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求F(X)=SIN(2X-π/4)-2√2SIN^2A的最小正周期及值域
人气:342 ℃ 时间:2020-09-27 22:19:21
解答
函数是下面的吧?F(x)=sin(2x-π/4)-2√2(sinx)^2=sin2xcos(π/4)-cos2xsin(π/4)-√2(1-cos2x)=(√2/2)sin2x-(√2/2)cos2x+√2cos2x-√2=(√2/2)sin2x+(√2/2)cos2x-√2=sin(2x+π/4)-√2所以最小正周期是T=2π/2=π...
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