## Precalculus (6th Edition) Blitzer

The required value of the ${{\cos }^{2}}\alpha$ is $\frac{1+\cos \,2\alpha }{2}$.
In order to find the value of ${{\cos }^{2}}\alpha$, the power reducing formula is used that is as shown below: $\cos 2\alpha =2\,{{\cos }^{2}}\alpha -1$ The aforementioned formula is proved in the earlier problems. Solve the aforementioned formula on the right side for the ${{\cos }^{2}}\alpha$: \begin{align} & 2{{\cos }^{2}}\alpha =1+\cos 2\alpha \\ & {{\cos }^{2}}\alpha =\frac{1+\cos 2\alpha }{2} \end{align} Hence, the required value of the ${{\cos }^{2}}\alpha$ is $\frac{1+\cos \,2\alpha }{2}$.