sin²a+cos²a=1
所以cos(π/4-x)=12/13
原式=(cos²x-sin²x)/(cosπ/4cosx-sinπ/4sinx)
=(cosx+sinx)(cosx-sinx)/[√2/2*(cosx-sinx)]
=√2(cosx+sinx)
=√2*√2sin(x+π/4)
=2cos(π/4-x)
=24/13
sin²a+cos²a=1
所以cos(π/4-x)=12/13
原式=(cos²x-sin²x)/(cosπ/4cosx-sinπ/4sinx)
=(cosx+sinx)(cosx-sinx)/[√2/2*(cosx-sinx)]
=√2(cosx+sinx)
=√2*√2sin(x+π/4)
=2cos(π/4-x)
=24/13