Trignometric Horrors I

If we take $\theta=\dfrac{2\pi}{7}$ , then we need to evaluate $\alpha=\tan\theta+\tan2\theta+\tan4\theta$ we observe that $\theta+2\theta+4\theta=2\pi$ . So, we have $\cos(\theta+2\theta+4\theta)=1\Rightarrow \cos\theta\cos2\theta\cos4\theta(1-\alpha)=1$ Now, $\cos\theta\cos2\theta\cos4\theta=\dfrac{\sin 8\theta}{8\sin\theta}=1/8$ . Hence we get $\alpha=1-8=-7\Rightarrow |\alpha|=\boxed{7}$

Consider the equation $\cos(7z) = -1$ , by the Chebyshev Polynomial of the First Kind , letting $y = \cos(z)$ . We have

$64y^7 - 112y^5 + 56y^3 - 7y + 1 = 0$

Because $y=-1$ is a solution, we can factor out the expression, and knowing that $\cos \left ( \frac {\pi}{7} \right ), \cos \left ( \frac {2\pi}{7} \right ), \cos \left ( \frac {3\pi}{7} \right ) , \ldots , \cos \left ( \frac {6 \pi}{7} \right )$ are roots of the sextic equation. Finally the equation factors to

$(y+1)(8y^3 - 4y^2 - 4y + 1)^2 = 0$

The cubic equation has a multiplicity of $2$ because $\cos \left ( \frac {\pi}{7} \right ) = -\cos \left ( \frac {6\pi}{7} \right ), \cos \left ( \frac {2\pi}{7} \right ) = -\cos \left ( \frac {5\pi}{7} \right ), \cos \left ( \frac {3\pi}{7} \right ) = -\cos \left ( \frac {4\pi}{7} \right )$

The cubic equation below has roots $\cos \left ( \frac {\pi}{7} \right ), \cos \left ( \frac {2\pi}{7} \right ), \cos \left ( \frac {3\pi}{7} \right )$

$8y^3 - 4y^2 - 4y + 1 = 0$

With a further substitution of $x = \frac {1}{y}$ and some simplification. We have another cubic equation with roots $\sec \left ( \frac {\pi}{7} \right ), \sec \left ( \frac {2\pi}{7} \right ), \sec \left ( \frac {3\pi}{7} \right )$

$x^3 - 4x^2 - 4x + 8 = 0$

Denote $f(x) = x^3 - 4x^2 - 4x + 8$ , then $f(\sqrt{x}) = 0$ has roots $\sec^2 \left ( \frac {\pi}{7} \right ), \sec^2 \left ( \frac {2\pi}{7} \right ), \sec^2 \left ( \frac {3\pi}{7} \right )$

$x \sqrt{x} - 4x - 4\sqrt{x} + 8 = 0$

$\Rightarrow x(x-4)^2 = (4x-8)^2$

$\Rightarrow x^3 - 24x^2 + 80x - 64 = 0$

By Vieta's Formula, we have

$\sec^2 \left ( \frac {\pi}{7} \right ) + \sec^2 \left ( \frac {2\pi}{7} \right ) + \sec^2 \left ( \frac {3\pi}{7} \right ) = -(-24) = 24$

Using the trigonometric identity: $\tan^2 (A) + 1 = \sec^2 (A)$ , we obtain

$\tan^2 \left ( \frac {\pi}{7} \right ) + \tan^2 \left ( \frac {2\pi}{7} \right ) + \tan^2 \left ( \frac {3\pi}{7} \right ) = 21$

Similarly, by Vieta's Formula

$\sec^4 \left ( \frac {\pi}{7} \right ) + \sec^4 \left ( \frac {2\pi}{7} \right ) + \sec^4 \left ( \frac {3\pi}{7} \right ) = 24^2 - 2 \cdot 80 = 416$

$\Rightarrow \left ( \tan^2 \left ( \frac {\pi}{7} \right ) + 1 \right )^2 + \left ( \tan^2 \left ( \frac {2\pi}{7} \right )+ 1 \right )^2 + \left ( \tan^2 \left ( \frac {3\pi}{7} \right ) + 1 \right )^2 = 416$

$\Rightarrow \tan^4 \left ( \frac {\pi}{7} \right ) + \tan^4 \left ( \frac {2\pi}{7} \right ) + \tan^4 \left ( \frac {3\pi}{7} \right ) = 371$

Now, we apply the trigonometric identities: $\tan(A) = -\tan( \pi - A)$ and $\tan(A) = -\tan(\pi + A)$ , then we have the two equations

$\tan^2 \left ( \frac {8\pi}{7} \right ) + \tan^2 \left ( \frac {2\pi}{7} \right ) + \tan^2 \left ( \frac {4\pi}{7} \right ) = 21$

$\tan^4 \left ( \frac {8\pi}{7} \right ) + \tan^4 \left ( \frac {2\pi}{7} \right ) + \tan^4 \left ( \frac {4\pi}{7} \right ) = 371$

Note that the sum of the three numbers $\frac {8\pi}{7}, \frac {2\pi}{7}, \frac {4\pi}{7}$ is equals to $2\pi$ . By expanding the compound angle formula $\tan(A+B+C) = 0$ , we get

$\tan(A) + \tan(B) + \tan(C) = \tan(A) \tan(B) \tan(C)$

Now consider a cubic equation $g(x)$ with roots $\tan \left ( \frac {8\pi}{7} \right ) , \tan \left ( \frac {2\pi}{7} \right ) , \tan \left ( \frac {4\pi}{7} \right )$

Using Newton's Sum method, denote $S_n$ and $P_n$ as the $n^{\text{th} }$ symmetric sum of roots of $g(x)$ and sum of $n^\text{th}$ powers of roots of $g(x)$ respectively

Then $S_1 = P_1 , 21 = P_2 = S_1 ^2 - 2S_2$ and $S_1 = S_3$ from above

Consider the expansion

$a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(ab+ac+bc)^2 + 4abc(a+b+c)$

$\Rightarrow P_4 = P_2 ^2 - 2 S_2 ^2 + 4 S_1 ^2$

$\Rightarrow 371 = 21^2 - 2 S_2 ^2 + 4 S_1 ^2$

$\Rightarrow 35 = S_2 ^2 - 2 S_1 ^2$

Solving the two equations $21 = S_1 ^2 - 2S_2, 35 = S_2 ^2 - 2 S_1 ^2$ simultaneously, denoting $P,Q$ as $S_1, S_2$ respectively

We have $21 = P^2 - 2Q, 35 = Q^2 - 2P^2$ , add twice the first equation to the second, we obtain $77 = Q^2 - 4Q \Rightarrow 81 = (Q-2)^2 \Rightarrow Q = 11,-7$

Hence $\alpha = Q = S_2$ is the sum of negative numbers because the tangent of both $\frac {4\pi}{7}$ and $\frac {8\pi}{7}$ yields a negative value.

This means that $\alpha = -7$ only. So the answer is $\boxed{7}$