已知函数f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).求tana=2时,f(a)若x属于〔π/12,π/2〕,求f(x)的取值范围

已知函数f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).求tana=2时,f(a)若x属于〔π/12,π/2〕,求f(x)的取值范围
已知函数f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).求tana=2时,f(a)
若x属于〔π/12,π/2〕,求f(x)的取值范围

已知函数f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).求tana=2时,f(a)若x属于〔π/12,π/2〕,求f(x)的取值范围
,而sin^2a+cos^2a=1,得sin^2a=4/5
f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).
=sinx(cosx+sinx)+2sin(x+π/4)cos(x+π/4)
=1/2sin2x+sin^2x+sin(2x+π/2)
=1/2sin2x+1/2-1/2cos2x+cos2x
=(sin2x+cos2x+1)/2
tana=sina/cosa=2,所以sin^2a=4cos^2a,即1-cos2a=4*（1+cos2a）得cos2a=-3/5
sin2a=2sinacosa/（sin^2a+cos^2a）=2tana/（tan^2a+1）=4/5
所以f(a) =1/2*（4/5-3/5+1）=3/5
（2）令f’（x）=1/2（cos2x-sin2x）>=0,接下来就是求单调区间,然后得到取值范围,具体不想再算了,主要是方法.
这种题一般先要化简,对于函数题可以求导然后数形结合得值域.