求函数y=sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期 值域
人气:108 ℃ 时间:2019-10-19 04:40:46
解答
y=sin^4x+cos^4x+4sin^2xcos^2x-1=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=sin^2(2x)/2=(1-cos4x)/4=1/4-cos4x/4周期T=2π/4=π/2值域是:[0,1/2]
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