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

来源：学生作业帮助网编辑：作业帮时间：2024/11/17 07:21:11

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

sinx+siny=1/2,cosx+cosy=（根号3）/3,求sin[(x-y)/2]
两式各自平方,得
(sinx)^2+2sinxsiny+(siny)^2=1/4
(cosx)^2+2cosxcosy+(cosy)^2=1/3
cos2a=cosa*cosa-sina*sina=2cosa*cosa-1=1-2sina*sina
两式相加,得 2＋2（sinxsiny+cosxcosy)=7/12
cos(x-y)=sinxsiny+cosxcosy=-17/24＝1－2*sin[(x-y)/2]*sin[(x-y)/2]
得
sin[(x-y)/2]＝正负根号（41/48)
由于cos(x-y)

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

sinx+siny=1/3,则siny-(cosx)^2的最大值为?最小值为? 已知sinx+siny=1/3求siny-(cosx)^2的最小值和最大值已知sinx+siny=1/3求siny-(cosx)^2的最小值已知1+cosx-siny+sinx*siny=0,1-cosx-cosy+sinx*cosy=0.求Sinx. 已知1+cosx-siny+sinx*siny=0,1-cosx-cosy+sinx*cosy=0,求sinx的值求y=(sinx)^2-cosx+2最值已知sinx+siny=1/3,求siny-(cosx)^2的最值 cosx+cosy=1,则sinx+siny的范围证明.（sinx+siny)/cosx-cosy=ctg(y-x)/2 cosx+cosy=1/2,sinx-siny=1/3,则cos(x+y)=? cosx-cosy=1/2,sinx-siny=1/3,sin(x+y)=RT 已知cosx+cosy=1/2,sinx-siny=1/3,则cos(x+y)= 已知cosx+cosy=1/2,sinx-siny=1/3,求cos(x+y) sinx+siny=-1/3 cosx+cosy=1/2,求sin(x+y)的值. 若cosx*cosy+sinx*siny=1/3 ,则cos(2x-2y)=? sinx-siny=1/3,cosx-cosy=2/3,sin(x+y)=? 已知sinx*cosy=1/2 求cosx*siny的取值范围若sinx*siny=1/2,则cosx*cosy取值范围是若sinx+siny=1,则(cosx+cosy)^2的最大值是