Question

第一步已经知道了，请写出第二步的详细过程

已知函数f（x）=（1+1/tanx）sin²X-2 sin(x+∏/4) sin(x-∏/4)
① 若tanx=2时，f(x)的值
②若x属于〔π/12,π/2〕,求f(x)的取值范围
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

Answer 1

f(x)=(sin2x+cos2x+1)/2
=√2/2sin(2x+π/4)+1/2
x属于〔π/12,π/2〕,2x+π/4属于〔5π/12,5π/4〕,
2x+π/4=π/2,sin(2x+π/4)max=1
2x+π/4=5π/4,sin(2x+π/4)min=-√2/2
f(x)max=√2/2+1/2
f(x)min=0
f(x)的取值范围[0,(√2+1)/2]

追问

不好意思，我再问一下f(x)=(sin2x+cos2x+1)/2
=√2/2sin(2x+π/4)+1/2
√2/2是怎么回事，不是应该是1/2吗？

追答

sin2x+cos2x
=√2(√2/2sin2x+√2/2cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
(sin2x+cos2x+1)/2
=[√2sin(2x+π/4)+1]/2
=√2/2sin(2x+π/4)+1/2