Lower bound for $2\sin^2(y\pi)$ I was trying to understand the proof of a theorem, and the author uses the fact that if $y \in \mathbb{Q} \cap(0, \frac{1}{2}]$, then 
$$2\sin^2(y\pi) \geq \frac{8}{n^2},$$
where $y=\frac{p}{q}$, with $\gcd(p,q)=1$ and $n \in \mathbb{N}, n \geq q$.
I can "see" it's true, because the closer $y$ is to $\frac{1}{2}$ the closer to $1$, $\sin^2(y\pi)$ gets, and the closer to $0$ $y$ is, the closer to $\pi\frac{p}{q}$ $\sin^2(y\pi)$ is, however, if $\pi\frac{p}{q}$ is close to $0$, then $q$ must be big, and then $\frac{4}{n^2}$ will be very small, smaller then $\sin^2(y\pi)$ because of the quadratic term...
But I can't prove this formally ... Can someone give a little help? :)
Thanks!

First, because $\sin \pi y$ is concave on $(0,1/2]$, we have
$$\sin \pi y\geq 2y,\quad\text{for }y\in(0,1/2].$$
This picture illustrates this step:

Then, because both functions are non-negative on $(0,1/2]$, squaring both sides yields to
$$\sin^2\pi y\geq 4y^2=\frac{4p^2}{q^2}\geq\frac{4}{q^2}\geq\frac{4}{n^2}.$$
Multiply both sides by $2$ and you are done.
