理得到:x = 2kπ-π/4-∮ 或 x = 2kπ-5π/4-∮因此,函数f(x)的周期为π/2,且∮=π/8。将x替换为(π/4-x),得到:f(π/4-x) = Asin(π/4+2x+π/8)
亲,接上段因为sin(π/4+θ) = cos(θ),所以:f(π/4-x) = Acos(2x)因此,函数f(π/4-x)是一个偶数。