y=cos2x+sin2x
=√2(√2/2cos2x+√2/2sin2x)
=√2(sinπ/4cos2x+cosπ/4sin2x)
=√2sin(π/4+2x)
当sin(π/4+2x)=1时,y最大=√2
当sin(π/4+2x)=-1时,y最小=-√2
函数y=cos2x+sin2x的值域{-√2<=y<=√2}
y=cos2x+sin2x
=(√2)[(√2/2)cos2x+(√2/2)sin2x]
=(√2)sin(2x+π/4)
∵x∈R
∴y∈[-√2,√2]
谢谢
y=cos2x+sin2x
=√2sin(2x+π/4)
-1≤sin(2x+π/4)≤1
-√2≤y≤√2
函数的值域为[-√2,√2]