Tangent Tango 2

Let's first convert all the angles to radians, $12^\circ \to \frac{\pi}{15}, \quad 24^\circ \to \frac{2\pi}{15}, \quad 48^\circ \to \frac{4\pi}{15}, \quad 96^\circ \to \frac{8\pi}{15}$

The equation $\cos(5X) = - \frac12$ has roots $\frac{2\pi}{15}, \frac{4\pi}{15}, \frac{8\pi}{15}, \frac{10\pi}{15}, \frac{14\pi}{15} .$

Using quintuple angle formula, $\cos(5X) = 16\cos^5(X) - 20\cos^3(X) + 5\cos(X)$ . Let $y = \cos(X)$ , the equation $\cos(5X) = - \frac12$ becomes $16y^5 - 20y^3 + 5y = -\frac12 \quad \Leftrightarrow \quad (2y + 1)(16y^4 - 8y^3 - 16y^2 + 8y + 1 ) = 0.$ The quintic factor must have roots $\cos( \tfrac{2\pi}{15}) , \cos( \tfrac{4\pi}{15}) , \cos( \tfrac{8\pi}{15}) , \cos( \tfrac{14\pi}{15})$ .

We have $g_1(Y) := 16Y^4 - 8Y^3 - 16Y^2 + 8Y + 1$ whose roots are $\cos( \tfrac{2\pi}{15}) , \cos( \tfrac{4\pi}{15}) , \cos( \tfrac{8\pi}{15}) , \cos( \tfrac{14\pi}{15})$ .

Likewise, $g_1 \left( \sqrt Y \right) = 0$ has roots $\cos^2( \tfrac{2\pi}{15}) , \cos^2( \tfrac{4\pi}{15}) , \cos^2( \tfrac{8\pi}{15}) , \cos^2( \tfrac{14\pi}{15}) = \cos^2( \tfrac{\pi}{15})$ .

$\begin{array} { r c l } g_1 \left( \sqrt Y \right) &=& 0 \\ 16Y^2 - 8Y \sqrt Y - 16Y + 8\sqrt Y + 1 &=& 0 \\ 16Y^2 - 16Y + 1 &=& 8\sqrt Y (Y - 1) \\ (16Y^2 - 16Y + 1)^2 &=& \left(8\sqrt Y \right)^2 (Y - 1)^2 \\ 256Y^4 - 576Y^3 + 416Y^2 - 96Y + 1 &=& 0 \end{array}$

And so $g_2(Y) := 256Y^4 - 576Y^3 + 416Y^2 - 96Y + 1$ has roots $\cos^2( \tfrac{2\pi}{15}) , \cos^2( \tfrac{4\pi}{15}) , \cos^2( \tfrac{8\pi}{15}) , \cos^2( \tfrac{\pi}{15})$ .

Furthermore, $g_3 (Y) := Y^4 \cdot g_2 (\tfrac 1Y) = Y^4 - 96Y^3 + 416Y^2 - 576Y + 256$ has roots $\sec^2( \tfrac{2\pi}{15}) , \sec^2( \tfrac{4\pi}{15}) , \sec^2( \tfrac{8\pi}{15}) , \sec^2( \tfrac{\pi}{15})$ .

Using the identity, we $\sec^2 (Z) - 1 = \tan^2(Z)$ , we have

$\begin{array} { r c l } f (Y) = g_3(Y+ 1) &=& (Y+1)^4 - 96(Y+1)^3 + 416(Y+1)^2 - 576(Y+1) + 256 \\ &=& Y^4 - 92Y^3 + 134Y^2 - 28Y + 1 \end{array}$ has roots $\tan^2( \tfrac{2\pi}{15}) , \tan^2( \tfrac{4\pi}{15}) , \tan^2( \tfrac{8\pi}{15}) , \tan^2( \tfrac{\pi}{15}).$ Hence, $(A,B,C,D) = (-92,134,28,1) \Rightarrow A+2B+3C+4D = \boxed{96} .$

Using the identity $\tan 3x = \tan (60° - x) \tan x \tan (60° + x)$ , if $x = 6°$ , then $\tan 18° = \tan 54° \tan 6° \tan 66°$ , and since $\tan 18° = \frac{1}{\tan 72°}$ and $\tan 54° = \frac{1}{\tan 36°}$ , this can be rearranged to $\tan 36° = \tan 6° \tan 66° \tan 72°$ .

Using the identity $\tan 5x = \tan (72° - x) \tan (36° - x) \tan x \tan (36° + x) \tan (72° + x)$ , if $x = 6°$ , then $\tan 30° = \tan 66° \tan 30° \tan 6° \tan 42° \tan 78°$ , and since $\tan 42° = \frac{1}{\tan 48°}$ and $\tan 78° = \frac{1}{\tan 12°}$ , this can be rearranged to $\tan 6° \tan 66° = \tan 12° \tan 48°$ .

Combining these two equations gives $\tan 36° = \tan 12° \tan 48° \tan 72°$ , so that $\tan 3 \theta = \tan \theta \tan 4\theta \tan 6\theta$ is true for $\theta = 12°$ .

Since $\tan 36° = \tan 6° \tan 66° \tan 72°$ is equivalent to $-\tan 36° = -\tan 6° \tan 66° \tan 72°$ , and since $-\tan 6° = \frac{1}{\tan 96°}$ and $-\tan 36° = \tan 144°$ , this can be rearranged to $\tan 72° = \tan 24° \tan 96° \tan 144°$ , so that $\tan 3 \theta = \tan \theta \tan 4\theta \tan 6\theta$ is true for $\theta = 24°$ .

Since $\tan 36° = \tan 12° \tan 48° \tan 72°$ is equivalent to $-\tan 36° = \tan 12° \tan 48° (-\tan 72°)$ , and since $-\tan 36° = \tan 144°$ , $\tan 12° = \tan 192°$ , and $-\tan 72° = \tan 288°$ , this can be rearranged to $\tan 144° = \tan 48° \tan 192° \tan 288°$ , so that $\tan 3 \theta = \tan \theta \tan 4\theta \tan 6\theta$ is true for $\theta = 48°$ .

Since $\tan 36° = \tan 6° \tan 66° \tan 72°$ is equivalent to $\tan 36° = (-\tan 6°) \tan 66° (-\tan 72°)$ , and since $\tan 36° = \tan 576°$ , $-\tan 6° = \frac{1}{\tan 96°}$ , $\tan 66° = \frac{1}{\tan 384°}$ , and $-\tan 72° = \tan 288°$ , this can be rearranged to $\tan 288° = \tan 96° \tan 384° \tan 576°$ , so that $\tan 3 \theta = \tan \theta \tan 4\theta \tan 6\theta$ is true for $\theta = 96°$ .

Using the tangent identity equations with $t = \tan \theta$ :

$\tan 2\theta = \cfrac{2\tan \theta}{1 - \tan^2 \theta} = \cfrac{2t}{1 - t^2}$

$\tan 3\theta = \tan (2\theta + \theta) = \cfrac{\tan 2\theta + \tan \theta}{1 - \tan 2\theta \tan \theta} = \cfrac{\frac{2t}{1 - t^2} + t}{1 - (\frac{2t}{1 - t^2}) t} = \cfrac{t(t^2 - 3)}{3t^2 - 1}$

$\tan 4\theta = \tan 2(2 \theta) = \cfrac{2 \tan 2\theta}{1 - \tan^2 2\theta} = \cfrac{2 (\frac{2t}{1 - t^2})}{1 - (\frac{2t}{1 - t^2})^2} = \cfrac{4t(1 - t^2)}{t^4 - 6t^2 + 1}$

$\tan 6\theta = \tan 2(3\theta) = \cfrac{2 \tan 3\theta}{1 - \tan^2 3\theta} = \cfrac{2 (\frac{t(t^2 - 3)}{3t^2 - 1})}{1 - (\frac{t(t^2 - 3)}{3t^2 - 1})^2} = \cfrac{2t(t^2 - 3)(3t^2 - 1)}{(1 - t)(1 + t)(t^2 - 4t + 1)(t^2 + 4t + 1)}$

Substituting these into $\tan 3\theta = \tan \theta \tan 4\theta \tan 6\theta$ gives $\cfrac{t(t^2 - 3)}{3t^2 - 1} = t \cdot \cfrac{4t(1 - t^2)}{t^4 - 6t^2 + 1} \cdot \cfrac{2t(t^2 - 3)(3t^2 - 1)}{(1 - t)(1 + t)(t^2 - 4t + 1)(t^2 + 4t + 1)}$ , which rearranges and simplifies to $t^8 - 92t^6 + 134t^4 - 28t^2 + 1 = 0$ . Substituting $x = t^2$ gives $x^4 - 92x^3 + 134x^2 - 28x + 1 = 0$ , which has the desired solutions of $x = \tan^2 12°$ , $x = \tan^2 24°$ , $x = \tan^2 48°$ , and $x = \tan^2 96°$ .

Therefore, $A = -92$ , $B = 134$ , $C = -28$ , $D = 1$ , and $A + 2B + 3C + 4D = \boxed{96}$ .