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I wish to find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ as part of a larger problem. I can see that the solution will involve De Moivre's theorem somehow, but I cannot see how to apply it. I have looked at solutions of $z^7 - 1 = 0$ but to no avail. Can anyone suggest a method for solving this problem?

wrb98
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2 Answers2

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Note that $\tan^2 \frac\pi7$, $\tan^2 \frac{2\pi}7$, and $\tan^2 \frac{3\pi}{7}$ are the roots of the polynomial equation $$x^3 - 21x^2 + 35x - 7 = 0.$$ If we label these roots $r$, $s$, and $t$, then Vieta's formulae tell us that $$\begin{cases}r + s + t = 21,\\ rs + rt + st = 35,\\ rst = 7.\end{cases}$$ From these, we would like to calculate $r^3 + s^3 + t^3$. We can do this by taking $$(r + s + t)^3 - 3 (r+s+t)(rs + rt + st) + 3rst = 21^3 - 3\cdot 21 \cdot 35 + 3\cdot 7 = 7077.$$

We can get the polynomial equation that made this solution work by observing that if $\theta$ is a multiple of $\frac\pi7$, then $(\cos\theta + i\sin\theta)^7 = 1$. Expanding this polynomial and taking the imaginary part yields $$7\cos^6\theta \sin \theta - 35 \cos^4 \theta \sin^3\theta + 21 \cos^2 \theta \sin^5\theta - \sin^7 \theta = 0$$ or $$\tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7\tan\theta = 0.$$ If we set $x = \tan\theta$, then the solutions to this are $0$, $\pm \tan \frac\pi7$, $\pm \tan \frac{2\pi}7$, and $\pm \tan \frac{3\pi}7$. So we divide by $x$ and cut all exponents in half to get only the roots we're interested in.

Misha Lavrov
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  • Thanks. I got the polynomial like you did and tried Vieta but it was the cubic factorization that made me stumble! – wrb98 Mar 21 '17 at 23:49
  • As a corollary, any sum of the form $\tan^6 \frac\pi{2k+1} + \tan^6 \frac{2\pi}{2k+1} + \dots + \tan^6 \frac{k\pi}{2k+1}$ can be obtained in this way, and is equal to $\binom{2k+1}{2}^3 - 3 \binom{2k+1}{2}\binom{2k+1}{4} + 3 \binom{2k+1}{6}$. – Misha Lavrov Mar 21 '17 at 23:58
  • Could you use a similar argument if the power were 6k, where k is a positive integer - the inductive step would be messy, no? – wrb98 Mar 22 '17 at 00:38
  • For any power, we could do a similar argument using the Newton–Girard formulae to relate the sum of powers to coefficients of the polynomial. But if the angle is fixed and we want to vary the power, another approach is to write down the linear recurrence whose characteristic polynomial is the polynomial whose roots we're looking at. – Misha Lavrov Mar 22 '17 at 00:43
  • I've never come across Newton-Gilard before, could you perhaps show how this would work in this case? – wrb98 Mar 22 '17 at 00:50
  • See https://en.wikipedia.org/wiki/Newton%27s_identities; these are basically the rules that tell us how to get $r_1^n + r_2^n + \dots + r_k^n$ from the coefficients of the polynomial with roots $r_1, r_2, \dots, r_k$. In the article, $e_i$ is the sum of products of the roots taken $i$ at a time, and $p_i$ is the sum of the $i$-th powers of the roots, so identities like $p_3 = e_1 p_2 + e_2 p_1 + 3 e_3$ tell us how to compute the sum of cubes using the polynomial's coefficients and the sum of squares. – Misha Lavrov Mar 22 '17 at 01:02
  • Let us continue this discussion in chat. – wrb98 Mar 22 '17 at 01:08
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Misha beat me on time, so I will give a slightly different point of view. It is well-known that $\cos\frac{2\pi}{7}$, $\cos\frac{4\pi}{7}$ and $\cos\frac{6\pi}{7}$ are algebraic conjugates, roots of the Chebyshev polynomial $$ p(x)=8x^3+4x^2-4x-1.$$ On the other hand $$ \tan^6\frac{\pi}{7}=\left(\frac{1}{\cos^2\frac{\pi}{7}}-1\right)^3 = \left(\frac{1-\cos\frac{2\pi}{7}}{1+\cos\frac{2\pi}{7}}\right)^2$$ hence $\tan^2\frac{\pi}{7}$, $\tan^2\frac{2\pi}{7}$, $\tan^2\frac{3\pi}{7}$ are algebraic conjugates as well, roots of the polynomial $$ q(x)=x^3-21x^2+35x-7 $$ with companion matrix $$ M = \begin{pmatrix} 0 & 0 & 7 \\ 1 & 0 & -35 \\ 0 & 1 & 21 \end{pmatrix}$$ whose third power is $$ M^3 = \left( \begin{array}{ccc} 7 & 147 & 2842 \\ -35 & -728 & -14063 \\ 21 & 406 & 7798 \\ \end{array} \right)$$ The wanted sum is just $\text{Tr}(M^3)$, hence it equals $7-728+7798=\color{red}{7077}$.

Jack D'Aurizio
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