化简函数f(x)=[(1+cos2x)²-2cos2x-1]/[sin(π/4+x)*sin(π/4-x)]

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化简函数f(x)=[(1+cos2x)²-2cos2x-1]/[sin(π/4+x)*sin(π/4-x)]

化简函数f(x)=[(1+cos2x)²-2cos2x-1]/[sin(π/4+x)*sin(π/4-x)]
化简函数f(x)=[(1+cos2x)²-2cos2x-1]/[sin(π/4+x)*sin(π/4-x)]

化简函数f(x)=[(1+cos2x)²-2cos2x-1]/[sin(π/4+x)*sin(π/4-x)]
(1+cos2x)²-2cos2x-1
=1+2cos(2x)+cos²(2x)-2cos(2x)-1
=cos²(2x)
=[cos²(x)-sin²(x)]²
=[cos(x)-sin(x)]² * [cos(x)+sin(x)]²

sin(π/4+x)=sin(π/4)cos(x)+cos(π/4)sin(x)=[cos(x)+sin(x)]/√2
sin(π/4-x)=sin(π/4)cos(x)-cos(π/4)sin(x)=[cos(x)-sin(x)]/√2

f(x)=[cos(x)-sin(x)]² * [cos(x)+sin(x)]² * [cos(x)-sin(x)] / [cos(x)+sin(x)]
= [cos(x)-sin(x)]³ * [cos(x)+sin(x)]
= [cos(x)-sin(x)]² * [cos²(x)-sin²(x)]
=[cos²(x)-2sin(x)cos(x)+sin²(x)] * cos(2x)
=[1-sin(2x)]cos(2x)

(1+cos2x)²-2cos2x-1
=1+2cos(2x)+cos²(2x)-2cos(2x)-1
=cos²(2x)
sin(π/4+x)=sin(π/4)cos(x)+cos(π/4)sin(x)=[cos(x)+sin(x)]/√2
sin(π/4-x)=sin(π/4)cos(x)-cos(π/4)sin(x)=[cos(x)-sin(x)]/√2
sin(π/4+x)*sin(π/4-x)=1/2*[cos²(x)-sin²(x)]=1/2cos2x
所以原式=1/2[cos2x]^3