The angles \(\dfrac{\pi}{3}~\&~\dfrac{\pi}{6}\) are in "one-half" of an equilateral triangle.
That is a right triangle with sides of measures \(1,~2,~\&~\sqrt 3\)
So \(\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}~\&~\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}\).