已知x^2+y^2=1,z^2+w^2=1,xz+yw=0,求xy+zw的值
问题描述:
已知x^2+y^2=1,z^2+w^2=1,xz+yw=0,求xy+zw的值
我不会,帮一下啦
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