y=sin2x+cos2x=√2(sin2x*√2/2+cos2x*√2/2)
=√2[sin2xcos(π/4)+cos2xsin(π/4)]
=√2sin(2x+π/4)
因为 sinx∈[-1,1]
sin(2x+π/4)∈[-1,1]
所以 √2sin(2x+π/4)∈[-√2,√2]
y∈[-√2,√2]
即函数值域是[-√2,√2] π/4
y=sin2x+cos2x=√2(sin2x*√2/2+cos2x*√2/2)
=√2[sin2xcos(π/4)+cos2xsin(π/4)]
=√2sin(2x+π/4)
因为 sinx∈[-1,1]
sin(2x+π/4)∈[-1,1]
所以 √2sin(2x+π/4)∈[-√2,√2]
y∈[-√2,√2]
即函数值域是[-√2,√2] π/4