已知cos(x-y/2)=-1/9,sin(x/2-y)=2/3,0<x<π,0<y<π/2求cos(x+y)的值

cos(x-y/2)=-1/9,sin(x/2-y)=2/3,0<x<π,0<y<π/2求cos(x+y)

因为cos(x-y/2)=-1/9 0<x<π,0<y<π
所以 π/2<(x-y/2)<π
sin(x-y/2)=4*√5/9
因为sin(x/2-y)=2/3,0<x<π,0<y<π/2
所以 0<(x-y/2)<π/2
cos(x/2-y)=√5/3

cos[（x-y/2）-(x/2-y)]=cos（x-y/2）*cos(x/2-y)+sin（x-y/2）*sin(x/2-y)=-1/9*√5/3+4*√5/9*2/3=7*√5/27

cos(x+y)=cos2[（x-y/2）-(x/2-y)]=2cos^2[（x-y/2）-(x/2-y)]-1=-239/729