## Algebra 2 (1st Edition)

Using the given formula and the known values: $\sin\frac{23\pi}{12}=\sin3\pi/4\cos7\pi/6+\sin7\pi/6\cos3\pi/4=\frac{1}{\sqrt2}(-\frac{\sqrt3}{2})+(-\frac{1}{2})(-\frac{1}{\sqrt2})=-\frac{\sqrt3-1}{2\sqrt2}$