a∈(0,π/2),a+π/6∈(π/6,2π/3),cos(a+π/6)∈(-1/2,√3/2)
f(a)=sin(a+π/6)=2/3
∴cos(a+π/6)=√5/3
f(-a)=sin(-a+π/6)=sin(π/3-(a+π/6))
=√3/2cos(a+π/6)-1/2sin(a+π/6)
=√3/2*√5/3-1/2*2/3
=(√15-2)/6