If $q_1

Question

When I was studying on a conjecture, I noticed this problem:

• If $q_1<q_2\in(0,\frac{1}{2})\subset\mathbb{Q}$, then $f(q_1,q_2)=\cos(q_1\pi)-\cos(q_2\pi)$ is bijective between the domain and the image set?

I find the only counter example with the help of wolfram mathematica: $$ \cos \left(\frac{2 \pi }{7}\right)-\cos \left(\frac{3 \pi }{7}\right)=\cos \left(\frac{\pi }{7}\right)-\cos \left(\frac{\pi }{3}\right) $$ It is confused why there has the only counter example.

This reflects a somehow special property of trigonometric functions, because it does not always hold in other functions, for example:

• If $q_1<q_2\in\mathbb{Q}$, then $f(q_1,q_2)=q_2^2-q_1^2$ is not bijective because there are counter examples : $25^2-20^2=39^2-36^2$and so on.

More details: I find another conjectures with the help of wolfram mathematica.

$\textbf{Conjecture 1.}$ If integer $n\geq 9,n=2^a p^b, a,b \geq 0$, p is a prime and $a\geq2$ if $b=1$, then when there are linear relationships in $\left\{ \cos(\frac{k\pi}{n}):k\in \left[1,\left\lfloor \frac{n-1}{2}\right\rfloor \right]\subset \mathbb{Z}\right\}$, the number of $\cos(\frac{k\pi}{n})$ in each set of linear relationships is $p$.

For example, if $n = 40$, we have $$ \cos \left(\frac{3 \pi }{8}\right)+\cos \left(\frac{\pi }{40}\right)-\cos \left(\frac{7 \pi }{40}\right)-\cos \left(\frac{9 \pi }{40}\right)+\cos \left(\frac{17 \pi }{40}\right)=0\\ -\cos \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{20}\right)-\cos \left(\frac{3 \pi }{20}\right)+\cos \left(\frac{7 \pi }{20}\right)+\cos \left(\frac{9 \pi }{20}\right)=0\\ -\cos \left(\frac{\pi }{8}\right)+\cos \left(\frac{3 \pi }{40}\right)-\cos \left(\frac{11 \pi }{40}\right)+\cos \left(\frac{13 \pi }{40}\right)+\cos \left(\frac{19 \pi }{40}\right)=0 $$ each set of linear relationships is $5$, $5$ is the largest prime factor of $40$.

I'm just an undergraduate student at the university and I don't have any idea to prove these two conjectures. It would be very useful if someone could give a counter example or provide some ideas for proof.

Answer 1

Here is the case when the rational numbers have denominators a prime (a similar approach may work for the general case). That is, let $p$ be a prime, so that

$$\cos(\frac{2\pi k_1}p)-\cos(\frac{2\pi k_2}p)=\cos(\frac{2\pi\ell_1}p)-\cos(\frac{2\pi \ell_2}p)$$ for integers $k_1,k_2,\ell_1,\ell_2\in(0,\frac{p-1}2]$. Now, since the Galois orbit of $\zeta_p=e^{2\pi i/p}\in\mathbb Q(\zeta_p)$ is a normal basis (see this), the Galois orbit of $$\zeta_p+\zeta_p^{-1}=2\cos(\frac{2\pi}p)\in\mathbb Q(\zeta_p+\zeta_p^{-1})=\mathbb Q(\cos(\frac{2\pi }p))$$ is a normal basis. That is, $\{\cos(\frac{2\pi k}p):k\in(\mathbb Z/p)^\times/\{\pm1\}\}$ are linearly independent, since $\mathbb Q(\cos(\frac{2\pi}p))/\mathbb Q$ has Galois group $(\mathbb Z/p)^\times/\{\pm1\}$.

Thus, we must have that $k_1=k_2$, so $\ell_1=\ell_2$ as well.

Answer 2

I find a counter example: $$ \cos \left(\frac{2 \pi }{7}\right)-\cos \left(\frac{3 \pi }{7}\right)=\cos \left(\frac{\pi }{7}\right)-\cos \left(\frac{\pi }{3}\right) $$