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

I'm trying to show this, but I can't seem to get it right. Graphically this equality seems to be intuitive, but I can't see how to prove it.
I tried setting $f(x)=\sin x$ and $g(x)=x$ and looking at the way that $f(x)$ decreases along $(0,\pi)$ by looking at its second derivative and comparing that to $g(x)$, but it didn't seem to work that way. Although I was close, it doesn't seem like to me that that method will work. Does anyone have a better way (preferably one that doesn't involve having to physically draw graphs?)
 A: $$\sin^2{\frac{\pi}{12}} = \frac{2-\sqrt{3}}{4} = \frac14 \frac1{2+\sqrt{3}} \gt \frac1{16}$$
because $2 \gt \sqrt{3}$.  The result follows.
A: If one can use $\sin \frac{\pi}{6} = \frac{1}{2}$, the double-angle formula gives
$$\sin \frac{\pi}{12} = \frac{\sin \frac{\pi}{6}}{2\cos \frac{\pi}{12}} = \frac{1}{4\cos \frac{\pi}{12}},$$
and all that remains is to argue that $0 < \cos \frac{\pi}{12} < 1$.
A: Using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html, $$\sin\dfrac\pi6-\sin\dfrac\pi{12}=\cdots>0$$ as sine,cosine ratios are positive in the first quadrant.
A: \begin{align}
   \cos\dfrac{\pi}{12} &<1 \\
   \sin \dfrac{\pi}{12} \cos\dfrac{\pi}{12}&<\sin \dfrac{\pi}{12} \\
   \frac 12\left(2\sin \dfrac{\pi}{12} \cos \dfrac{\pi}{12}\right)
      &<\sin \dfrac{\pi}{12} \\
   \frac 12\sin \dfrac{\pi}{6}&<\sin \dfrac{\pi}{12} \\
   \frac{1}{4}&<\sin \dfrac{\pi}{12} \\
\end{align}
A: Or , you could try this:
$$ \frac {\pi}{12} = \frac {\pi}{4} -\frac {\pi}{6}$$
using the trig identity,
$$ sin(x + y) = sin(x)cos(y) + sin(y)cos(x)$$
$$ sin(\frac {\pi}{4} + \frac {-\pi}{6}) = sin(\frac {\pi}{4})cos(\frac {-\pi}{6}) +sin(\frac {-\pi}{6})cos(\frac {\pi}{4})$$
$$\implies sin(\frac {\pi}{12})  = \frac {\sqrt{3}-\sqrt{2}}{4}$$
It is obvious that $\sqrt{3} < \sqrt{2}+1 \implies  \frac {\sqrt{3}-\sqrt{2}}{4} < \frac{1}{4}$.
