A RATIONAL VALUE of COSINE // مقدار گویای کسینوس

               In the magazine GAZETA MATEMATICĂ No 4, 1978, on covers 3-4 the following Problem appears (as a solution to the one proposed in GMB 11/1977, page 455 Problem#3) :

          Problem 3 page 455 / No. 11/1977  : Prove that if   $\alpha \in \mathbb{R}$  and 

      $\cos(\alpha \pi)=\frac{1}{3}$  then  $\alpha$  is irrational. (The angle  $\alpha \pi$ is considered in radians)

(We will return to this issue at the end)

               In the past, in my youth, I was more diligent. In the magazine GMB 10/2014 (so 35 years after the one mentioned in the preamble), in the -recently established- Column PROBLEMS for NATIONAL EXAMS, in the 12th grade, page 478, we see problems 27 and 28. (We made a system of cards with them, by grade, etc.; that's why I said "diligence")

This statement

Solving

As you can see, in Problem 2c), the question arises whether at the value  $q\pi$  for which  $\cos q\pi=\frac{3}{5}$, we have  $q\in\mathbb{Q}$  or not. When resolving this point, I remembered the problem in the Preamble. The answer is NO.

So, $\cos q\pi=\frac{3}{5}\;\Rightarrow\; q\notin \mathbb{Q}$.

         We assume the opposite (see line (1) in the third picture) : 

$\exists q\in \mathbb{Q}\;\;s.t.\;\;\cos q\pi=\frac{3}{5}\tag{1}$

Let  $q=\frac{m}{n},\;m\in\mathbb{Z},\;n\in \mathbb{N}$, with their greatest common divisor  $(m,n)=1$. Then

 the set of values ​​of the string  $\{\cos kq\pi;\;q\in\mathbb{Z}\}$ is finite :                           (2)

$\pm\cos 0q\pi,\;\pm\cos q\pi,\;\pm\cos 2q\pi,\dots,\;\pm\cos (n-1)q\pi \tag{3}$

     Indeed, we will show that, whatever  $k\in\mathbb{Z},\;\cos kq\pi$  has one of the values  (3). Writing, with Euclid's division lemma,  $k=n\cdot t+l,\;\;t\in\mathbb{Z},\;l\in\{o,\;1,\;\dots,\;n-1\}$,  we have

$\cos kq\pi=\cos k(\frac{m}{n})\pi=\cos (nt+l)\frac{m}{n}\pi=\cos (mt+l\frac{m}{n})\pi=\cos (mt\pi+lq\pi)=\pm\cos lqt$

qed$\square$

       On the other hand, we will show that if  $k$ is a power of 2, then  $\cos k\pi$  takes on an infinity of values. More precisely, we show, by mathematical induction, the equality :

$\cos 2^kq\pi=\frac{a_k}{5^{2^k}},\;\;\;a_k\in\mathbb{Z},\;\;\;5\not{\mid}a_k \tag{4}$

For $k=0,\;\;\cos 2^0q\pi=\cos q\pi=\frac{3}{5}=\frac{3}{5^{2^0}}$, and we have  $a_0=3,\;5\not{\mid }a_0.$ Then  $\cos 2q\pi=2\cos ^2q\pi-1=2(\frac{3}{5})^2-1=-\frac{7}{25}=\frac{-7}{5^{2^1}}$, so  $a_1=-7\;and\;5\not{\mid}a_1$.

Assuming (4) true for a fixed  $k$, we have

$\cos 2^{k+1}q\pi=\cos 2(2^kq\pi)=2\cos^2(2^kq\pi)-1=2\cdot (\frac{a_k}{5^{2^k}})^2-1=\frac{2a_k^2-5^{2^k\cdot 2}}{5^{2^k\cdot 2}}=\frac{2a_k^2-5^{2^{k+1}}}{5^{2^{k+1}}}$

and taking  $a_{k+1}=2a_k^2-5^{2^{k+1}}$  we have  $5\not{\mid}a_{k+1}$  because  $5\not{\mid}a_k$. So (4) is thrue for  $k+1$. Moreover, the values ​​given by (4) are all distinct, because if we had  $\frac{a_k}{5^{2^k}}=\frac{a_l}{5^{2^l}},\;k<l$, it would result  $a_l=5^{2^l-2^k}\cdot a_k$  so  $5\mid a_l$ - false.

$\blacksquare$

               The same argument is used in the original problem in the Preamble, where, from  $\cos \alpha \pi=\frac{1}{3}$,  results :

$\cos 2^k\alpha \pi=\frac{r}{3^{2^k}},\;\;r\in\mathbb{Z},\;\;\;3\not{\mid}r$