已知x*x+y*y=1,z*z+w*w=1,xz+yw=0,求xy+zw的值
问题描述：
已知x*x+y*y=1,z*z+w*w=1,xz+yw=0,求xy+zw的值
答
令x=cosa y=sina
z=cosb w=sina
xz+yw=cos(a-b)=0
a-b=π/2
xy+zw=sina*cosa+sinb*cosb
=(sin2a+sin2b)/2
=(sin2a+sin(2a-π))/2
=0