Question

函数y=sin^6x+cos^6x的最小正周期为

写下过程谢谢

Answer 1

y=√2 sin(6x+π/4)
所以T=2π/6=π/3

Answer 2

y=(sinx)^6+(cosx)^6=[(sinx)^2+(cosx)^2][(sinx)^4-(sinxcosx)^2+(cosx)^4]
=[(sinx)^2+(cosx)^2]^2-3(sinxcosx)^2=1-3Sin^2Cos^2x=1-3/4(2SinxCosx)^2=1-3/4Sin^2x=1-3/8(1-Cos4x)=5/8=3/8Cos4x

Answer 3

y=(sinx)^6+(cosx)^6=[(sinx)^2+(cosx)^2][(sinx)^4-(sinxcosx)^2+(cosx)^4]
=[(sinx)^2+(cosx)^2]^2-3(sinxcosx)^2
=-3(sinxcosx)^2
=-(3/4)(sin2x)^2
=(3/8)(1-cos4x)
T=2兀/4=兀/2