Question

函数f(x）=sin^4x+cos^2x的最小正周期

Answer 1

f(x)=sin^4x+cos^2x

=1-sin^2x+sin^4x

=sin^2x(sin^2-1)+1

=-sin^2xcos^2x+1

=-(1/2sin2x)^2+1

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

=7/8+1/8cos4x

最小正周期为1/2π