## Algebra 2 (1st Edition)

$\dfrac{\sqrt{2-\sqrt 3}}{2}$
Double -angle Theorem can be defined as: $\cos \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1+ \cos \theta}{2}}$ $\cos (-75 ^{\circ})=\cos \dfrac{(-150^{\circ})}{2}$ Thus, $\sqrt {\dfrac{1+ \cos (-150^{\circ})}{2}}= \sqrt {\dfrac{1+ \cos (150^{\circ})}{2}}$ and $\sqrt {\dfrac{1- (\sqrt 3/2)}{2}}=\sqrt {\dfrac{2-\sqrt 3}{4}}$ Hence, $\cos(-75 ^{\circ})=\dfrac{\sqrt{2-\sqrt 3}}{2}$