# Convex polygons - For $n>3$, $\sum_{i=1}^{n} \cos\theta_i < \cos\left(\sum_{i=1}^{n} {\theta_i}\right)$

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.

• 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. Aug 30, 2022 at 19:38
• @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 Aug 30, 2022 at 21:04

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.