已知函数f(x)=sin(x-π/6)+sin(x-π/6)+cosx+a(a属于R,a是常数)

题目

已知函数f(x)=sin(x-π/6)+sin(x-π/6)+cosx+a(a属于R,a是常数)
)（2）若x属于[-π/2,π/2]时,f(x)的最大值为1,求a的值.

答案

f(x)=sinxcosπ/6-cosxsinπ/6+sinxcosπ/6+cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=2sin(x+π/6)+a
f(x)=2sin(x+π/6)+a
-π/2