Question

已知函数f(x)=sin(π/3+wx)+cos(wx-π/6)(w＞0）,f(x)多少最小正周期为π (1)求f(x)？

Answer 1

f(x)=sin(π/3+wx)+cos(wx-π/6)
=sinπ/3coswx+cosπ/3sinwx+coswxcosπ/6+sinwxsinπ/6
=(√3/2)coswx+(1/2)sinwx+(√3/2)coswx+(1/2)sinwx
=√3coswx+sinwx
=2[(√3/2)coswx+(1/2)sinwx]
=2sin(wx+π/3)
∴f(x)的周期T=2π/w
f(x)在区间(-5π/(6w)+2kπ/w,π/6w+2kπ/w)单调增加；
在区间(π/6w+2kπ/w,7π/(6w)+2kπ/w,)单调减少,1,已知函数f(x)=sin(π/3+wx)+cos(wx-π/6)(w＞0）,f(x)多少最小正周期为π (1)求f(x)的解析式 (2)求f(x)单