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
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![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](/uploads/image/z/1989225-9-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%281%2B1%2Ftanx%29sin%5E2-2sin%28x%2B%CF%80%2F4%29sin%28x-%CF%80%2F4%29.%E6%B1%82tana%3D2%E6%97%B6%2Cf%28a%29%E8%8B%A5x%E5%B1%9E%E4%BA%8E%E3%80%94%CF%80%2F12%2C%CF%80%2F2%E3%80%95%2C%E6%B1%82f%28x%29%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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.