Prove that $\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ)+1}=44.5$

Question

So I've been stuck on this problem for a while now, and to be honest I have no idea how to start. I stumbled upon it while practicing for an Olympiad, and it is a past Olympiad problem (for teens) itself.

The problem is as follows:

Given is the function $f(x) = \frac{1}{x^3+1}$

Prove that:

$$\sum_{k=1}^{89} f(\tan(k^\circ)) = 44.5$$

In other words, prove that:

$$\frac{1}{\tan^3(1^\circ)+1} + \frac{1}{\tan^3(2^\circ)+1} + \cdots + \frac{1}{\tan^3(89^\circ)+1} = 44.5 $$

The question was multiple choice at first, but I managed to figure out 44.5 with my calculator. I still wondered how one would come up with a proof. Any guidelines or straight up solutions are welcome!

Answer 1

We can write $\tan(90^\circ - k^\circ) = \frac{1}{\tan(k^\circ)}$. \begin{align*} \sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ) + 1} &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=46}^{89} \frac{1}{\tan^3(k^\circ) + 1} + \frac{1}{\tan^3(45^\circ) + 1} \\ &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{1}{\tan^3(90 - k^\circ) + 1} + \frac{1}{2} \\ &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{1}{\frac{1}{\tan^3(k^\circ)} + 1} + \frac{1}{2} \\ &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{\tan^3(k^\circ)}{\tan^3(k^\circ) + 1} + \frac{1}{2} \\ &= \sum_{k=1}^{44} \frac{\tan^3(k^\circ) + 1}{\tan^3(k^\circ) + 1} + \frac{1}{2} \\ &= 44.5 \end{align*}

Answer 2

Use $f(x)+f(1/x)=1$, $\tan(90^\circ-x)=1/\tan x$, and $\sum_{k=1}^{89}a_k=\frac12\sum_{k=1}^{89}(a_k+a_{90-k})$ with $a_k:=f(\tan k^\circ)$.