Question

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。。 求具体过程正确答案！！！！！！！！！！！！！！！！！ 谢谢！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！

Answer 1

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<=x<=π/2
-π/3<=x+π/6<=2π/3
x+π/6=π/2，f(x)最大=2sin(π/2)+a=2+a
所以2+a =1
a=-1.

Answer 2

f(x)=sinxcosπ/6-cosxsinπ/6+sinxcosπ/6+cosxsinπ/6+cosx+a.
=2sinxcosπ/6+cosx+a.
=√3sinx+cosx+a.
=2sin(x+π/6)+a.
则f(x)=2sin(x+π/6)+a.
又因为-π/2<x<π/2.
-π/3<x+π/6<2π/3.
x+π/6=π/2，f(x)最大=2sin(π/2)+a=2+a.
所以2+a =1.
即a=-1.