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