若x^2+y^2+2x+2y+1=0则x+y的取值范围是?

(x+1)^2+(y+1)^2=1
令x+1=cosa
则(y+1)^21-(cosa)^2=(sina)^2
sina的值域关于原点对称
所以不妨令y+1=sina
所以x=cosa-1,y=sina-1
x+y=sina+cosa-2=√2sin(x+π/4)-2
-1<=sin(x+π/4)<=1
所以-√2-2<=x+y<=√2-2