Functional Complex Trignometry

For angles $\theta_1, \theta_2, \theta_3, \ldots, \theta_{52}$ , tangent of their sum is

$\tan(\theta_1 + \theta_2 +\theta_3 + \ldots +\theta_{52}) = \frac{S_1 - S_3 + S_5 - S_7 + \ldots -S_{49} + S_{51}}{1 - S_2 + S_4 - S_6 + S_8 + \ldots -S_{50} + S_{52}}$

where $S_n$ is the sum of tangents of angles taking $n$ at a time. Let's reduce our function to the above equation.

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Putting $x = i$ in $f(x)$ .

$f(i) = (i)^{52} + a_1(i)^{51} + a_2(i)^{50} + \ldots + a_{51}i + a_{52}$ $\boxed{f(i) = 1 - a_1i - a_2 + a_3i + \ldots + a_{51}i +a_{52}}$

Putting $x = -i$ in $f(x)$ .

$f(-i) = (-i)^{52} + a_1(-i)^{51} + a_2(-i)^{50} + \ldots + a_{51}(-i) + a_{52}$ $\boxed{f(-i) = 1 + a_1i - a_2 - a_3i + \ldots - a_{51}i +a_{52}}$

Subtracting $f(i)$ from $f(-i)$

$f(-i) - f(i) = 2a_1i - 2a_3i + 2a_5i - 2a_7i + \ldots - 2a_{51}i$ $\frac{f(-i) - f(i)}{2i} = a_1 - a_3 + a_5 - a_7 + \ldots - a_{51} \ldots \ldots (1)$

Adding $f(i)$ and $f(-i)$

$f(-i) + f(i) = 2 - 2a_2 + 2a_4 - 2a_6 + \ldots - 2a_{50} + 2a_{52}$ $\frac{f(-i) + f(i)}{2} = 1 - a_2 + a_4 - a_6 + \ldots - a_{50} + a_{52} \ldots \ldots (2)$

Dividing $(2)$ from $(1)$

$\frac{1}{i} \frac{f(-i) - f(i)}{f(-i) + f(i)} = \frac{a_1 - a_3 + a_5 - a_7 + \ldots - a_{51}}{1 - a_2 + a_4 - a_6 + \ldots - a_{50} + a_{52}}$

Now, for a polynomial, $a_{2n + 1} = -S_{2n+1}$ and $a_{2n} = S_{2n}$ for each natural $n$ . Putting it

$\frac{1}{i} \frac{f(-i) - f(i)}{f(-i) + f(i)} = -\frac{S_1 - S_3 + S_5 - S_7 + \ldots - S_{51}}{1 - S_2 + S_4 - S_6 + \ldots - S_{50} + S_{52}}$

$i\frac{f(-i) - f(i)}{f(-i) + f(i)} = \frac{S_1 - S_3 + S_5 - S_7 + \ldots - S_{51}}{1 - S_2 + S_4 - S_6 + \ldots - S_{50} + S_{52}}$

As roots of $f(x)$ are $\tan\theta_1, \tan\theta_2, \tan\theta_3, \ldots, \tan\theta_{52}$ , So

$\boxed{\tan(\theta_1 + \theta_2 +\theta_3 + \ldots +\theta_{52}) = i\frac{f(-i) - f(i)}{f(-i) + f(i)}}$

Putting values of $f(-i) = i(2 + \sqrt3) + 1$ and $f(i) = i(2 + \sqrt3) - 1$

$\tan(\theta_1 + \theta_2 +\theta_3 + \ldots +\theta_{52}) = i\frac{(i(2 + \sqrt3) + 1) - (i(2 + \sqrt3) - 1)}{(i(2 + \sqrt3) + 1) + (i(2 + \sqrt3) - 1)}$

$\tan(\theta_1 + \theta_2 +\theta_3 + \ldots +\theta_{52}) = 2 - \sqrt3$

$\color{#3D99F6}{\boxed{\theta_1 + \theta_2 +\theta_3 + \ldots +\theta_{52} = 15\degree}}$