Question

sinx+siny=2分之根号2，则cosx+cosy得取值范围是？

Answer 1

cos(x-y)=cosxcosy+sinxsiny
sinx+siny=√2/2
若cos(x-y)=-1，则x-y=pi+2k*pi,k为整数
此时sinx=sin(y+pi+2k*pi)=sin(y+pi)=-siny，
所以sinx+siny=0，不满足sinx+siny=√2/2
显然cos(x-y)∈[-1,1]是错误的，最起码cos(x-y)取不到-1
令cos(x-y)=cos(pi-z)=-cosz
则z越趋近于0时，cosz越趋近于1，于是cos(x-y)越趋近于-1
sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]=√2/2
得sin[(x+y)/2]cos[(x-y)/2]=√2/4
取x-y=pi-z,即x=y+pi-z,
有sin[(x+y)/2]cos[(x-y)/2]
=sin[y+(pi-z)/2]cos[(pi-z)/2]
=sin[y+(pi-z)/2]sin(z/2)=√2/4
现考虑大于0且接近于0的z，有sin(z/2)>0
由-1=<sin[y+(pi-z)/2]<=1知，0<(√2/4)/sin(z/2)<=1
sin(z/2)>=√2/4
故满足此条件的z，可使cos(x-y)取得最小值
最小值cos(x-y)=-cosz=2sin(z/2)^2-1=-3/4
令f(x)=cos(x-y)，对f(x)关于x求导知，
x=y时，cos(x-y)可取到最大值1
综上cos(x-y)∈[-3/4,1] 转载

Answer 2

sinx+siny=√2/2...⑴
设cosx+cosy=k ...⑵
(1)^2+(2)^2: 2+2(sinxsiny+cosxcosy)=(√2/2)^2+k^2
cos(x-y)=[(√2/2)^2+k^2-2]/2
∵cos(x-y)∈[-1,1]
∴[(√2/2)^2+k^2-2]/2∈[-1,1]
最后算得:k∈[-√14/2,√14/2]