## Precalculus (6th Edition) Blitzer

The required value is $\frac{\sqrt{6}-\sqrt{2}}{4}$
We know that the identity $\cos \left( x+y \right)$ is equal to the product of cosine of the first angle and the cosine of the second angle minus the product of the sine of the first angle and second angle. $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ Therefore, the expression can be written in terms of the sum formula: \begin{align} & \cos \left( {{45}^{\circ }}+{{30}^{\circ }} \right)=\cos {{45}^{\circ }}\cos {{30}^{\circ }}-\sin {{45}^{\circ }}\sin {{30}^{\circ }} \\ & =\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2} \\ & =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ & =\frac{\sqrt{6}-\sqrt{2}}{4} \end{align} Hence, the required value is $\frac{\sqrt{6}-\sqrt{2}}{4}$