Question

函数Y=sin^4x+cos^2x的最小正周期为????祥细过程

Answer 1

两种解法：
法一：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-(1/4)*(1-cos4x)/2=1-1/8(1-cos4x) =7/8+1/8cos4x （因为(sin2x)^2=(1-cos4x)/2）
最小正周期是2π/4= 1/2 π
法二：Y=sin^4x+cos^2x =(sin²x)²-sin²x+1 =[sin²x-1/2]²+3/4 =(1/4)cos²2x+3/4=7/8+1/8cos4x （因为(cos2x)^2=(1+cos4x)/2 )
最小正周期是2π/4= 1/2 π

Answer 2

Y=sin^4x+cos^2x
=(sin²x)²-sin²x+1
=[sin²x-1/2]²+3/4
=(1/4)cos²2x+3/4
最小正周期为 2π/2=π

Answer 3

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2π/4= 1/2π

Answer 4

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