y=cos^4x-sin^4x的周期是 3Q

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y=cos^4x-sin^4x的周期是3Qy=cos^4x-sin^4x的周期是3Qy=cos^4x-sin^4x的周期是3Qcos^4x-sin^4x=(cos^2x-sin^2x)(cos^2x+

y=cos^4x-sin^4x的周期是 3Q
y=cos^4x-sin^4x的周期是 3Q

y=cos^4x-sin^4x的周期是 3Q
cos^4x-sin^4x
=(cos^2x-sin^2x)(cos^2x+sin^2x)
=cos^2x-sin^2x
=2cos^2x-1
=cos2x
所以周期为∏

y=(cosx)^4-(sinx)^4
=[(cosx)^2-(sinx)^2][(cosx)^2+(sinx)^2]
=[(cosx)^2-(sinx)^2]*1
=cos2x
所以最小正周期为2π/2=π

y=cos^4x-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)=cos4x
周期T=2π/4=π/2