已知函数f(x)=2√3sinxcosx+cos^2x-sin^2x-1
问题描述:
已知函数f(x)=2√3sinxcosx+cos^2x-sin^2x-1
求周期和递增区间
答
f(x)=2√3sinxcosx+cos^2x-sin^2x-1
f(x)=√3sin2x+cos2x-1
=sin(2x+π/6)-1
所以周期为:T=2π/w=π.