f(x)=sinx+cosx+2sinxcosx,x∈[0,π],求f(x)的值域

y=(sinx+cosx)+(sinx+cosx)^2-(sinx^2+cosx^2)
=(sinx+cosx)+(sinx+cosx)^2-1
设sinx+cosx=t ,t=√2sin(x+π/4)∈[-√2,√2]
y=t^2+t-1
转化为关于t的二次函数,
最大值在t=√2时取到,y max=1+√2

sinx+cosx=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)
y = sinx+cosx+2sinxcosx
= sinx+cosx+2sinxcosxsin^2x+cos^2x-1
= sin^2x+2sinxcosx+cos^2x + sinx+cosx - 1
= (sinx+cosx)^2+(sinx+cosx)-1
= [√2sin(x+π/4)]^2 + √2sin(x+π/4) - 1
= 2[sin(x+π/4)]^2 + √2sin(x+π/4) - 1
= 2[sin(x+π/4)+√2/4]^2 - 5/4
-1≤sin(x+π/4)≤1
(-4+√2)/4 ≤ sin(x+π/4)+√2/4 ≤ (4+√2)/4
0 ≤ [sin(x+π/4)+√2/4)^2 ≤ (9+4√2)/8
0 ≤2 [sin(x+π/4)+√2/4)^2 ≤ (9+4√2)/4
-5/4 ≤2 [sin(x+π/4)+√2/4)^2 - 5/4 ≤ 1+√2
y的最大值 1+√2

y+1=sin^2x+2sinxcosx+cos^2x+sinx+cosx
y=(sinx+cosx)^2+(sinx+cosx)-1
令t=sinx+cosx
y=t^2+t-1
当x=0时
t=1
y=1+1-1
y=1
当x=-π时
t=sin(-π)+cos(-π)
t=cosπ-sinπ
t=-cos0+sin0
t=-1
y=1-1-1
y=-1
-1=