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As the title suggests, I'm trying to find the sum $$\tan^21^\circ+\tan^2 2^\circ+\cdots+\tan^2 89^\circ$$ I'm looking for a solution that doesn't involve complex numbers, or any other advanced branch in maths. The solution can involve techniques such as induction, telecoping, etc, but preferably only ideas from precalculus, e.g. trig identities, polynomials, etc.

EDIT: I know that the sum is a rational number.

Michael Hardy
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Ninja Boy
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    Are you sure the sum is rational? How do you know? – TonyK Sep 29 '14 at 19:44
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    But how would your calculator have given an irrational sum? Wolfram Alpha gives $265.0173714620087662428704848470169312673078166104585039010061...$, which looks fairly irrational to me, in this context. – TonyK Sep 29 '14 at 19:48
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    I converted to radians, for Wolfram Alpha's benefit. The sum is unchanged. – TonyK Sep 29 '14 at 19:49
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    Well, one of us is going to look silly here. Let us know when you find your calculator. – TonyK Sep 29 '14 at 19:52
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    That's better! Now we get $5310.333333333333333333333333333333333333333333333333333333333...$. Now there's a rational number if ever I saw one. – TonyK Sep 29 '14 at 19:57
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    Funnily enough, Wolfram Alpha doesn't get it. If you go here and click on "Exact form", you get a horrible mess instead of $15931/3$. – TonyK Sep 29 '14 at 19:59
  • i get $\frac{15931}3$ with a calculator. – Ali Caglayan Sep 29 '14 at 20:00
  • Most likely the $15931/3$ is an error. It may be a number very close to it but pocket calculators don't seem to differentiate. WA would pick up something like this and it didn't. Usually a good indicator that this is a coincidence. – Ali Caglayan Sep 29 '14 at 20:02
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    One thing to try: instead of doing the "special" case where you're looking at degrees, can you find a formula for $\sum_{k=1}^{n-1} \tan^2 {k \pi \over 2n}$, where the angles are in radians? Your question is the $n = 90$ case of this. If you try small integer values of $n$ it's not terribly hard to conjecture a formula. – Michael Lugo Sep 29 '14 at 20:03
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    Also, possibly relevant SE links: http://math.stackexchange.com/questions/217240/reference-for-a-tangent-squared-sum-identity, http://math.stackexchange.com/questions/2339/prove-that-sum-k-1n-1-tan2-frack-pi2n-fracn-12n-13?rq=1 , although I'm not sure if any of the proofs here satisfy your requirements - this problem seems very naturally to live in the complex numbers. – Michael Lugo Sep 29 '14 at 20:04
  • @Alitzer: Try this then: instead of clicking on "Exact form" in my link (here, click on "More digits" until you get bored. Do you still think it's irrational? I'm not claiming a proof, but it does suggest that your faith in WA might be misplaced. – TonyK Sep 29 '14 at 20:06
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    $15931 = 89 \times 179$, by the way. – TonyK Sep 29 '14 at 20:11
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    The 89 pairs of numbers $\pm \tan 1^\circ, \pm \tan 2^\circ, \ldots, \pm \tan 89^\circ$ are the roots of the polynomial $$\frac{(1+it)^{180} - (1-it)^{180}}{2i^{179}t} = \binom{180}{1} t^{178} - \binom{180}{3} t^{176} + \cdots$$ Even though the LHS of this formula include complex number, the RHS didn't. If you can justify the RHS, then you are done. – achille hui Sep 29 '14 at 20:54
  • @achillehui, Did you mean the following ? – lab bhattacharjee Sep 30 '14 at 03:55
  • @labbhattacharjee yes (even though I derive the formula by other mean). – achille hui Sep 30 '14 at 04:47
  • See this for an easier method https://math.stackexchange.com/a/3739719/545151 – user600016 Jun 30 '20 at 06:57

3 Answers3

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From Sum of tangent functions where arguments are in specific arithmetic series,

$$\tan180x=\frac{\binom{180}1t-\binom{180}3t^3+\cdots+\binom{180}{177}t^{177}-\binom{180}{179}t^{179}}{1-\binom{180}2t^2+\cdots-\binom{180}{178}t^{178}+\binom{180}{180}t^{180}}$$ where $t=\tan x$

If we set $\tan180x=0,180x=180^\circ r$ where $r$ is any integer

$\implies x=r^\circ$ where $0\le r<180$

So, the roots of $\binom{180}1t-\binom{180}3t^3+\cdots+\binom{180}{177}t^{177}-\binom{180}{179}t^{179}=0$ $\iff\binom{180}{179}t^{179}-\binom{180}{177}t^{177}+\cdots+\binom{180}3t^3-\binom{180}1t=0$ are $\tan r^\circ$ where $0\le r<180,r\ne90$ ($r=90$ corresponds to the denominator $=\infty$)

So, the roots of $\binom{180}{179}t^{178}-\binom{180}{177}t^{176}+\cdots+\binom{180}3t^2-\binom{180}1=0$ are $\tan r^\circ$ where $0<r<180,r\ne90$

So, the roots of $\binom{180}{179}u^{89}-\binom{180}{177}u^{88}+\cdots+\binom{180}3u-\binom{180}1=0$ are $\tan^2r^\circ$ where $0<r<90$

Using Vieta's formula, $\sum_{r=1}^{89}\tan^2r^\circ=\dfrac{\binom{180}{177}}{\binom{180}{179}}$

Community
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lab bhattacharjee
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The function $$ \frac{180/z}{z^{180}-1} $$ has residue $1$ at each root of $z^{180}-1$ $\left(\text{i.e. }e^{k\pi i/90}\text{ for }k=0\dots179\right)$ and residue $-180$ at $z=0$.

On $|z|=1$, $$ \tan(\theta/2)=-i\frac{z-1}{z+1} $$ Integrating $$ f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{180/z}{z^{180}-1} $$ around a large circle is $0$ since the integrand is approximately $|z|^{-181}$. Thus, the sum of residues is $$ 2\sum_{k=0}^{89}\tan^2\left(\frac{k\pi}{180}\right)+\operatorname*{Res}_{z=0}f(z)+\operatorname*{Res}_{z=-1}f(z)=0 $$ Since $\operatorname*{Res}\limits_{z=0}f(z)=180$ and $\operatorname*{Res} \limits_{z=-1}f(z)=-\frac{32402}{3}$, we get $$ \begin{align} \sum_{k=0}^{89}\tan^2\left(\frac{k\pi}{180}\right) &=\frac12\left(\frac{32402}{3}-180\right)\\ &=\frac{15931}{3} \end{align} $$

This same method also gives $$ \sum_{k=0}^{89}\tan^4\left(\frac{k\pi}{180}\right)=\frac{524560037}{45} $$ and $$ \sum_{k=0}^{89}\tan^6\left(\frac{k\pi}{180}\right)=\frac{4855740968543}{135} $$

robjohn
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  • Ah, I just saw the part of the question that mentions no complex methods. I will leave this, anyway, as a comparison. $\dfrac{n/z}{z^n-1}$ is a useful function to evaluate sums on $\mathbb{Z}/n\mathbb{Z}$. – robjohn Oct 02 '14 at 06:01
  • I've found this excellent answer via an answer to my question at http://math.stackexchange.com/questions/1877014/a-series-involving-csc4/1877743#1877743 -- handy trick, thanks! However, it seems this works only since $z\mapsto\tan(\arg z)$ is analytic (outside of its poles), which doesn't generalize to other sums, for instance involving $\sin$. Am I wrong? – heiner Aug 01 '16 at 17:02
  • @heiner: $\sin(\theta)=\frac1{2i}\left(z-\frac1z\right)$ on the unit circle, so you could use that. – robjohn Aug 01 '16 at 17:37
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The following function $$r_m(x)=\frac{(1+ix)^m-(1-ix)^m}{(1+ix)^m+(1-ix)^m}$$ has the easy property $$r_m(\tan\theta)=i\tan(m\theta).$$ If we set $r_{180}(x)=0$, cancelling $x$, we find lab's equation again: $$180x^{178}-{180\choose 3}x^{176}+...=0$$ with roots $\small\tan1^\circ,...,\tan89^\circ,\tan91^\circ,...,\tan179^\circ$. Setting $x^2=t$ and Vieta's formula concludes.

Bob Dobbs
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