# Show that sin &#x2061;<!-- ⁡ --> ( &#x03C0;<!-- π --> 12 </

Show that $\mathrm{sin}\left(\frac{\pi }{12}\right)>\frac{1}{4}$
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talhekh
If one can use $\mathrm{sin}\frac{\pi }{6}=\frac{1}{2}$, the double-angle formula gives
$\mathrm{sin}\frac{\pi }{12}=\frac{\mathrm{sin}\frac{\pi }{6}}{2\mathrm{cos}\frac{\pi }{12}}=\frac{1}{4\mathrm{cos}\frac{\pi }{12}},$
and all that remains is to argue that $0<\mathrm{cos}\frac{\pi }{12}<1$
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Augustus Acevedo
$\begin{array}{rl}\mathrm{cos}\frac{\pi }{12}& <1\\ \mathrm{sin}\frac{\pi }{12}\mathrm{cos}\frac{\pi }{12}& <\mathrm{sin}\frac{\pi }{12}\\ \frac{1}{2}\left(2\mathrm{sin}\frac{\pi }{12}\mathrm{cos}\frac{\pi }{12}\right)& <\mathrm{sin}\frac{\pi }{12}\\ \frac{1}{2}\mathrm{sin}\frac{\pi }{6}& <\mathrm{sin}\frac{\pi }{12}\\ \frac{1}{4}& <\mathrm{sin}\frac{\pi }{12}\end{array}$