# Irrationality of the trigonometric functions

Let

$$z_n:=x_ny_n,$$ where $x_n$, $y_n$ $\in \mathbb{R}^{+}$. If $$z_n\notin\mathbb{Q}$$ show that $$\left(\frac{x_n-y_n}{\sqrt{(x^2_n+1)(y^2_n+1)}-x_ny_n-1}\right)^2\notin\mathbb{Q}$$ or $$\left(\frac{x_n+y_n}{\sqrt{(x^2_n+1)(y^2_n+1)}-x_ny_n+1}\right)^2\notin\mathbb{Q}$$

I think that I can prove this statement, but I don't know if my proof is right. Could you help me? My solution is below:

Let(where $sen(x)=sin(x)$) $$\label{eq:identidade_1} \hbar_{1}(a_n,b_n):=\left(sen^2\left(\frac{a_n+b_n}{2}\right)+sen^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_2} \hbar_{2}(a_n,b_n):=\left(sen^2\left(\frac{a_n+b_n}{2}\right)-sen^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_3} \hbar_{3}(a_n,b_n):=\left(cos^2\left(\frac{a_n+b_n}{2}\right)+cos^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_4} \hbar_{4}(a_n,b_n):=\left(cos^2\left(\frac{a_n+b_n}{2}\right)-cos^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_5} \hbar_{5}(a_n,b_n):=\left(sen^2\left(\frac{a_n+b_n}{2}\right)+cos^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_6} \hbar_{6}(a_n,b_n):=\left(sen^2\left(\frac{a_n+b_n}{2}\right)-cos^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_7} \hbar_{7}(a_n,b_n):=\left(cos^2\left(\frac{a_n+b_n}{2}\right)+sen^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$

$$\label{eq:identidade_8} \hbar_{8}(a_n,b_n):=\left(cos^2\left(\frac{a_n+b_n}{2}\right)-sen^{2}\left(\frac{a_n-b_n}{2}\right)\right)^2$$ We get:

$$\label{eq:relacao1} \hbar_{5}(a_n,b_n)-\hbar_{6}(a_n,b_n)=sen^2(a_n)+2sen(a_n)sen(b_n)+sen^2(b_n)$$

Note that:

$$\label{eq:relacao2} \hbar_{7}(a_n,b_n)-\hbar_{8}(a_n,b_n)=sen^2(a_n)-2sen(a_n)sen(b_n)+sen^2(b_n)$$ So: $$\label{eq:U} \hbar_{5}(a_n,b_n)-\hbar_{6}(a_n,b_n)-(\hbar_{7}(a_n,b_n)-\hbar_{8}(a_n,b_n))=4sen(a_n)sen(b_n)$$

In the other hand:

$$\label{eq:V} -(\hbar_{1}(a_n,b_n)-\hbar_{2}(a_n,b_n))+(\hbar_{3}(a_n,b_n)-\hbar_{4}(a_n,b_n))=4cos(a_n)cos(b_n)$$ Make the substitution $\displaystyle x_n=cot(a_n), y_n=cot(b_n)$ so that we have $\displaystyle cot(a_n) cot(b_n)$ irrational.

We have that exists at least one $i\in\{1,\ldots,8\}$ such that $\hbar_{i}(a,b)$ is irrational. This implies that one of the following expression $\displaystyle sen^2\frac{a_n+b_n}{2},sen^2\frac{a_n-b_n}{2},cos^2\frac{a_n+b_n}{2},cos^2\frac{a_n-b_n}{2}$ is irrational. The results imply that $\displaystyle cot^2\frac{\alpha+\beta}{2}$ or $\displaystyle cot^2\frac{\alpha-\beta}{2}$ is irrational, since that $\displaystyle cot\frac{x}{2}=\frac{senx}{1-cosx}$.

Doing $\displaystyle x=a_n+b_n$ yields

$\displaystyle cot\frac{a_n+b_n}{2}=\frac{sen(a_n+b_n)}{1-cos(a_n+b_n)}=\frac{sena_ncosb_n+senb_ncosa_n}{1-cosa_ncosb_n+senasenb_n}=\frac{cotb_n+cota_n}{csca_ncscb_n-cota_ncotb_n+1}=\frac{cotb_n+cota_n}{\sqrt{(cot^2a_n+1)(cot^2b_n+1)}-cota_ncotb_n+1}$

Similarly, doing $\displaystyle x=a_n-b_n$ yields

$\displaystyle cot\frac{a_n-b_n}{2}=\frac{sen(a_n-b_n)}{1-cos(a_n-b_n)}=\frac{sena_ncosb_n-senb_ncosa_n}{1-cosa_ncosb_n-sena_nsenb_n}=\frac{cotb_n-cota_n}{csca_ncscb_n-cota_ncotb_n-1}=\frac{cotb_n-cota_n}{\sqrt{(cot^2a_n+1)(cot^2b_n+1)}-cota_ncotb_n-1}$

$\blacksquare$

Someone has a simpler solution?

• Just as a possible approach here, you can prove that one of two numbers is irrational if the product is irrational. – Paul Feb 7 '18 at 23:44
• Thank's @shamisen for your edition! – Israel Meireles Chrisostomo Feb 8 '18 at 0:30
• What does that fancy $h$ function mean? How do you write that in LATEX? – Feeds Feb 19 '18 at 16:12
• In latex is \hbar. – Israel Meireles Chrisostomo Feb 21 '18 at 21:20