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It seems that for any convex polygon $P$ with $n>3$ sides and $n$ interior angles $\theta_i$, $$\sum_{i=1}^{n} \sin\theta_i > \sin\left(\sum_{i=1}^{n} {\theta_i}\right)$$ $$\sum_{i=1}^{n} \cos\theta_i < \cos\left(\sum_{i=1}^{n} {\theta_i}\right)$$

The first inequality is really straightforward to prove. As $\sum_{i=1}^{n} {\theta_i}=(n-2)180^°$, thus $\sin\left(\sum_{i=1}^{n} {\theta_i}\right)=0 \space \forall n>2$. The fact that $\sin \theta_i>0 \space \forall \theta_i\neq (k*180)^°$ concludes the proof.

However, I have not been able to prove the second inequality. We have that $\cos\left(\sum_{i=1}^{n} {\theta_i}\right)=-1 \space \forall n=2m+1, \space m\in \mathbb N$ and $\cos\left(\sum_{i=1}^{n} {\theta_i}\right)=1 \space \forall n=2m, \space m\in \mathbb N$. Also, $\cos \theta_i\geq 0 \space \forall \theta_i\leq 90^°$ and $\cos \theta_i< 0 \space \forall 90^°<\theta_i<180^°$.

Any hint, help or reference proving the second inequality would be welcomed.

Thanks!

EDIT

After working on a proof of the second inequality using induction, by trial-and-error experiments, it seems that, for triangles, $\sum_{i=1}^{3} {\cos\theta_i}\leq 1.5$; for quadrilaterals, $\sum_{i=1}^{4} {\cos\theta_i}< 0.5$; for pentagons, $\sum_{i=1}^{5} {\cos\theta_i}< -0.5$; for hexagons, $\sum_{i=1}^{6} {\cos\theta_i}< -1.5$; for heptagons, $\sum_{i=1}^{7} {\cos\theta_i}< -2.5$, ... Thus, the second inequality could be improved to affirm that for any convex polygon $P$ with $n$ sides and $n$ interior angles $\theta_i$, $$\sum_{i=1}^{n} \cos\theta_i \leq 4.5-n$$, with equality only when $n=3$.

The pattern arises as any triangle maximizes the sum of the cosines of interior angles when it has three acute angles equal to $60^º$; any quadrilateral maximizes the sum of the cosines of interior angles when it has three acute angles tending below to $60^º$, and one obtuse angle tending above to $180^º$; and for $n>3$, this pattern continues: any convex polygon maximizes the sum of the cosines of interior angles when it has three acute angles tending below to $60^º$, and the rest of angles are obtuse angles tending above to $180^º$.

A proof by induction could then be constructed, proving the triangle case, and showing that adding an additional vertex to form some convex $(n+1)$-polygon anexes a triangle of angles $\alpha,\beta$ and $(180^º-(\alpha+\beta))$ with one shared side with the $n$-gon, so the sum of the cosines of interior angles of the $(n+1)$-gon is $\cos\theta_{1}+\cos\theta_{2}+...+\cos(\theta_{n-1}+\alpha)+\cos(\theta_{n}+\beta)+\cos(180^º-(\alpha+\beta))$.

Applying the trigonometric identities, we need to show that

$$\cos\theta_{1}+\cos\theta_{2}+...+(\cos\theta_{n-1}\cos\alpha-\sin\theta_{n-1}\sin\alpha)+(\cos\theta_{n}\cos\beta-\sin\theta_{n}\sin\beta)-\cos(\alpha+\beta)\leq \cos\theta_{1}+\cos\theta_{2}+...+\cos\theta_{n-1}+\cos\theta_{n}-1$$ Or, simplifying, $$(\cos\theta_{n-1}\cos\alpha-\sin\theta_{n-1}\sin\alpha)+(\cos\theta_{n}\cos\beta-\sin\theta_{n}\sin\beta)-\cos(\alpha+\beta)\leq \cos\theta_{n-1}+\cos\theta_{n}-1$$, where at most one of $\alpha,\beta$ or $(\alpha+\beta)$ is equal or greater than $90^º$.

Proving the above inequality would finish the proof. However, I have not been able to prove it yet.

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    $\begingroup$ I don't got the time to work on the details but if this is true, then I would attempt to prove odd implies even and then prove inductively the odd case separately. Also, the fact that a convex polygon can have at most $3$ acute angles should help since intuitively this means $n-3$ of the $\cos$ terms are negative. $\endgroup$
    – dezdichado
    Aug 30, 2022 at 19:38
  • $\begingroup$ @dezdichado thanks for your comment! The fact you mention is what makes me guess that the inequality is generally true; I will try to proceed according to your suggestion $\endgroup$ Aug 30, 2022 at 21:04

1 Answer 1

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Finally, I think I got the complete proof of the following statement, stated at the "EDIT" section of the OP: for any convex polygon $P$ with $n$ sides and $n$ interior angles $\theta_i$, $$\sum_{i=1}^{n} \cos\theta_i \leq 4.5-n$$ with equality only when $n=3$.

Any triangle maximizes the sum of the cosines of interior angles when it has three acute angles equal to $60^º$. A proof of this is not difficult and can be found for instance here. Therefore, for any triangle, we have that $\sum_{i=1}^{3} \cos\theta_i \leq 4.5-3$.

Applying the inductive hypothesis, we have that, for some convex polygon $P$ with $n$ sides and $n$ interior angles, $$\sum_{i=1}^{n} \cos\theta_i \leq 4.5-n$$ Adding an additional vertex to form some convex $(n+1)$-gon anexes a triangle of angles $\alpha, \beta$ and $(180^°-(\alpha+\beta))$ with one shared side with the $n$-gon, so the sum of the cosines of interior angles of the $(n+1)$-gon is

$$\cos\theta_{1}+\cos\theta_{2}+...+\cos(\theta_{n-1}+\alpha)+\cos(\theta_{n}+\beta)+\cos(180^º-(\alpha+\beta))$$ So it is needed to prove that $$cos(\theta_{n-1}+\alpha)+\cos(\theta_{n}+\beta)+\cos(180^º-(\alpha+\beta))\leq \cos\theta_{n-1}+\cos\theta_n-1$$

Thanks to @dezdichado contribution here, proving that, for $a,b\geq 0$ and $0\leq a+b\leq 180^°$, $\cos(a)+\cos(b)-\cos(a+b)\geq 1$, we have that $$cos(\theta_{n-1}+\alpha)+\cos(\theta_{n}+\beta)+\cos(180^º-(\alpha+\beta))\leq (\cos\theta_{n-1}+\cos\alpha-1)+(\cos\theta_{n}+\cos\beta-1)-(\cos\alpha+\cos\beta-1)$$ And cancelling terms, $$cos(\theta_{n-1}+\alpha)+\cos(\theta_{n}+\beta)+\cos(180^º-(\alpha+\beta))\leq \cos\theta_{n-1}+\cos\theta_n-1$$

Which finishes the proof.

Therefore, my initial conjecture that $$\sum_{i=1}^{n} \cos\theta_i < \cos\left(\sum_{i=1}^{n} {\theta_i}\right)$$ for any convex polygon $P$ with $n>3$ sides is false for $n=3$ and $n=5$, and true for $n=4$ and $n>5$.

Special thanks to @dezdichado for the useful comment, and cited contribution.

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