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函数f(x)=sin^4x+cos^2x的最小正周期
人气:127 ℃ 时间:2019-09-23 09:53:29
解答
f(x)=sin^4x+cos^2x
=1-sin^2x+sin^4x
=sin^2x(sin^2-1)+1
=-sin^2xcos^2x+1
=-(1/2sin2x)^2+1
=-1/8(1-cos4x)+1
=7/8+1/8cos4x
最小正周期为1/2π
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