y=(sinx+cosx)²+2cos²x在[0,π/2]上的值域

y=(sinx+cosx)²+2cos²x在[0,π/2]上的值域
y=(sinx+cosx)²+2cos²x在[0,π/2]上的值域

y=(sinx+cosx)²+2cos²x在[0,π/2]上的值域
y=(sinx+cosx)²+2cos²x
=sin²x+cos²x+2sinxcosx+2cos²x
=1+sin2x+cos2x-1
=sin2x+cos2x
=√2(sin2x·√2/2+√2/2·cos2x)
=√2sin(2x+π/4)
x∈[0,π/2]
0≤2x≤π
π/4≤2x+π/4≤5π/4
所以
当2x+π/4=π/2时,sin(2x+π/4)=1,取得最大值√2
当2x+π/4=5π/4时,sin(2x+π/4)=-√2/2,取得最小值-√2/2×√2=-1
值域为[-1,√2]

解
y=（sinx）^2+2sinxcosx+（cosx）^2+2（cosx）^2
=sin2x+2（cosx）^2+1
=sin2x+cos2x+2
=√2sin（2x+π/4）+2
0<=x<=π/2
0<=2x<=π
π/4<=2x+π/4<=5π/4
画sinx，x∈【π/4，5π/4】
的草图有

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解
y=（sinx）^2+2sinxcosx+（cosx）^2+2（cosx）^2
=sin2x+2（cosx）^2+1
=sin2x+cos2x+2
=√2sin（2x+π/4）+2
0<=x<=π/2
0<=2x<=π
π/4<=2x+π/4<=5π/4
画sinx，x∈【π/4，5π/4】
的草图有
-√2/2<=sin（2x+π/4）<=1
-1<=√2sin（2x+π/4）<=√2
1<=√2sin（2x+π/4）+2<=2+√2
所以值域为{y|1<=y<=2+√2}

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