## Algebra 2 (1st Edition)

$\dfrac{\sqrt{2-\sqrt 3}}{2}$
Double -angle Theorem can be defined as: $\sin \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{2}}$ $\sin (-\dfrac{11 \pi}{2})=\sin (-\dfrac{11 \pi}{2}+2 \pi)=\sin (\dfrac{ \pi}{12})$ and $\sin (\dfrac{ \pi}{12})=\sin (\dfrac{ \pi/6}{2})$ or, $=\pm \sqrt {\dfrac{1- \cos (\pi/6)}{2}}$ or, $= \sqrt {\dfrac{1-(-\sqrt 3/2)}{2}}$ or, $=\dfrac{\sqrt{2-\sqrt 3}}{2}$