Show that $\mathrm{sin}\left(\frac{\pi}{12}\right)>\frac{1}{4}$

auto23652im
2022-07-07
Answered

Show that $\mathrm{sin}\left(\frac{\pi}{12}\right)>\frac{1}{4}$

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talhekh

Answered 2022-07-08
Author has **15** answers

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 $0<\mathrm{cos}\frac{\pi}{12}<1$

$\mathrm{sin}\frac{\pi}{12}=\frac{\mathrm{sin}\frac{\pi}{6}}{2\mathrm{cos}\frac{\pi}{12}}=\frac{1}{4\mathrm{cos}\frac{\pi}{12}},$

and all that remains is to argue that $0<\mathrm{cos}\frac{\pi}{12}<1$

Augustus Acevedo

Answered 2022-07-09
Author has **4** answers

$\begin{array}{rl}\mathrm{cos}{\displaystyle \frac{\pi}{12}}& <1\\ \mathrm{sin}{\displaystyle \frac{\pi}{12}}\mathrm{cos}{\displaystyle \frac{\pi}{12}}& <\mathrm{sin}{\displaystyle \frac{\pi}{12}}\\ \frac{1}{2}(2\mathrm{sin}{\displaystyle \frac{\pi}{12}}\mathrm{cos}{\displaystyle \frac{\pi}{12}})& <\mathrm{sin}{\displaystyle \frac{\pi}{12}}\\ \frac{1}{2}\mathrm{sin}{\displaystyle \frac{\pi}{6}}& <\mathrm{sin}{\displaystyle \frac{\pi}{12}}\\ \frac{1}{4}& <\mathrm{sin}{\displaystyle \frac{\pi}{12}}\end{array}$

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