# Evaluate $\sin(\frac{19\pi}{12})$

## 1 Answer

\[=\frac{1-\cos (\frac{\pi}{6})}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}=\frac{2-\sqrt{3}}{4}.\]

Since $0< \frac{19\pi}{12} < \pi $, $\sin (\frac{19\pi}{12})>0$. Hence

\[\sin (\frac{19\pi}{12})=\sqrt{\frac{2-\sqrt{3}}{4}}.\]

Erdos

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