sinα+sinβ=2分之根号2,
令cosα+cosβ=k
分别平方并相加
由sin²x+cos²x=1
所以2+2(cosαcosβ+sinαsinβ)=1/2+k²
cos(α-β)=cosαcosβ+sinαsinβ=(k²-3/2)/2
-1<=cos(α-β)<=1
-1<=(k²-3/2)/2<=1
-1/2<=k²<=7/2
即0<=k²<=7/2
所以-√14/2<=cosα+cosβ<=√14/2