Recall that
[tex]\sin^2x=\dfrac{1-\cos2x}2[/tex]
So you have
[tex]\sin^2\dfrac{11\pi}{12}=\dfrac{1-\cos\frac{11\pi}6}2=\dfrac{1-\frac{\sqrt3}{2}}2=\dfrac{2-\sqrt3}4[/tex]
This means
[tex]\sin\dfrac{11\pi}{12}=\dfrac{\sqrt{2-\sqrt3}}2[/tex]
where you take the positive square root because [tex]\sin x[/tex] must be positive for angles [tex]x[/tex] in the interval [tex]0<x<\pi[/tex].
