y=(sinx)^4＋(cosx)^4
=(sin²x＋cos²x)²－2sin²xcos²x
=1－0.5[2sinxcosx]²
=1－0.5sin²2x
=1－0.25[1－cos4x]
=0.75＋0.25cos4x
最小正周期是T=2π/4=π/2
y=(sinx)^4＋(cosx)^4
=(sin²x＋cos²x)²－2sin²xcos²x
=1－(1/2)[2sinxcosx]²
=1－(1/2)sin²2x
=1－(1/4)[1－cos4x]
=(3/4)＋(1/4)cos4x
最小正周期是2π/4=π/2
f(x)=(sinx)^4+(cosx)^2
=(sinx)^4-(sinx)^2+1
=(sinx)^2[(sinx)^2-1]+1
=-(sinx)^2*(cosx)^2+1
=-(sin2x)^2/4+1
=(cos4x-1)/8+1
=====> T=2π/4=π/2
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024