Let n>6 positive integer and $A_1A_2...A_n$ a convex polygon. Prove that there exist i and j such that $$|\cos(A_i)-\cos(A_j)|<{1\over 2(n-6)}$$
My trials: I tried by contradiction but it didn’t work. I also tried supposing WLOG that angles A1>A2>A3>...>An. I also tried factoring cos(Ai)-cos(Aj) but it didn’t work for me either.
Would a pigheon principle or anything be useful here? Please help.
The angles $A_i$ are between $0$ and $\pi$ and we know that $\sum(\pi-A_i)=2\pi$. If the claim is false, then for each integer $k$, there is at most one $i$ with $$1-\frac{k+1}{2(n-6)}<\cos A_i\le 1-\frac{k}{2(n-6)}.$$ It follows that the $k$th largest angle has cosine $\le 1-\frac{k-1}{2(n-6)}$. Then the six smallest angles have cosine $\le 1-\frac{n-1}{2(n-6)}$, $\le1-\frac{n-2}{2(n-6)}$, $\le1-\frac{n-3}{2(n-6)}$, $\le1-\frac{n-4}{2(n-6)}$, $\le1-\frac{n-5}{2(n-6)}$, and $\le1-\frac{n-6}{2(n-6)}=\frac 12$, respectively. We conclude that the $6$th smallest angle is $\le \frac23\pi$ and the five smaller angles are $<\frac23\pi$. Therefore, $$ \sum(\pi-A_i)>6\cdot \frac\pi3=2\pi,$$ contradiction