若sinx+cosy=1/3,cosx+siny=1/6,则sin(x+y)=?要有过程

Question

若sinx+cosy=1/3,cosx+siny=1/6,则sin(x+y)= ?
要有过程

Answer 1

一、因为sinx+cosy=1/3,cosx+siny=1/6,所以(sinx+cosy)^2=(sinx)^2+(cosy)^2+2sinxcosy=1/9;（cosx+siny）^2=(cosx)^2+(siny)^2+2cosxsiny=1/36二、两式相加,得2+2（sinxcosy+cosxsiny）=2+2sin(x+y)= 5/36三、解得,sin(x+y)= -67/72