已知的a平方+b的平方=1,c的平方+d的平方=1,ac+bd=0,求证ab+cd=0.

证法1:
ab+cd
=ab(c^2+d^2)+cd(a^2+b^2)(因为a^2+b^2=1,c^2+d^2=1)
=abc^2+abd^2+cda^2+cdb^2
=(abc^2+cda^2)+(abd^2+cdb^2)
=ac(bc+ad)+bd(ad+bc)
=(ac+bd)(ad+bc)
=0
证法2:
设a=sinx,b=cosx,c=siny,d=cosy
则sinxsiny+cosxcosy=cos(x-y)=0
ab+cd=sinxcosx+sinycosy
=1/2(sin2x+sin2y)
=1/2*2sin(x+y)cos(x-y)
=0