Question

高中数学三角函数已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cos+a的最大值为1.？

Answer 1

(1)先化简
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a
=sinxcosπ/6+sinπ/6cosx+sinxcosπ/6-sinπ/6cosx+cosx+a
=√3倍sinx+cosx+a
=2sin(x+π/6)+a
当sin(x+π/6)=1时,f(x)最大,最大值为2+a
∴2+a=1 ∴a=-1
(2)由（1）可得f(x)=2sin(x+π/6)-1
f(x)>=0即2sin(x+π/6)-1>=0
∴sin(x+π/6)>=1/2
∴2kπ+π/6,1,高中数学三角函数已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cos+a的最大值为1.
已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cos+a的最大值为1.
(1)求常数a的值
(2)求使f(x)>=0成立的取值 ***
(3)若x€[0, π],求函数f(x)的值域