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

Evaluate the exact value of the expression sin($\frac{19π}{12} $ ) .

1 Answer

\[\sin^2 (\frac{19\pi}{12})=\frac{1-\cos 2(\frac{19\pi}{2})}{2}=\frac{1-\cos \frac{38 \pi}{2}}{2}=\frac{1-\cos (\frac{2\pi}{12})}{2}\]
\[=\frac{1-\cos (\frac{\pi}{6})}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}=\frac{2-\sqrt{3}}{4}.\]

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 Erdos
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