你帮我看看哪一步做错了sin（π／4-x)=5/13、x在（0、π／2）内 则cos2x／cos（π／4+x）=?稍等上图

没错

sin(π/4-x)=sin(π/2 - π/4 -x)=cos( π/4 +x)=5/13
∵0∴π/4 ∴sin( π/4 +x)=√[1- cos²( π/4 +x)]=12/13
∴cos2x/cos(π/4+x)
=sin(π/2 + 2x)/cos(π/4+x)
=2sin(π/4+x)cos(π/4+x)/cos(π/4+x)
=2sin(π/4+x)
=2*12/13
=24/13

cos(π/4+x)=sin[π/2-(π/4+x)]=sin(π/4-x)=5/13
而因为0

cos(π／4+x)=cos(π／2-π／4+x)=sin(π／4-x)=5/13;
sin(π／4+x)=cos(π／4-x)=12/13;
cos2x=cos(π／4+x-π／4+x)=cos(π／4+x)cos(π／4-x)+sin(π／4+x)sin(π／4-x)
=5/13*12/13+5/13*12/13=120/169
cos2x／cos（π／4+x）=24/13