令cosx+cosy=t 则(sinx+siny)^2+(cosx+cosy)^2=(根号2/2)^2+t^2=t^2+1/2
(sinx+siny)^2+(cosx+cosy)^2=(sinx)^2+(cosx)^2+(siny)^2+(cosy)^2+2(sinxsiny+cosxcosy)
=2+2cos(x-y)
2+2cos(x-y)=t^2+1/2
所以 t^2-3/2=2cos(x-y)
所以 -2≤t^2-3/2≤2 即 -1/2≤t^2≤7/2
又因为 t^2≥0
所以 0≤t^2≤7/2
所以 -(根号14)/2≤t≤(根号14)/2
即 -(根号14)/2≤cosx+cosy≤(根号14)/2