已知0<x<pai/4,sin(pai/4-x)=5/13,求cos2x/cos(pai/4+x)的值。写出详细的过程，谢谢

画一个直角三角形ABC
AB＝5，BC＝12，AC=13
最小的角是pai/4-x
把5的边BA延长到12，D
ADC＝x

0<x<π/4
0<π/4-x<π/4
cos(π/4-x)>0
sin(π/4-x)=5/13
因为(sina)^2+(cosa)^2=1
所以cos(π/4-x)=12/13

cos2x=sin(π/2-2x)=sin[2(π/4-x)]
=2sin(π/4-x)cos(π/4-x)
=120/169

cos(π/4+x)
=sin[π/2-(π/4+x)]
=sin(π/4-x)
=5/13

所以cos2x/cos(pai/4+x)
=(120/169)/(5/13)
=24/13