Question

若-π/2≦x≦π/2，则有函数f（x）＝sinx＋√3cosx的最大值和最小值分别是多少

我已经知道答案了分别是2，－1 求过程

Answer 1

答：
f（x）＝sinx＋√3cosx

=2（sinxcosπ/3+cosxsinπ/3)
=2sin(x+π/3)
-π/2≦x≦π/2

-π/6≦x+π/3≦5π/6

所以：-1/2<=sin(x+π/3)<=1
所以：-1<=f(x)<=2