Question

函数y=√3sinx+cosx,x∈[-π/2,π/2]的最大值是

y=√3sinx+cosx
=2sin(x+A)，(其中tanA=1/√3，即A=π/6)
=2sin(x+π/6)

因为x∈[-π/2,π/2]
所以(x+π/6)∈[-π/3,(2π)/3]
所以y(max)=2
但是我最后怎么算出来一会是等于1，一会又等于√3呢？求解答

Answer 1

(x+π/6)∈[-π/3,(2π)/3]

sinx在(-π/3,π/2)递增
(π/2,2π/3)递减
所以最大是sinπ/2=1
所以最大值是2×1=2

Answer 2

(x+π/6)∈[-π/3,(2π)/3]

当 x+π/6=π/2 时
sin(x+π/6)最大值=1
2sin(x+π/6)最大值=2
不是取端点值