Leland Morrow

2022-07-01

Justify that $\mathrm{tan}{1}^{\circ }$ is an irrational number.

trajeronls

Here is a more algebraic approach not given in Minus One-Twelfth's link. Let $c$ and $s$ denote $\mathrm{cos}{1}^{\circ }$ and $\mathrm{sin}{1}^{\circ }$, respectively. We have
$\left(c+is{\right)}^{180}={e}^{\pi i}=-1,$ so
$\mathrm{\Im }\left(c+is{\right)}^{180}=0.$ That is
$\sum _{k=1}^{90}\left(-1{\right)}^{k-1}\left(\genfrac{}{}{0}{}{180}{2k-1}\right){s}^{2k-1}{c}^{180-2k+1}=0.$ Note that $s,c\ne 0$. If $t$ is $\mathrm{tan}{1}^{\circ }$, we have
$\sum _{k=1}^{90}\left(-1{\right)}^{k-1}\left(\genfrac{}{}{0}{}{180}{2k-1}\right){t}^{2\left(k-1\right)}=\frac{\sum _{k=1}^{90}\left(-1{\right)}^{k-1}\left(\genfrac{}{}{0}{}{180}{2k-1}\right){s}^{2k-1}{c}^{180-2k+1}}{s{c}^{179}}=0.$
If $t$ is a rational number, then $t=\frac{p}{q}$ for some positive integers p,q such that $gcd\left(p,q\right)=1$. By the rational root theorem, $p,q\mid 180$, so we can write $t=\frac{n}{180}$ for some integer $n$. Because
$\frac{\pi }{180}<\mathrm{tan}\frac{\pi }{180}<\frac{\mathrm{tan}\left(\pi /4\right)}{45}=\frac{1}{45},$
where the first inequality is due to the inequality $\mathrm{tan}x>x$ for all $x\in \left(0,\pi /2\right)$, and the second inequality is true by convexity of $\mathrm{tan}$ on $\left(0,\pi /2\right)$. However, this means
$\pi
but this is a contradiction (no integers lie strictly between $\pi$ and $4$). So $t=\mathrm{tan}{1}^{\circ }$ cannot be rational.

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