已知sin[(π/4)-x)]=5/13,0<x<π/4.求cos2x/(cosπ/4+x)

1.∵sin[(π/4)-x)]=5/13,0<x<π/4
∴cos[(π/4)-x)]=12/13
cos2x=cos[(π/2)-2x]=2sin[(π/4)-x)]*cos[(π/4)-x)]=2×5/13×12/13=120/169
cos[(π/4)+x]=sin[π/2-(π/4+x)]=sin(π/4-x)=5/13
∴cos2x/cos[(π/4)+x]=120/169 /5/13 =24/13

2.∵0<β<π/2<α<π
又∵cos[α-(β/2)]=-1/9,sin[(α/2)-β]=2/3
∴sin[α-(β/2)]=根号80/9,cos[(α/2)-β]=根号5/3
∴sin[(α+β)/2]=sin{[α-(β/2)]-[(α/2)-β]}=sin[α-(β/2)]cos[(α/2)-β]－sin[(α/2)-β]cos[α-(β/2)]=22/27
cos[(α+β)/2]=根号75/27
cos(α+β)=-根号239/27

3.∵tan(α-β)=1/2
∴tan(2α-2β)=4/3
∴tan(2α-β)=tan(2α-β+β)=1
∵α,β∈(0,π)且tan(α-β)=1/2,tanβ=-1/7
∴β∈(π/2,π)，α∈(0,π/4)，
∴2α-β∈(-π,0)
∴2α-β=-3π/4

你要给我加分,太难打了!!!!!!!!

已知sin[(π/4)-x)]=5/13,0<x<π/4. 求cos2x/[cos(π/4)+x]
解:sin[(π/4)-x)]=5/13 得cos[(π/4)-x)]=12/13
sin[(π/4)-x)]=cos{(π/2)-[(π/4)-x]}=cos[(π/4)+x] =5/13

cos2x=sin[(π/2)-2x)]=sin{2[(π/4)-x)]}
=2sin[(π/4)-x)]cos[(π/4)-x)