sinx+siny=1/2,cosx+cosy=（根号3）/3,求sin[(x-y)/2]

两式平方，得；
1+2sinxsiny=1/4
1+2cosxcosy=1/3
故sinxsiny=-3/8
cosxcosy=-1/3
cos(x-y)=cosxcosy+sinxsiny=-17/24
sin[(x-y)/2]=｛[1-cos(x-y)]/2｝开方=(65/48)开方

A: sin(x)+sin(y) = 1/2;
B: cos(x)+cos(y) = 1/sqrt(3);
A^2+B^2 = 2+2*(sin(x)sin(y)+cos(x)cos(y))=7/12;
cos(x-y) = -17/24;
设 T = x-y;
cos(T) = 2*cos^2(T/2) - 1;
cos(T/2) = sqrt(7/48);
sin((x-y)/2) = sqrt(41/48);
正负的讨论你自己来吧……

两式各自平方,得(sinx)^2+2sinxsiny+(siny)^2=1/4(cosx)^2+2cosxcosy+(cosy)^2=1/3cos2a=cosa*cosa-sina*sina=2cosa*cosa-1=1-2sina*sina两式相加,得 2＋2（sinxsiny+cosxcosy)=7/12cos(x-y)=sinxsiny+cosxcosy=-17/...