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How can I get the closed-form of the above expression.I tried to find it but I can't do.

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  • $\begingroup$ WA cannot find a closed form. $\endgroup$ – lhf Oct 29 '14 at 13:03
  • $\begingroup$ @lhf: A case of praising with faint damnation? I'm sure WA could find it if the question were posed in just the right way. $\endgroup$ – hardmath Oct 29 '14 at 13:31
  • $\begingroup$ See casus irreducibilis. $\endgroup$ – Lucian Oct 30 '14 at 17:43
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Let $x=\tan(\pi/9)$.

First, $$\tan 3a=\frac{\tan a+\tan 2a}{1-\tan a\tan 2a}=\frac{\tan a+\frac{2\tan a}{1-\tan^2 a}}{1-\frac{2\tan^2 a}{1-\tan^2a}}=\frac{3\tan a-\tan^3a}{1-3\tan^2a}$$

Since $\tan(\pi/3)=\sqrt 3$ you have

$$\sqrt 3=\frac{3x-x^3}{1-3x^2}$$

Therefore, $\tan(\pi/9)$ is one of the roots of the equation $$X^3-3\sqrt 3X^2-3X+\sqrt 3=0$$

If you are able to solve it, you will eventually get that 'closed form'. Cubic equations have a method to solve them, but it is far from simple. Alternatively you may use a computer program to solve the equation, like Maxima, Mathematica, Maple, etc.

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  • $\begingroup$ The closed form of that root is $$x = \sqrt{3}+\frac{1-i \sqrt{3}}{\sqrt[3]{½\left(-\sqrt{3}+i\right)}}+\sqrt[3]{½ \left(-\sqrt{3}+i\right)} \cdot \left(1+i \sqrt{3}\right)$$ $\endgroup$ – Alice Ryhl Oct 29 '14 at 15:00
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Do you allow complex radicals in your "closed form"? (If you apply Cardano's formulas in ajotatxe's solution you will get them.) If so, why not just use radicals like $$ (-1)^{1/9} $$ so that you can write $$ \tan\frac{\pi}{9} = \frac{-2 \left( -1 \right) ^{5/9}+4 \left( -1 \right) ^{4/9}-2 \left( -1 \right) ^{2/9}-2 (-1)^{1/9}}{\sqrt{3}}+\sqrt {3} $$ ??

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  • $\begingroup$ WA does suggest that. $\endgroup$ – lhf Oct 29 '14 at 13:55

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