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/2]上时,求函数f(x)的值域
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![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/2]上时,求函数f(x)的值域](/uploads/image/z/5196187-19-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%5E2wx%2B%E6%A0%B93sinwxsin%28wx%2B%CF%80%2F2%29+1%29%E6%B1%82f%28x%29%E7%9A%84%E5%80%BC+2%29x%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%CF%80%2F12%2C%CF%80%2F2%5D%E4%B8%8A%E6%97%B6%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F)
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/2]上时,求函数f(x)的值域
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/2]上时,求函数f(x)的值域
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/2]上时,求函数f(x)的值域
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.
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