## Precalculus (6th Edition) Blitzer

The exact value of $\cos \left( 120{}^\circ -45{}^\circ \right)$ is $\frac{-1+\sqrt{3}}{2\sqrt{2}}$.
Use the difference formula of cosines and evaluate the term as, $\cos \left( 120{}^\circ -45{}^\circ \right)=\cos 120{}^\circ \cos 45{}^\circ +\sin 120{}^\circ \sin 45{}^\circ$ Substitute the values $\cos 120{}^\circ =-\frac{1}{2},\text{ }\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\sin 120{}^\circ =\frac{\sqrt{3}}{2},\text{ and }\sin 45{}^\circ =\frac{1}{\sqrt{2}}$. \begin{align} & \cos \left( 120{}^\circ -45{}^\circ \right)=\left( -\frac{1}{2}\times \frac{1}{\sqrt{2}} \right)+\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right) \\ & =\frac{-1+\sqrt{3}}{2\sqrt{2}} \end{align} Hence, the exact value of $\cos \left( 120{}^\circ -45{}^\circ \right)$ is equivalent to $\frac{-1+\sqrt{3}}{2\sqrt{2}}$.