若sinx+sinx的平方=1则cosx的平方+cos的四次方

cos^2x+cos^4x
=1-sin^2x+(1-sin^2x)^2
=1-sin^2x+1-2sin^2x+sin^4x
=sin^4x-3sin^2x+2
=(sin^2x-1)(sin^2x-2)……………[因为sinx+sin^2x=1,所以sin^2x＝1-sinx]
=(1-sinx-1)(1-sinx-2)
=-sinx(-sinx-1)
=sinx(sinx+1)
=sinx+sin^2x
=1