Question

sinX的6次方+cosX的6次方的周期怎么算?

Answer 1

y=sin^6 x+cos^6 x

=(sin^2 x+cos^2 x)*(sin^4 x+cos^4 x-

sin^2 x*cos^2 x)

=sin^4 x+cos^4 x+2sin^2 x*cos^2 x-3sin^2 x*cos^2 x

=(sin^2 x+cos^2 x)^2-3*sin^2 x*cos^2 x

=1-3*sin^2 x*cos^2 x

=-3*[(1-cos2x)/2]*[(1+cos2x)/2]+1

=-3*[(1-cos^2 2x)/4]+1

=-3/4*sin^2 2x+1

=-3/4*[(1-cos4x)/2]+1

=-3/8*(1-cos4x)+1

=3/8*cos4x+5/8

则T=2兀/4=兀/2