问题标题:
函数y=【cos(3x+π/4)-sin(3x+π/4)】/[cos(3x+π/4)+sin(3x+π/4)]的定义域,最小正周期、值域是?
问题描述:
函数y=【cos(3x+π/4)-sin(3x+π/4)】/[cos(3x+π/4)+sin(3x+π/4)]的定义域,最小正周期、值域是?
江祥春回答:
上下同除以cos(3x+π/4)得:原式=【1-tan(3x+π/4)】/[1+tan(3x+π/4)]
用tan的和分角公式,得原式=[1-tan3x-1-tan3x]/[1-tan3x+1+tan3x]
化简,得:原式=-2tan3x/2,即,原式=-2tan3x
∴,定义域为[-π/6+kπ/3,π/6+kπ/3]
最小正周期π/3
值域为R
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