求函数f(x)=√3cosx+sinx的单调递增区间_数学解答

求函数f(x)=√3cosx+sinx的单调递增区间

人气：259 ℃　时间：2020-02-04 07:51:37

解答

f(x)=√3cosx+sinx=2sin(x+π/3)
2kπ-π/2f(x)=√3cosx+sinx=2sin(x+π/3)WHY√3cosx+sinx=2[sinx*(1/2)+cosx*(√3/2)]=2(sinxcosπ/3+cosxsinπ/3)=2sin(x+π/3)