Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 959: 8

$\dfrac{\sqrt{2+\sqrt 3}}{2}$

Double -angle Theorem can be defined as: $\sin \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{2}}$ $\sin (5 \pi/12)=\sin \dfrac{5 \pi/6}{2}$ or, $=\pm \sqrt {\dfrac{1- \cos (5 \pi/6)}{2}}$ Thus, $\sqrt {\dfrac{1- \cos (5\pi/6)}{2}}= \sqrt {\dfrac{1-(-\sqrt 3/2)}{2}}$ or, $=\dfrac{\sqrt{2+\sqrt 3}}{2}$