I want to show $\tan 3^{\circ}$ is irrational I have looked at other similar questions but they all have different patterns to approach trignometric values like $\tan3$ so what I am thinking of is assuming $tan 3$ is rational that is to say $\tan 3 = a/b$ so this means $\sin 3^{\circ} / \cos 3^{\circ} = a/b$ so we have $\sin 3^{\circ}b= \cos 3^{\circ} a$ so from here I want to find a contradiction but I need help.


Assume $\tan(3^{\circ})$ is rational that is $\exists a,b\in\mathbb{N}$, $b\neq0$ and $\gcd(a,b)=1$ such that $\displaystyle\tan(3^{\circ})=\frac{a}{b}$. Therefore : $$ \tan(6^{\circ})=\frac{2\tan(3^{\circ})}{1-\tan^{2}(3^{\circ})} =\displaystyle\frac{\displaystyle\frac{2a}{b}}{\displaystyle\frac{b^{2}-a^{2}}{b^{2}}} =\frac{2ab}{b^{2}-a}\in\mathbb{Q} $$ Repeat this until you observe that $\tan(12^{\circ})\in\mathbb{Q}\implies\tan(24^{\circ})\in\mathbb{Q}$. However notice that : $$ \tan(30^{\circ})=\tan(24^{\circ}+6^{\circ})=\frac{\tan(24^{\circ})+\tan(6^{\circ})}{1-\tan(24^{\circ})\tan(6^{\circ})} $$ Since $\tan(24^{\circ}) ,\tan(6^{\circ})\in\mathbb{Q}$, then $\tan(30^{\circ})\in\mathbb{Q}$. However : $$ \tan(30^{\circ})=\frac{\sqrt{3}}{3}\notin\mathbb{Q} $$ Contradiction is thus achieved. Hence, $\tan(3^{\circ})\notin\mathbb{Q}$

  • 2
    $\begingroup$ Alright thank you for this nice approach. $\endgroup$ Feb 8 '21 at 5:05
  • 1
    $\begingroup$ +1 Really nice approach, I now have new method in my toolkit :) $\endgroup$ Feb 8 '21 at 8:44
  • $\begingroup$ I am so glad to hear $\text{:-)}$ $\endgroup$ Feb 8 '21 at 8:57

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