## Precalculus (6th Edition) Blitzer

The exact value of $\cos \left( 240{}^\circ +45{}^\circ \right)$ is $\frac{\sqrt{3}-1}{2\sqrt{2}}$.
Use the sum formula of cosines and evaluate the expression as, $\cos \left( 240{}^\circ +45{}^\circ \right)=\cos 240{}^\circ \cos 45{}^\circ -\sin 240{}^\circ \sin 45{}^\circ$ Rewrite the expression for sine and cosine of $240{}^\circ$ as the sum of two angles as, \begin{align} & \sin 240{}^\circ =\sin \left( 60{}^\circ +180{}^\circ \right) \\ & \cos 240{}^\circ =\cos \left( 60{}^\circ +180{}^\circ \right) \\ \end{align} Use the sum formula of sines and cosines simultaneously to evaluate the modified expressions as, \begin{align} & \cos \left( 240{}^\circ +45{}^\circ \right)=\left( \cos 60{}^\circ \cos 180{}^\circ -\sin 60{}^\circ \sin 180{}^\circ \right)\cos 45{}^\circ \\ & -\left( \sin 60{}^\circ \cos 180{}^\circ +\cos 60{}^\circ \sin 180{}^\circ \right)\sin 45{}^\circ \end{align} Substitute the values, $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 60{}^\circ =\frac{1}{2},\text{ }\cos 180{}^\circ =-1,\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\sin 60{}^\circ =\frac{\sqrt{3}}{2}$ , and $\sin 180{}^\circ =0$. \begin{align} & \cos \left( 240{}^\circ +45{}^\circ \right)=\left( \left( \frac{1}{2}\times \left( -1 \right) \right)-\left( \frac{\sqrt{3}}{2}\times 0 \right) \right)\frac{1}{\sqrt{2}}-\left( \left( \frac{\sqrt{3}}{2}\times \left( -1 \right) \right)+\left( \frac{1}{2}\times 0 \right) \right)\frac{1}{\sqrt{2}} \\ & =-\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}} \\ & =\frac{\sqrt{3}-1}{2\sqrt{2}} \end{align} Hence, the exact value of $\cos \left( 240{}^\circ +45{}^\circ \right)$ is equivalent to $\frac{\sqrt{3}-1}{2\sqrt{2}}$.