已知函数f(x)=sin(wx+π⼀6)+sin(wx-π⼀6)-2cos^2 wx⼀2, w>0

2025-01-08 00:02:50
推荐回答(1个)
回答1:

f(x)=sin(wx+π/6)+sin(wx-π/6)-2cos^2 wx/2
= sinwxcosπ/6+coswxsinπ/6 +sinwxcosπ/6-coswxsinπ/6-2cos^2 wx/2
=2 sinwxcosπ/6-2cos^2 wx/2
=√3sinwx-(1+coswx)
=2sin(wx-π/6)-1,
所以此函数图像与直线y=-1的两个相邻交点间距离恰好为半个周期。
即π/w=π/2,w=2.
f(x)= 2sin(2x-π/6)-1,
2kπ-π/2≤2x-π/6≤2kπ+π/2,
kπ-π/6≤x≤kπ+π/3.(k∈Z).
所以函数的单调增区间为[kπ-π/6, kπ+π/3](k∈Z).