Chapter 7 - Analytic Trigonometry - 7.5 Sum and Difference Formulas - 7.5 Assess Your Understanding - Page 487: 17

$\frac{\sqrt 6+\sqrt 2}{4}$

$sin(\frac{5\pi}{12})=sin(\frac{3\pi}{12}+\frac{2\pi}{12})=sin(\frac{\pi}{4}+\frac{\pi}{6})=sin(\frac{\pi}{4})cos(\frac{\pi}{6})+cos(\frac{\pi}{4})sin(\frac{\pi}{6})=(\frac{\sqrt 2}{2})(\frac{\sqrt 3}{2})+(\frac{\sqrt 2}{2})(\frac{1}{2})=\frac{\sqrt 6+\sqrt 2}{4}$