Answer
$\dfrac{-\sqrt 6+\sqrt 2}{4}$
Work Step by Step
By using the Sum and Difference formulas: $\cos (a+b)=\cos a \cos b -\sin a \sin b$ and $\cos (a-b)=\cos a \cos b +\sin a \sin b$
$\cos 105^{\circ}=\cos (135^{\circ}-30^{\circ})$
or, $=\cos 135^{\circ} \cos 30^{\circ} +\sin 135^{\circ} \sin 30^{\circ}$
or, $=(\dfrac{-\sqrt 2}{2})(\dfrac{\sqrt 3}{2})+(\dfrac{\sqrt 2}{2})(\dfrac{1}{2})$
or, $=\dfrac{-\sqrt 6+\sqrt 2}{4}$
