函数y=sin^4x+cos^4x的最小正周期
人气:106 ℃ 时间:2019-10-08 19:49:56
解答
y=sin^4x+cos^4x
y=[(sinx)^2+(cosx)^2]-2(sinxcosx)^2
=1-(2sinxcosx)^2/2
=1-(sin2x)^2/2
=0.75+(1-2(sin2x)^2)/4
=0.75+(cos4x)/4
最小正周期pi/2
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