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.

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成立的取值集合 (3)若x€[0, π],求函数f(x)的值域

Answer 1

解:(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<=x+π/6<=2kπ+5π/6
解得{x|2kπ<=x<=2kπ+2π/3,k∈Z}
(3)∵x€[0,π],∴x+π/6∈[π/6,7π/6]
∴f(x)的最大值为1，最小值为f(π)=2sin(7π/6)-1=-2
∴值域为【-2，1】