①f(x)=sin^2wx+√3sinwxsin(wx+π/2)
=1/2(1-cos2wx) +√3sinwx coswx
=1/2(1-cos2wx) +√3/2sin2wx
=√3/2sin2wx-1/2cos2wx+1/2
=sin(2wx-π/6) +1/2
所以函数值域是[-1/2,3/2].
②x∈[-π/12,π/2]时,
本小题还需告诉另外的条件求出w的值,然后求出2wx-π/6的范围,
才可以进一步求出函数值域.
已知函数f(x)=sin^2wx+根3sinwxsin(wx+π/2) 1)求f(x)的值 2)x在区间[-π/12,π
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