Prove that $|\sin n|+|\sin (n+1)| > 2\sin(1/2)$ for all $n\in \mathbb N$ 
Show that
  $$|\sin{(n)}|+|\sin{(n+1)}|>2\sin{\dfrac{1}{2}},n \ge 1,n\in \mathbb N$$

My try: let
$$F(n)=|\sin{(n)}|+|\sin{(n+1)}|$$
then
$$F(n+\pi)=|\sin{(n+\pi)}|+|\sin{(n+\pi+1)}|=|\sin{(n)}|+|\sin{(n+1)}|=F(n)$$
and
$$F(\pi-n)=|\sin{(\pi-n)}|+|\sin{(\pi-n+1)}|=|\sin{n}|+|\sin{(n-1)}|\neq F(n)$$
so we must prove when $n\in (0,\pi)$,
have 
$$F(n)>2\sin{\dfrac{1}{2}}$$
when
$n\in (0,\pi-1)$,then
$$F(n)=\sin{n}+\sin{(n+1)}=\sin{n}(1+\cos{1})+\sin{1}\cos{n}$$
and
$n\in(\pi-1,\pi)$,then
$$F(n)=\sin{n}-\sin{(n+1)}$$
How prove it this two case have $F(n)>2\sin{\dfrac{1}{2}}$? Thank you
and I know this well know inequality
$$|\sin{x}|+|\sin{(x+1)}|+|\sin{(x-1)}|\ge 2\sin{1},x\in \mathbb R$$
 A: Observe that
$$
f(x)=\big|\sin x\big|+\big|\sin (x+1)\big| =\left\{
\begin{array}{rll}
\sin x+\sin(x+1) & \text{if} & x\in[0,\pi-1], \\
\sin x-\sin(x+1) & \text{if} & x\in[\pi-1,\pi], \\
-\sin x -\sin (x+1) & \text{if} & x\in [\pi,2\pi-1], \\
-\sin x+\sin(x+1) & \text{if} & x\in [2\pi-1,2\pi].
\end{array}
\right.
$$
Also note that
$$
\sin x+\sin(x+1)=2\sin(x+½)\cos(½),\quad
\sin x-\sin(x+1)=-2\sin(½)\cos(½+x).
$$
Then show a corresponding inequality in each subinterval. 
Unfortunately, the inequality does not hold (for $n=0$ is not true!). The one that does hold is
$$
\big|\sin x\big|+\big|\sin (x+1)\big|\ge \sin 1.
$$
A: The inequality is going to be really hard to prove.  Consider $n=22$:
$$\sin(22)=-0.00885130929,\quad\sin(23)=-0.84622040417,\quad 2\sin(1/2)=0.9588510772$$
This example comes from the approximation $\pi=22/7$.  In general, $k\pi$ is occasionally close to an integer $n$, so that $|\sin(n)|\approx0$ while $|\sin(n+1)|\approx|\sin(1)|=0.8414709848$. 
A: Using induction and the formula $$\sin(nx)=2^{n−1}\Pi_{k=0}^{n−1}\sin(x+\frac{\pi k}{n})$$
which is not very hard to prove by induction (again) for instance, $\sin(2x)=2\sin(x)\cos(x)=2\sin(x)\sin(x+\pi/2)$.
1). $n=1$ case is shown above $|\sin{(1)}|+|\sin{(1+1)}|=\sin(1)(1+2\sin(1+\frac{\pi}{2})>2\sin{\frac{1}{2}}$;
2). $n=k$ assumption case, $|\sin{(k)}|+|\sin{(k+1)}|=2^{k-1}|\sin(1)||\sin(1+\frac{\pi}{k})|\dots|\sin(1+\frac{\pi(k-1)}{k})|(1+2|\sin(1+\frac{k\pi}{k+1}))>2\sin{\frac{1}{2}}$
3). $n=k+1$ then 
$$|\sin{(k+1)}|+|\sin{(k+2)}|=2^k|\sin(1)||\sin(1+\frac{\pi}{k})|\dots|\sin(1+\frac{\pi(k-1)}{k})||(1+2\sin(1+\frac{(k+1)\pi}{k+2}))|$$=$\left\{|\sin{(k)}|+|\sin{(k+1)}|\right\}\left\{2\frac{|\sin(1+\frac{k\pi}{k+1})|+2|\sin(1+\frac{(k+1)\pi}{k+2})|}{1+2|\sin(1+\frac{k\pi}{k+1})|}\right\}>
....>2\sin{\dfrac{1}{2}}$
Here $\dots$ means we need to prove $\left\{2\frac{|\sin(1+\frac{k\pi}{k+1})|+2|\sin(1+\frac{(k+1)\pi}{k+2})|}{1+2|\sin(1+\frac{k\pi}{k+1})|}\right\}>1$.
Note $$\left\{2\frac{|\sin(1+\frac{k\pi}{k+1})|+2|\sin(1+\frac{(k+1)\pi}{k+2})|}{1+2|\sin(1+\frac{k\pi}{k+1})|}\right\}=2\left\{\frac{1+2\frac{|\sin(1+\frac{(k+1)\pi}{k+2})|}{|\sin(1+\frac{k\pi}{k+1})|}}{\frac{1}{|\sin(1+\frac{k\pi}{k+1})|}+2}\right\}>2\left\{\frac{1+2}{\frac{1}{|\sin(1+\frac{k\pi}{k+1})|}+2}\right\}>1$$.
Here $\frac{|\sin(1+\frac{(k+1)\pi}{k+2})|}{|\sin(1+\frac{k\pi}{k+1})|}>1$ and $\min(\sin(1+\frac{k\pi}{k+1}))>\frac{1}{4}$ are used due to sin function's increasing characteristics from $[0,\frac{\pi}{2}]$ where $1$ is within it.
Suppose now, it is proved $|\sin{(n)}|+|\sin{(n+1)}|>2\sin{\dfrac{1}{2}},n \ge 1,n\in \mathbb N$.
