Question

高中数学三角函数

已知函数f（x）=sin[x+(6/π)]+sin[x-(6/π)]+acosx+b (a b ∈R )

若f(x)在区间[-π/3，0]上单调递减，且恰好能取到f(x)的最小值2，求a和b的值

Answer 1

(1)f[x]=sin[x+π/6]+sin[x-π/6]+cosx+a 

=sinxcos[π/6]+cosxsin[π/6]+sinxcos[π/6]-cosxsin[π/6]+cosx+a 

=2sinxcos[π/6]+cosx+a 

=√3sinx+cosx+a 

=2sin[x+π/6]+a 

当sin[x+π/6]=1,f[x]取到最大值为f[x]max=2+a=1 

a=-1 

(2)f[x]=2sin[x+π/6]-1 

令u=x+π/6，则sinu 的单调增区间是[-π/2+2kπ，π/2+2kπ] (k∈Z) 

则sin(x+π/6]的单调增区间是[-π/2+2kπ-π/6,π/2+2kπ-π/6](k∈Z) 

则f[x]的单调增区间为[-2π/3+2kπ，π/3+2kπ] ，(π∈Z) 

（3）f[x]=2sin[x+π/6]-1>=0 

sin[x+π/6]>=1/2 

如图在[0,2π]一个周期里满足sin[x+π/6]>=1/2的x范围是 

π/6<= x+π/6<=5π/6 

则所有周期内 x满足 π/6+2kπ<= x+π/6<=5π/6+2kπ 

化简得 2kπ<=x<=2π/3+2kπ，(k∈Z) 

则f[x]>=0成立的X的取值集为{x|2kπ<=x<=2π/3+2kπ，(k∈Z)}

Answer 2

f(x)=sin(x+ π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数）
=2sinxcosπ/6+acosx+b
=√3sinx+acosx+b
=√(3+a^2)sin(x+θ)+b (sinθ=a/√(3+a^2),cosθ=√3/√(3+a^2),
f（x)在区间[-π/3,0]上单调递增，且恰好能够取到f(x)的最小值2
∴-π/3+θ=-π/2,b-√(3+a^2)=2
θ=-π/6
a=-1,b=2+2=4

Answer 3