Question

函数Y=sin^4x+cos^2x的最小正周期为_______

Answer 1

Y=sin^4x+cos^2x=(1-cos2x)²/4 +(1+cos2x)/2
=(3+cox²2x)/4
=(7+cox4x)/8
周期T=2π/4=π/2

Answer 2

Y=sin^2x(1-cos^2x)+cos^2x
=sin^2x-sin^2xcos^2x+cos^2x
=sin^2x+cos^2x-sin^2xcos^2x
=1-(sinxcosx)^2
=1-(1/2sin2x)^2
=1-1/4(sin2x)^2 ①
因为(sin2x)^2=(1-cos4x)/2
所以①=1-(1/4)*(1-cos4x)/2=1-1/8(1-cos4x)
=7/8+1/8cos4x
最小正周期是2∏/4= 1/2∏

Answer 3

同学你写得不规范，我觉得写成y=(sinx)^4+(cosx)^2比较恰当。
化简一下：Y=(sinx)^4+(cosx)^2=((1-cos2x)/2)^2+(1+cos2x)/2=(cos4x)/8+7/8.
周期为T=2π/4=π/2.