sequences and series - $text{Prove that}$ $frac{sin(frac{n+1}2)*cos(frac n2)}{sinfrac 12} gefrac n2$

Prove that$$\frac{\sin\left(\frac{n+1}2\right)\times\cos\left(\frac n2\right)}{\sin\left(\frac 12\right)} \ge\frac n2$$

So far I've switched up the problem and gotten it down to all sin functions. I have $$\frac {\sin\left(\frac{2n+1}2\right)+\sin\left(\frac {1}2\right)}{2\sin\left(\frac {1}2\right)}\ge\frac n2$$

So far this is the farthest I've got that seems to make sense. I graph the function each time I do it to make sure that the move I made was a legal move. From the graph I can see that the graph has a maximum at 1.5429

So am I going to have to use induction to prove this statement? or am I going about this all the wrong way?

Answer

look: $$
\frac {1}{sin(\frac {1}2)}\ge\frac{\sin(\frac{n+1}2)\cdot\cos(\frac n2)}{\sin(\frac 12)} \ge\frac n2
$$
so:
$$
\frac {1}{\sin(\frac {1}2)}\ge\frac n2 \iff n\le \frac{2}{\sin(\frac {1}2)} \approx 4.2
$$
that can easily be falsified by choosing an $n$ suffieciently large (like $5$ for example)

EDIT

to help you for your "ultimate goal":
$$
\sum_{k=0}^n|\cos(k)|\ge\frac n2 \iff \frac 2n\sum_{k=0}^n|\cos(k)|\ge1
$$
we can prove very easily that this new statement is true for sufficiently large $n$:
$$
\lim_{n\to +\infty}\frac 2n\sum_{k=0}^n|\cos(k)|=2\Big(\frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(x)|\mbox{ d}x\Big)=\frac{4}{\pi} \ge1 \mbox{(true)}
$$

to get that integral use this fact:
$\{n\ \ \mbox{ mod }\ \ \pi\ \ \ |\ n\in \mathbb{N}\}$ is dense in $]0,\pi[$
then use the definition of riemann sum and successively exploit the fact that $\cos(x)$ is a circular function

to give a complete proof you could use numerical calculations to prove that it works in $0\le n\le N_h$ then this other proof to show that above $N_h$ it also works