[tex]\it sin 105^o=sin(60^o+45^o)=sin60^ocos45^o+sin45^ocos60^o=\\ \\ =\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6+\sqrt2}{4}\\ \\ \\ cos 105^o= cos(60^o+45^o)= cos60^ocos45^o-sin60^osin45^o=\\ \\ =\dfrac{1}{2}\cdot\dfrac{\sqrt2}{2}-\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt2-\sqrt6}{4}=-\dfrac{\sqrt6-\sqrt2}{4}[/tex]
