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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?

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    $\begingroup$ Is the question true for $n=3$ ? $\endgroup$
    – inequal
    Sep 20, 2014 at 14:05
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    $\begingroup$ This is wrong because the left side is bounded, while the right side is not!. In fact this is true only for $n=0,1,2$. $\endgroup$
    – Omran Kouba
    Sep 20, 2014 at 14:09
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    $\begingroup$ This seems to be true if $-\infty \lt n \le 2$ $\endgroup$
    – Claude Leibovici
    Sep 20, 2014 at 14:13
  • $\begingroup$ 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. $\endgroup$
    – Fmonkey2001
    Sep 20, 2014 at 14:22
  • $\begingroup$ i edited my answer to help you with you original proof for: $\sum_{k=0}^n|\cos(k)|\ge\frac n2$ @Fmonkey2001 $\endgroup$
    – Frank
    Sep 20, 2014 at 14:44

4 Answers 4

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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

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  • $\begingroup$ 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? $\endgroup$
    – Fmonkey2001
    Sep 20, 2014 at 16:08
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    $\begingroup$ 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) $\endgroup$
    – Frank
    Sep 20, 2014 at 17:18
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    $\begingroup$ 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! $\endgroup$
    – Fmonkey2001
    Sep 20, 2014 at 17:50
  • $\begingroup$ 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? $\endgroup$
    – Fmonkey2001
    Sep 21, 2014 at 0:43
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    $\begingroup$ 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 $n \lt N_h$ $\endgroup$
    – Fmonkey2001
    Sep 21, 2014 at 13:44
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Hint: Show that the claim is false by noting that $\sin$ and $\cos$ are bounded above by $1$.

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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.

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

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

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