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Because of convenient trigonometric identities, we can find the exact value of things like $\tan70^{\circ}$ because $70^{\circ}=2\cdot35^{\circ}$ (tangent double-angle identity), $35^{\circ}=36^{\circ}-1^{\circ}$ (tangent angle difference identity), $3^{\circ}=3\cdot1^{\circ}$ (tangent triple-angle identity) and $3^{\circ}=18^{\circ}-15^{\circ}$ (tangent angle difference identity), $18^{\circ}=\frac{1}{2}\cdot36^{\circ}$ (tangent half-angle identity), and $15^{\circ}=\frac{1}{2}\cdot30^{\circ}$, leaving $\tan70^{\circ}$ in terms of the more well-known $\tan30^{\circ}$ and $\tan36^{\circ}$. I am fairly sure that we can find exact values by following similar procedures for all $\tan\theta$ ($\theta$ in degrees) as long as $\theta\in\mathbb{Q}$.

So, can we do the same for finding the exact values of inverse trigonometric functions? Specifically, I am looking for a method of finding the exact value of $\tan^{-1}\frac{1}{2}$ in terms of things like integers, radicals, and complex numbers (which are required in the case of $\tan70^{\circ}$, for example), but not other trigonometric functions or infinite series. I am not asking for the exact value itself as I am sure no one would like to burden themselves finding it, but rather just the method so that I may.

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    $\begingroup$ What is the exact value of $\tan 1{^\circ}?$ Since this angle is not constructible, I would be surprised to learn that its tangent can be computed exactly. $\endgroup$
    – saulspatz
    Jun 12, 2018 at 3:56
  • $\begingroup$ I am fairly sure that we can find exact values by following similar procedures for all $\tan\theta$ ($\theta$ in degrees) as long as $\theta\in\mathbb Q$. I rather doubt this, although I suppose it depends somewhat on what you mean by "exact values" and "things like...". For example could you find an exact expression for $\tan20^\circ$? $\endgroup$
    – mweiss
    Jun 12, 2018 at 3:57
  • $\begingroup$ @saulspatz I think you and I are thinking along the same lines. $\endgroup$
    – mweiss
    Jun 12, 2018 at 3:58
  • $\begingroup$ @mweiss Exactly so. $\endgroup$
    – saulspatz
    Jun 12, 2018 at 4:00
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    $\begingroup$ @saulspatz $60^\circ = 3 × 20^\circ$? $\endgroup$
    – For the love of maths
    Jun 12, 2018 at 4:58

1 Answer 1

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Just an illustration (I cannot post this as comment):

$$t_3=\tan3^\circ=\frac{-2 \sqrt{2 \left(\sqrt{5}+3\right)}+\sqrt{6 \sqrt{5}+15}+1}{\sqrt{3}-\sqrt{2 \sqrt{5}+5}-2 \sqrt{4 \sqrt{5}-\sqrt{2 \left(\sqrt{5}+3\right)}-\sqrt{66 \sqrt{5}+150}+12}}$$

$$t_1=\tan1^\circ=-\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{t_3^3+\sqrt{-\left(t_3^2+1\right)^2}+t_3}+\frac{i \left(\sqrt{3}+i\right) \left(t_3^2+1\right)}{2 \sqrt[3]{t_3^3+\sqrt{-\left(t_3^2+1\right)^2}+t_3}}+t_3 $$

So, yes, you can calculate $\tan N^\circ$ for any natural number N exactly.

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  • $\begingroup$ $tan 1^o ≈ sin 1^o ≈ \frac{1}{2\pi} Rad.$. You probably could find $\pi$ in terms of some certain mathematical constants or relations containing integers. $\endgroup$
    – sirous
    Jun 12, 2018 at 13:09
  • $\begingroup$ And you may do the half-angle identity with $\tan1^{\circ}$ to get $\frac{1}{2}^{\circ}$ or the triple-angle identity again for $\frac{1}{3}^{\circ}$. This and the fact that we can get exact values for all natural numbers leads me to believe we can for all rational numbers too. $\endgroup$
    – Davis Rash
    Jun 12, 2018 at 14:48
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    $\begingroup$ @DavisRash: Yes, you can get halves of degree because you can solve quadratic equations and thirds because you can solve qubic equations. But how would you calculate $\tan\frac{1}{5}^{\circ}$ or $\tan\frac{1}{7}^{\circ}$? $\endgroup$
    – Oldboy
    Jun 12, 2018 at 15:06
  • $\begingroup$ @Oldboy You got me there. It looks like we can do $\frac{1}{5}^{\circ}+\frac{1}{5}^{\circ}+\frac{1}{5}^{\circ}+\frac{1}{5}^{\circ}+\frac{1}{5}^{\circ}=1^{\circ}$, but this requires solving a quintic equation. $\endgroup$
    – Davis Rash
    Jun 12, 2018 at 15:36
  • $\begingroup$ @Oldboy Well, it turns out we can use De Moivre's theorem to get $\sin\frac{1}{5}^{\circ}=\frac{1}{2i}\sqrt[150]{\frac{\sqrt{3}}{2}+\frac{i}{2}}-\frac{1}{2i}\sqrt[150]{\frac{\sqrt{3}}{2}-\frac{i}{2}}$, as well as other forms depending on what angle you want. Like in that instance I tried to make the angles $30^{\circ}$, but we can do $45^{\circ}$, say. I'm not sure how useful these expressions are given that they are in super high-order radicals, but I believe this shows it works for all rational angles now. $\endgroup$
    – Davis Rash
    Jun 12, 2018 at 18:16

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