y=sin^4x+cos^4x.求函数周期RTRTRTRTRTRTRTRRTR
问题描述:
y=sin^4x+cos^4x.求函数周期
RTRTRTRTRTRTRTRRTR
答
y=sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-sin^2 (2x)/2
=cos(4x)/2
最小正周期π/2
答
y=sin^4x+cos^4x
=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-sin^2 (2x)/2
=1-{[1-cos(4x)]/2}/2
=cos(4x)/4 +3/4
所以T=2π/4=π/2