Question

y=(sinx)四次方+（cosx）四次方的周期

Answer 1

y=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2

=1-2*[(sin2x)/2]^2

=1-(sin2x)^2/2

=1-(1/2)*(1-cos4x)/2

=1-1/4+(1/4)cos4x

=(1/4)cos4x+3/4

所以最小正周期T=2π/4=π/2