根據(jù)題意,在f(x+y)+f(x-y)=2f(x)f(y)中,令y=0可得,
2f(x)=2f(x)f(0),又由f(x)不是常函數(shù),即f(x)=0不恒成立,則f(0)=1,依次分析4個命題可得:對于①、在f(x+y)+f(x-y)=2f(x)f(y)中,令x=y=
,
可得
,結合f(0)=1,f(t)=0,可得
,則可得
,故①錯誤.
對于②、在f(x+y)+f(x-y)=2f(x)f(y)中,令x=0,可得f(y)+f(-y)=2f(0)f(y)=2f(y),f(y)+f(-y)=0不恒成立,f(x)不是奇函數(shù),故②錯誤.
對于③、在f(x+y)+f(x-y)=2f(x)f(y)中,令y=t可得,在f(x+t)+f(x-t)=2f(x)f(t)=0,即f(x+t)=-f(x-t),則f(x+3t)=-f(x+t)=f(x-t),即f(x+3t)=f(x-t),則f(x)是周期函數(shù)且一個周期為4t,③正確.
對于④、根據(jù)題意,無法判斷f(x)的單調性,則④錯誤.故答案為③.