Question

已知f(x)=sin(x+π/6)+sin(x-π/6)+acosx+b,(a,b∈R,且均为常数).（1）求函数f(x

Answer 1

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