设函数f(x)=sin(wx+2π/3)+sin(wx-2π/3)(w>0)的最小正周期为π,求函数的单调区间?
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设函数f(x)=sin(wx+2π/3)+sin(wx-2π/3)(w>0)的最小正周期为π,求函数的单调区间?
设函数f(x)=sin(wx+2π/3)+sin(wx-2π/3)(w>0)的最小正周期为π,求函数的单调区间?
设函数f(x)=sin(wx+2π/3)+sin(wx-2π/3)(w>0)的最小正周期为π,求函数的单调区间?
f(x)=sin(wx+2π/3)+sin(wx-2π/3)
=sinwxcos(2π/3)+coswxsin(2π/3)+sinwxcos(2π/3)-coswxsin(2π/3)
=sinwxcos(2π/3)+sinwxcos(2π/3)
=2sinwxcos(2π/3)
=-sinwx
∵T=π
∴2π/w=π
w=2
∴函数为f(x)=-sin2x
∴函数的单调增区间为:(π/2+kπ,3π/2+kπ)k∈Z
函数的单调减区间为:(-π/2+kπ,π/2+kπ) k∈Z
f(x)=2sinwx-----------w=2
所以与的sinx单调区间一样,在(0,kT/2)上增,(kt/2,3kt/2)上减
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