Are you truly irrational?
Let $\pi=a/b$, the quotient of two positive coprime integers.
Let's define the polynomials.
$f(x)=\frac{x^{n}(a+bx)^{n}}{n!}$ and
$\mathbf{F(x)}=f(x)-f^{2}(x)+f^{4}(x)-...
+(-1)^{n}f^{(2n)}(x)$
We will specify the positive integer $n$ later.
$n!f(x)$ has integer coefficient and it is a function of $x$ of degree no fewer than $n$, $f(x)$ and its derivative $f^{(i)}(x)$ have integral values for $x=0$ and also for $x=\pi=a/b$, since $f(x)=f(\frac{a}{b}-x)$ considering elementary calculus, we have
\mathbf{F}(x)cosx\}=$
${\mathbf{F}^{''}}(x)sinx+\mathbf{F}(x)sinx=f(x)sinx$ and
[\mathbf{F}^{'}(x)sinx-\mathbf{F}(x)cosx]^{\pi}_{0}$
$\mathbf{F}(\pi) = \mathbf{F}(0)$ (1)
We have $\mathbf{F}(\pi) = \mathbf{F}(0)$ is an integer since $f^{(i)}(\pi)$
and $f^{(i)}(0)$ are integers. However for $ 0 < x < {{\pi}} $;
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