Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 959: 7

$\dfrac{\sqrt{2+\sqrt 2}}{2}$

Double -angle Theorem can be defined as: $\cos \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1+ \cos \theta}{2}}$ $\cos (\pi/8)=\cos \dfrac{\pi/4}{2}$ Thus, $\sqrt {\dfrac{1+ \cos (\pi/4)}{2}}= \sqrt {\dfrac{1+(\sqrt 2/2)}{2}}$ and $ \sqrt {\dfrac{1+(\sqrt 2/2)}{2}}=\dfrac{\sqrt{2+\sqrt 2}}{2}$ Hence, $\cos (\pi/8)=\dfrac{\sqrt{2+\sqrt 2}}{2}$