# $\text{Prove that}$ $\frac{\sin(\frac{n+1}2)*\cos(\frac n2)}{\sin\frac 12} \ge\frac n2$

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?

• Is the question true for $n=3$ ? Sep 20, 2014 at 14:05
• This is wrong because the left side is bounded, while the right side is not!. In fact this is true only for $n=0,1,2$. Sep 20, 2014 at 14:09
• This seems to be true if $-\infty \lt n \le 2$ Sep 20, 2014 at 14:13
• My ultimate goal is to prove that $\sum_{k=0}^n|\cos(k)|\ge\frac n2$ So I was using the fact that $\sum_{k=0}^n\cos(k)= \frac {\sin(\frac {n+1}2)*\cos(\frac n2)}{\sin(\frac 12)}$ I'm going back through the proof and I've used the Dirichlet kernel and found that it also equals $1+\cos(1)+\cos(2)+....+\cos(n)$ I just don't know if I'm on the right track or not. Sep 20, 2014 at 14:22
• i edited my answer to help you with you original proof for: $\sum_{k=0}^n|\cos(k)|\ge\frac n2$ @Fmonkey2001 Sep 20, 2014 at 14:44

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

• Would it matter to the integral you constructed that my "goal" limited my n to be a positive integer? Or since you used n belonging to the natural numbers is that okay? Sep 20, 2014 at 16:08
• i too assumed n to be a positive integer, the integral is just a tool to get to the value of that limit and to show that it is above 1, or maybe i didn't understand what you just asked? i'm here to help (if i can) Sep 20, 2014 at 17:18
• Oh wow! Nevermind! I see! I might have to ask some questions on the fact that it's dense in [0,$\pi$] and the riemann sum. I'm going over some old notes and trying to remember everything. So right now I'm good, but I will be back when I need help! Thanks so much! Sep 20, 2014 at 17:50
• So it's not enough to just prove the above proof? I'm also going to have to use another proof to make it complete? Why does this one proof not suffice? Sep 21, 2014 at 0:43
• So this proof that you did will take me up to $N_h$ and then I'll have to find a numerical proof to take me up to some $n \lt N_h$ Sep 21, 2014 at 13:44

Hint: Show that the claim is false by noting that $\sin$ and $\cos$ are bounded above by $1$.

You have said that $$\frac{\sin(\frac{n+1}2)*\cos(\frac n2)}{\sin\frac 12} \ge\frac n2$$ reaches a maximum of $1.5429$. Now let $n=4$. Then we have $$1.5429 \geq \frac{\sin(\frac{3}{2})*\cos(\frac{2}{1})}{\sin\frac 12} \ge\frac{4}{2} \\ \implies 1.5429 \geq \frac{4}{2}=2$$ Obvious contradiction.

The statement is actually false. Simply trying $n=10$:

$$\frac{\sin((10+1)/2)\cos(10/2)}{\sin(1/2)} \approx -0.42 < 10/2 = 5$$