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Just I made a curious conjecture when I was playing with my calculator. We will use $\displaystyle\prod_{i=1}^n p_i$ where $p_i$ is the $i$-th prime. Then I have noted that,

$$\left\lvert\cos \left(\displaystyle\prod_{i=1}^3 p_i\right)\right\rvert=\left\lvert\cos \left(\displaystyle\prod_{i=1}^4 p_i\right)\right\rvert=\left\lvert\cos \left(\displaystyle\prod_{i=1}^5 p_i\right)\right\rvert=\dfrac{\sqrt{3}}{2}$$

The typical thing is that if in case of the product of consecutive primes we take any composite number then the equality doesn't hold. For example $\left\lvert\cos (2\cdot 3 \cdot 5 \cdot 6)\right\rvert \neq \dfrac{\sqrt{3}}{2}$, $\left\lvert\cos (2\cdot 3 \cdot 5 \cdot 7 \cdot 8)\right\rvert \neq \dfrac{\sqrt{3}}{2}$. Now since I am calculating on a 12-digit calculator, I don't have plenty of evidence.

Basing upon this very little evidence I am putting forward a conjecture.

Conjecture

$$\displaystyle\lim_{n\to \infty}\left\lvert\cos \left(\displaystyle\prod_{i=1}^n p_i\right)\right\rvert=\dfrac{\sqrt{3}}{2}$$

I think that both the conjectures are probably false but I can't prove or disprove anyone of them. Though counterexamples are welcome but in that case I will be glad to know if there are infinitely many counterexamples or not.

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  • $\begingroup$ Perhaps you can use somehow the fact that $|\cos x| = \sqrt{3}/2$ iff $x = 30 + 180N$ or $x = 150 + 180N$ for some integer $N$. $\endgroup$ – Yuval Filmus Nov 26 '14 at 6:42
  • $\begingroup$ @YuvalFilmus: It seems that but the product of primes is very erratic and asymptotic bounds isn't helping me at all. $\endgroup$ – user 170039 Nov 26 '14 at 6:47
  • $\begingroup$ Interesting to see $p_i$ instead of $\pi$ occuring here :) $\endgroup$ – Hagen von Eitzen Nov 26 '14 at 7:10
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As I note in my comment, $|\cos x| = \sqrt{3}/2$ iff $x=30+180N$ or $x=150+180N$ for some integer $N$. Now suppose that $x = 2\cdot 3\cdot 5\cdot y = 30y$. In this case the criterion is simpler: $|\cos x| = \sqrt{3}/2$ iff $y=1+6N$ or $y=-1+6N$ for some integer $N$, or more succinctly $y \equiv \pm 1 \pmod{6}$. If $y$ satisfies this relation then for any number $z$ satisfying $z \equiv \pm 1 \pmod{6}$ we will have $yz \equiv \pm 1 \pmod{6}$, and so $|\cos xz| = \sqrt{3}/2$. This means that $|\cos(2\cdot3\cdot5\cdot A)| = \sqrt{3}/2$ whenever $A$ is a product of integers which are equivalent to $\pm 1$ modulo $6$. All primes larger than $3$ are in this class, thus verifying your second conjecture. Your first conjecture is completely bogus: in fact, your conditions hold iff $a \equiv \pm1 \pmod{6}$.

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  • $\begingroup$ The previous version of the question did have two conjectures; the first was deleted. $\endgroup$ – Yuval Filmus Nov 26 '14 at 17:17

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