函数Y=SIN2X+COS2X的值域?解: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] (为什么 sinx∈[-1,1)谢谢sin(2x+π/4)∈[-1,1] 所以 √2sin(2x+π/4)

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函数Y=SIN2X+COS2X的值域?解: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] (为什么 sinx∈[-1,1)谢谢sin(2x+π/4)∈[-1,1] 所以 √2sin(2x+π/4)
函数Y=SIN2X+COS2X的值域?
解: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] (为什么 sinx∈[-1,1)谢谢
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

函数Y=SIN2X+COS2X的值域?解: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] (为什么 sinx∈[-1,1)谢谢sin(2x+π/4)∈[-1,1] 所以 √2sin(2x+π/4)
这样表述可能更好理
-1≤ sin(2x+π/4) ≤ 1