## Precalculus (6th Edition) Blitzer

The exact value of $\cos \left( 135{}^\circ +30{}^\circ \right)$ is $-\frac{\left( 1+\sqrt{3} \right)}{2\sqrt{2}}$.
Use the sum formula of cosines and evaluate the expression as, $\cos \left( 135{}^\circ +30{}^\circ \right)=\cos 135{}^\circ \cos 30{}^\circ -\sin 135{}^\circ \sin 30{}^\circ$ Rewrite the expression for sine and cosine of $135{}^\circ$ as the sum of two angles as, \begin{align} & \sin 135{}^\circ =\sin \left( 45{}^\circ +90{}^\circ \right) \\ & \cos 135{}^\circ =\cos \left( 45{}^\circ +90{}^\circ \right) \\ \end{align} Use the sum formula of sines and cosines simultaneously to evaluate the modified expressions as, \begin{align} & \cos \left( 135{}^\circ +30{}^\circ \right)=\left( \cos 45{}^\circ \cos 90{}^\circ -\sin 45{}^\circ \sin 90{}^\circ \right)\cos 30{}^\circ \\ & -\left( \sin 45{}^\circ \cos 90{}^\circ +\cos 45{}^\circ \sin 90{}^\circ \right)\sin 30{}^\circ \end{align} Substitute the values, $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 30{}^\circ =\frac{\sqrt{3}}{2},\text{ }\cos 90{}^\circ =0,\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\sin 30{}^\circ =\frac{1}{2}$ , and $\sin 90{}^\circ =1$. \begin{align} & \cos \left( 135{}^\circ +30{}^\circ \right)=\left( \left( \frac{1}{\sqrt{2}}\times 0 \right)-\left( \frac{1}{\sqrt{2}}\times 1 \right) \right)\frac{\sqrt{3}}{2}-\left( \left( \frac{1}{\sqrt{2}}\times 0 \right)+\left( \frac{1}{\sqrt{2}}\times 1 \right) \right)\frac{1}{2} \\ & =-\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} \\ & =-\frac{\left( 1+\sqrt{3} \right)}{2\sqrt{2}} \end{align} Hence, the exact value of $\cos \left( 135{}^\circ +30{}^\circ \right)$ is equivalent to $-\frac{\left( 1+\sqrt{3} \right)}{2\sqrt{2}}$.