f(x)=(1+√2*cos(2x-π/4))/sin(π/2-x)
=[1+√2*cos(2x-π/4)]/cosx
=[1+√2*(cos2xcosπ/4+sin2xsinπ/4)]/cosx
=[1+√2*(√2/2*cos2x+√2/2*sin2x)]/cosx
=(1+cos2x+sin2x)/cosx
=(1+2cos²x-1+2sinxcosx)/cosx
=(2cos²x+2sinxcosx)/cosx
=2(cosx+sinx)
=2√2*(√2/2*cosx+√2/2*sinx)
=2√2*(sinπ/4cosx+cosπ/4sinx)
=2√2*sin(π/4+x)如果改成f(x)=(1-√2*cos(2x-π/4))/sin(π/2-x)呢，改怎么化简