设sin(45-x)=5/13,0<x<45,求cos2x/cos(45+x)
cos(45+x)=sin[90-(45+x)]=sin(45-x)=5/13
sin(x+45)=√[1-cos^2(45+x)]=12/13
cos2x=cos^2x-sin^2x=(cosx+sinx)(cosx-sinx)
=1/2*sin(x+45)sin(45-x)=30/169
cos2x/cos(45+x)=(30/169)/(5/13)=6/13
cos(45+x)=sin[90-(45+x)]=sin(45-x)=5/13
sin(x+45)=√[1-cos^2(45+x)]=12/13
cos2x=cos^2x-sin^2x=(cosx+sinx)(cosx-sinx)
=1/2*sin(x+45)sin(45-x)=30/169
cos2x/cos(45+x)=(30/169)/(5/13)=6/13