已知函数f(x)=(4cos^4x-2cos2x-1)/[tan（π/4+x)sin^2(π/4-x)] 求f(-17π

4(cosx)^4-2cos2x-1=[2(cosx)^2]^2-1-2cos2x=[(2(cosx)^2-1]*[2(cosx)^2+1]-2cos2x
=cos2x*[2(cosx)^2+1]-2cos2x=cos2x*[2(cos)^2-1]=(cos2x)^2
tan(π/4+x)·[sin(π/4-x)]^2=(1+tanx)/(1-tanx)*[√2/2*cosx- √2/2*sinx]^2
=(cosx+sinx)/(cosx-sinx)*1/2*(cosx-sinx)^2
=1/2*(cosx+sinx)*(cosx-sinx)=1/2*[(cosx)^2-(sinx)^2]
=1/2*cos2x
f(x)=(cosx)^2/[1/2*cos2x]=2cos2x
(1), f(-17π/12)=2cos(-17π/6)=2cos[-2π-5π/6]=2cos(-5π/6)=2cos(-π+π/6)
=-2cosπ/6=-2*√3/2=-√3