Question

已知函数f（x）=asin（x-π/3）+asin（x+π/3）-2sin²x,其中x∈［0,π］,a为常数.

Answer 1

x∈［0,π］

-π/3≤x-π/3≤2π/3

sin（x-π/3）=1/2,

x=π/6,

f（π/6）=asin（π/6-π/3）+asin（π/6+π/3）-2sin²(π/6)

=-1/2 a+a-1/2

=(a-1)/2.