第一步已经知道了,请写出第二步的详细过程
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