Question

（1）求f（x）=sin^4x-cos^4x的最小正周期

Answer 1

解

f(x)=sin^4x-cos^4x

=(sin²x+cos²x)(sin²x-cos²x)

=sin²x-cos²x

=-cos2x

∴最小正周期为：T=2π/2=π

f(x)=2cos²x+2√3sinxcosx

=√3sin2x+(2cos²x-1)+1

=√3sin2x+cos2x+1

=2(√3/2sin2x+1/2cos2x)+1

=2(sin2xcosπ/6+cos2xsinπ/6)+1

=2sin(2x+π/6)+1

∴周期为：T=2π/2=π