I want to show the inequality $\lvert\sin(z)\rvert > \frac{2}{\pi}$ for $z$ on the circle of radius $(n+1/2) \pi$. What I have gotten so far is using the fact that
$\lvert\sin(x+iy)\rvert^2 = \sin^2(x)+ \sinh^2(y)$
to get
$\sin^2((n+\frac{1}{2})\pi \cos(\theta)) + \sinh^2((n+\frac{1}{2})\pi \sin(\theta)) > \frac{4}{\pi^2}$ for $\theta \in [0,2\pi]$
I verified on a graphing calculator that such statement is true, but have not been able to present any proper argument.
Please help me, thanks in advance.
WLOG we can consider only $x, y \ge 0$, so $y = \sqrt{\left(n+\frac{1}{2}\right)^2 \pi^2 - x^2}$
Observe (by staring at the graph) that
$f(x) := \sin^2{x} + \sinh^2{y} = \sin^2{x} + \sinh^2{\sqrt{\left(n+\frac{1}{2}\right)^2 \pi^2 - x^2}}$ is decreasing from $x = 0$ to $x = \left(n+\frac{1}{2}\right)\pi $. To prove it, I will use the following two inequalities (you can come up with a proof):
(1) $\sin{x} \le x$
(2) $\sinh{x} \ge x$
Now for $x \ge 0$, we have
$\begin{align} f'(x) &= 2\sin{x}\cos{x} + 2\sinh{y}\cosh{y} \cdot \frac{-x}{y} \\ &= \sin{2x} - 2x \cdot \frac{\sinh{2y}}{2y} \\ &\le 2x\left(1 - \frac{\sinh{2y}}{2y}\right) \text{ ...... by (1)} \\ &\le 2x\left(1 - 1\right) \text{ ...... by (2)} \\ &\le 0. \end{align}$
Therefore $f(x)$ must attain minimum at $x = \left(n+\frac{1}{2}\right)\pi$, and we actually proved a stronger bound:
$|\sin{z}| \ge \sqrt{f\left(\left(n+\frac{1}{2}\right)\pi\right)} = 1$