Verified answer

[sin[(π/6)-∆x]-(1/2)]/∆x = [sin(π/6)cos∆x -- cos(π/6)sin∆x]/∆x

= cos∆x/2∆x -- {sqrt(3)/2}sin∆x/∆x -- 1/2∆x

now taking limit as ∆x tends to zero it becomes -- sqrt(3)/2