Question

设函数f(x)=√3*sinxcosx-cos^2x 求f(x)的最小正周期。请带上过程 谢谢！

Answer 1

f(x)=√3*sinxcosx-cos^2x
=(√3/2)*sin2x-[(cos2x-1)/2]
=(√3/2)*sin2x-(1/2)*cos2x+(1/2)
=sin(2x-π/6)+(1/2)
最小正周期：T=2π/ω=2π/2=π

Answer 2

f(x)=√3*sinxcosx-cos^2x =√3/2*sin2x-(1+cos2x)/2
=sin(2x-π/6)-1/2
T=2π/2=π
想知道周期怎么来的，那就是：
f(x+T)=f(x)
sin(2(x+T)-π/6)=sin(2x-π/6)
2T=2π
所以T=2π/2