Question

X∈【0，π/2】 函数y=sinx+cosx的最大值和最小值

Answer 1

y=sinx+cosx
=√2sin(x+π/4)
x∈【0，π/2】
x+π/4∈【π/4，3π/4】
当x+π/4=π/4或3π/4时y最小值为1
当x+π/4=π/2时y最大值为√2

Answer 2

y=sinx+cosx
=√2(√2/2*sinx+√2/2*cosx)
=√2(sinxcosπ/4+cosxsinπ/4)
=√2sin(x+π/4)

x∈[0，π/2]
x+π/4∈[π/4，3π/4]

当x=π/4时,sin(x+π/4)=1
ymax=√2

当x=π/2时,sin(x+π/4)=√2/2
ymin=√2/2*√2=1

Answer 3

解：
y=sinx+cosx=√2sin（x+π/4)
x+π/4∈[π/4,3π/4]
则sin（x+π/4)∈[√2/2,1]
所以ymax=√2
ymin=1