已知函数fx根号3sinwxcoswx+cos^2wx-3⼀2(w>0),其最小正周期为兀⼀2,

f(x)解析式
2024-12-25 17:31:15
推荐回答(2个)
回答1:

f(x) = √3sinwxcoswx + cos^2wx - 3/2
= √3/2sin2wx + 1/2(cos2wx+1) - 3/2
= sin2wxcosπ/6 + cos2wxsinπ/6 - 1
= sin(2wx+π/6) - 1
w>0,最小正周期为π/2
2π/(2w)=π/2
w=2

f(x)= sin(4x+π/6) - 1

回答2:

f(x)=(√3/2)sin2wx+(1+cos2wx)/2-3/2
=(√3/2)sin2wx+(1/2)cos2wx-1
=sin(2wx+π/6)-1
最小正周期T=2π/2w=π/2
得:w=2
所以,f(x)=sin(4x+π/6)-1