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I have a set $S$ which contains angles (in degrees) from $1$ to $179$, excluding $90$.

I want to find a summation of $\tan A\tan B$, where $A,B\in S$, and $A\ne B$.

So basically, this summation contains:$\tan 1\tan 2 + \tan 1\tan 3... \tan1 \tan 179 + \tan 2\tan 3... \tan 2\tan 179.... \tan 178\tan 179 \text{, i.e., summation of all possible ways of selecting $2$ elements from that set.}$

The same thing is to be done for $3$ at a time as well. All the angles are in degrees.

How could I go about this?

A friend and I were randomly coding something and tried compiling this which gave us an answer of $-5310.\overline{3}$. We were intrigued by this because the answer ends with an endless repeating decimal, and hence is a rational number, whereas all values of $\tan x, x\in S$ are irrational excluding $45$ and $135$.

Does anyone have any clue how we could approach actually solving this?

Integreek
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Kavin Upreti
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  • Are you specifically excluding $\tan^2 1^\circ, \tan^2 2^\circ, \ldots$ from this sum? ¶ Also, $\tan 45^\circ = 1, \tan 135^\circ = -1$, which are clearly rational. – Brian Tung Sep 24 '24 at 05:26
  • Yes, I am specifically excluding cases where A = B. A,B have to be different.

    and my apologies for tan 45 and tan135, it was an error on my part

     – Kavin Upreti Sep 24 '24 at 05:48
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    $\tan1^\circ\tan2^\circ+\tan1^\circ\tan178^\circ=0$, so you are only summing $\tan1^\circ\tan179^\circ+\tan2^\circ\tan178^\circ+\ldots+\tan89^\circ\tan91^\circ=-\Sigma_{n=1}^{89}\tan^2n$ – Empy2 Sep 24 '24 at 05:56
  • Have you factored the numerator of your rational number ? – Empy2 Sep 24 '24 at 06:09
  • Oh, yeah it makes sense. Guess it just "happened" lol. Anyway thank you

    I'll try factoring and seeing if I get any clues

     – Kavin Upreti Sep 24 '24 at 07:05

1 Answers1

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In that case, we'll have to use $\tan(π-\theta)=-\tan(\theta)$.

In that case, $$\tan 1°(\tan 2°+\tan 3°+...\tan 178°+\tan 179°)=\tan 1°\tan 179°$$

In case of $\tan 2°$, we will get $\tan 2°(\tan 178°+\tan 179°)$ and so on.

Our final sum $S$ will look like $$S=\sum_{i=1°}^{n=89}\bigg(\tan(180°-i)\sum_{i=j}^{n=179°-i} \tan j\bigg)$$ The internal summation equals to $\tan i$ only, because of the above explained property.

By further simplification, we are actually calculating

$$S=\sum_{i=1°}^{n=89°}-\tan^2(i)$$ Which is what you calculated. Nothing cancels out here, so you need a calculator to find out the numerical value.

Edit This is a great answer by Lab Bhattacharjee which presents the reason for the answer to be rational.

Gwen
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