作业帮已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围
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![已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围](/uploads/image/z/2691462-30-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%281%2B1%2Ftanx%29sin%5E2x%2Bmsin%28x%2B%CF%80%2F4%29sin%28x-%CF%804%29%2C%E5%BD%93m%3D0%E6%97%B6%2C%E6%B1%82f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B%CF%80%2F8%2C3%CF%80%2F4%5D%E4%B8%8A%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围
已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围
已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围
m=0,所以f(x)=)=(1+1/tanx)sin^2x=sin^2x+sinxcosx=1-cos^2x+1/2sin2x=1-(cos2x+1)/2+1/2sin2x=-1/2cos2x+1/2sin2x+1/2=二分之根号二sin(2x-π/4)+1/2,下面的应该就好做了吧