Question

函数y=sin^4x+cos^4x的最小正周期？

Answer 1

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/2,5,y=sin^4x+cos^4x
=(sin²x+cos²x)²-2sin²xcos²x
=1-1/2*4sin²xcos²x
=1-sin²2x/2
最小正周期T=2*2π/2=2π,0,y=sin^4x+2sin²xcos²x+cos^4x-2sin²xcos²x
=(sin²x+cos²x)²-sin²(2x)
=1-sin²(2x)
∵T=2π/2=π
∴y的最小正周期T=π
您的问题已经被解答~~(>^ω^<)喵
如果采纳的话，我是很开心的哟(～ o ～)~zZ,0,y=sin^4x+cos^4x=(1-cos^2x)^2+cos^4x=
2cos^4x-2cos^2x+1
=1/2(cos2x+1)^2-2cos^2x+1
=1/2cos^2(2x)+cos2x-(cos2x+1)+1.5
=1/4(cos4x+1)+0.5,
所以最小正周期为派/2,0,y=sin^4x+cos^4x
=sin^4x+2sin^2xcos^2x+cos^4x-2sin^2xcos^2x
=(sin^2+cos^2)^2-(1/2)(2sinxcosx)^2
=1-(1/2)(sin2x)^2
=1-(1/4)(1-cos4x)
=(1/4)cos4x+3/4
故最小正周期为=2π/4=π/2,0,