Tangents Of Angle Progression

Relevant wiki: Proving Trigonometric Identities - Intermediate

First, we look up List of Trigonometric Identies to pull out this multiple angle Cosine expansion, for even $n$

$\cos\left( n\theta \right) =\displaystyle \sum _{ k=0 }^{ \frac{1}{2}n }{ \left( { \left( -1 \right) }^{ k }\left( \begin{matrix} n \\ 2k \end{matrix} \right) { \left( \cos\left( \theta \right) \right) }^{ n-2k }{ \left( \sin\left( \theta \right) \right) }^{ 2k } \right) }$

which, after dividing both sides by ${ \left( \cos\left( \theta \right) \right) }^{ n}$ , we have

$\dfrac { \cos\left( n\theta \right) }{ { \left( \cos\left( \theta \right) \right) }^{ n } } =\displaystyle\sum _{ k=0 }^{ \frac{1}{2} n }{ \left( { \left( -1 \right) }^{ k }\left( \begin{matrix} n \\ 2k \end{matrix} \right) x^{ k } \right) }$

where $x=\tan^2 \theta$

Letting $n=90$ , for every odd angle $\theta$ in degrees the left side of the equation

$\dfrac { \cos\left( n\theta \right) }{ { \left( \cos\left( \theta \right) \right) }^{ n } }$

is $0$ .

Hence $x=\tan^2 \theta$ for every odd angle $\theta$ in degrees are roots of the degree $45$ polynomial on the right

$\displaystyle\sum _{ k=0 }^{ \frac{1}{2} n }{ \left( { \left( -1 \right) }^{ k }\left( \begin{matrix} n \\ 2k \end{matrix} \right) x^{ k } \right) }$

and so using Vieta’s Formula , the sum of the $45$ such roots is the negative of the $2$ nd coefficient, or

$-{ \left( -1 \right) }^{ 1 }\left( \begin{matrix} 90 \\ 2\left( 1 \right) \end{matrix} \right) =4005$

nice solution.