## Precalculus (6th Edition) Blitzer

a) $\alpha =50{}^\circ \text{ and }\beta =5{}^\circ$ in the expansion $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ$. b) The expression $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ$ is equivalent to $\cos 45{}^\circ$. c) The exact value of $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ$ is $\frac{1}{\sqrt{2}}$.
(a) From the difference formula of cosines, $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ The expansion using the above identity can be written as, $\cos \left( 50{}^\circ -5{}^\circ \right)=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ$ Compare the identity with the above expansion to determine the value of $\alpha \text{ and }\beta$. Hence, $\alpha =50{}^\circ \text{ and }\beta =5{}^\circ$. (b) The expansion using the cosine difference formula can be solved as, \begin{align} & \cos \left( 50{}^\circ -5{}^\circ \right)=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \\ & \cos 45{}^\circ =\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \end{align} Hence, the cosine of an angle $45{}^\circ$ is equivalent to the expression $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ$. (c) The expansion using the cosine difference formula can be solved as: \begin{align} & \cos \left( 50{}^\circ -5{}^\circ \right)=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \\ & \cos 45{}^\circ =\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \\ & \frac{1}{\sqrt{2}}=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \end{align} Hence, the exact value of $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ$ is $\frac{1}{\sqrt{2}}$.