Directly related to: What is $\tan50^\circ$

$50^\circ = \frac{5\pi}{18} $ is a rational multiple of $\pi$. Therefore should be related to the ninth roots of unity $e^{\pi i /18}$, but how does one compute the exact value ?

EDIT I will settle for the extension $\mathbb{Q}[\tan 50^\circ]$ and the minimal polynomial over $\mathbb{Q}$...

Here it is done for the 5-th roots of unity: How to find the exact value of $ \cos(36^\circ) $?


One cannot express $\tan(50^\circ)$ purely in terms of real radicals. For if one could, then one could express $\cos(20^\circ)$ in terms of real radicals, and it is known that one cannot do that. (It is an instance of the casus irreducibilis of the cubic.)

As is implicitly pointed out in the post, one can express $\tan(50^\circ)$ in terms of a primitive ninth root of unity.

Added: The edited question asks for the minimal polynomial of $\tan(50^\circ)$. Let $x=\tan(50^\circ)$. Using the identity $\tan(3\theta)=\frac{3\tan\theta -\tan^3\theta}{1-3\tan^2\theta}$, we find that $\frac{3x-x^3}{1-3x^2}=-\frac{1}{\sqrt{3}}$. Square and simplify. We get a sextic in $x$, which is irreducible by the Eisenstein Irreducibility Criterion.

  • $\begingroup$ I was guessing maybe you take two cube roots in different places. $\endgroup$ – cactus314 Jun 13 '14 at 13:17
  • $\begingroup$ No combination of real (cube) roots will do it. $\endgroup$ – André Nicolas Jun 13 '14 at 13:19
  • $\begingroup$ You can still write down roots in cases of Causis irreducibilis, e.g. $x^3 - 15 x + 4 = (x-4)(x - (2+i))(x-(2-i))$. $\endgroup$ – cactus314 Jun 13 '14 at 14:35
  • 2
    $\begingroup$ Sure. But not for a cubic irreducible over the rationals, in particular the minimal polynomial of $\cos(20^\circ)$. $\endgroup$ – André Nicolas Jun 13 '14 at 15:56

We can find exact value of tan 50∘ (and other integer angles and polygons) using Precise-Rewritten method. The result will be in repeated nested radicals.

For this our target angle is 50∘. Make a table having three columns starting from 90∘ (called as Central):

First column is the Central and its halves. Each Central requires to add or subtract towards target angle (A, which is 50∘ in this case).

Second column is the cumulative of the first column excluding 90∘. We need to reach to the target angle A in this column.

Third column is just $\frac{previous Central}{current Central}$, which is, always, either +2 or -2. Let us see the calculation as:enter image description here

In the table, we can see there is repetition of +2-2+2. We are lucky, we are near to the exact radical values for Sin 50∘.

Now, gather the result from the third column as: $$+2 \overline {+2 - 2 + 2}$$.

Replace each by √(2 as: $$√(2 \overline {+√(2 -√(2 +√(2}$$.

Half of this is Sin 50∘ as $$\frac{1}{2}√(2 \overline {+√(2 -√(2 + √(2} ]$$ Subtract the whole thing after first radical, it will be Cos 50∘ as $$\frac{1}{2}√(2 - \overline {+√(2 -√(2 + √(2}]$$

Now,we can use classical formula for $\tan 50∘$, although Precise-Rewritten method prescribed something else. $$\tan 50∘ = \frac{\frac{1}{2}√(2 \overline {+√(2 -√(2 + √(2} ]}{\frac{1}{2}√(2 - \overline {+√(2 -√(2 + √(2}]}$$

Therefore, $$\tan 50∘ = \frac{√(2 \overline {+√(2 -√(2 + √(2} ]}{√(2 - \overline {+√(2 -√(2 + √(2}]}$$

Note: Closing brackets collapsed for easy only.

Source: Breaking Classical Rules in Trigonometry: Precise-Rewritten method


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