已知sin(π/4-x)=5/13,x属于(0,π/4),求(cos2x)/cos[(π/4)+x]

x属于(0,π/4)
0<π/4-x<π/4
所以cos(π/4-x)>0
[sin(π/4-x)]^2+[cos(π/4-x)]^2=1
所以cos(π/4-x)=12/13
sin(π/2-2x)=2sin(π/4-x)cos(π/4-x)=120/169
所以cos2x=sin(π/2-2x)=120/169
cos(π/4+x)=sin[π/2-(π/4+x)]=sin(π/4-x)=5/13
所以
(cos2x)/cos(π/4+x)
=(120/169)/(5/13)
=24/13