Question

已知f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1

Answer 1

1、

f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a

=√3sinx+cosx+a

=√[(√3)²+1²]sin(x+z)+a

=2sin(x+z)+a

其中tanz=1/√3,所以z=π/6

sin(x+z)最大=1

所以f(x)最大=2+a=1

a=-1

2、

f(x)=2sin(x+π/6)-1>=0

sin(x+π/6)>=1/2

所以2kπ+π/6