## Precalculus (6th Edition) Blitzer

The exact value of $\cos 105{}^\circ$ is $\frac{1-\sqrt{3}}{2\sqrt{2}}$.
Rewrite the expression for cosine of $75{}^\circ$ as the sum of two angles as, $\cos 105{}^\circ =\cos \left( 60{}^\circ +45{}^\circ \right)$ Use the sum formula of cosines and evaluate the modified expression as, $\cos \left( 60{}^\circ +45{}^\circ \right)=\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ$ Substitute the values, $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 60{}^\circ =\frac{1}{2},\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ and }\sin 60{}^\circ =\frac{\sqrt{3}}{2}$. \begin{align} & \cos \left( 60{}^\circ +45{}^\circ \right)=\left( \left( \frac{1}{2}\times \frac{1}{\sqrt{2}} \right)-\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right) \right) \\ & =\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}} \\ & =\frac{1-\sqrt{3}}{2\sqrt{2}} \end{align} Hence, the exact value of $\cos 105{}^\circ$ is equivalent to $\frac{1-\sqrt{3}}{2\sqrt{2}}$.