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Is it possible to find a set of numbers $(a_i)_{i\leq n} \in \mathbb {(N^{\star})}^n$ and another natural number $b$ such that $$\sum_{i=0}^n \tan a_i = \tan b $$?

In other words, given integral, strictly positive, angles in radians, can their tangents sum up to the tangent of another integer number?

Also, is it possible to have two distinct sets of integers $((a_i)_{i\leq n} \in \mathbb {(N^{\star})}^n$ and $(b_i)_{i\leq m} \in \mathbb {(N^{\star})}^m$ such that $$\sum_{i=0}^n \tan a_i = \sum_{i=0}^m \tan b_i$$ ?

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This is only vaguely related, but with $\text{arctan}$ there are many possibilities (en.wikipedia.org/wiki/Machin-like_formula). –  Joel Cohen Nov 30 '11 at 17:48

2 Answers 2

up vote 6 down vote accepted

By using the addition formula $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$ repeatedly, we can see that if $k$ is a positive integer, then $\tan kt$ is a rational function of $\tan t$, with integer coefficients. Thus if an equation of the type $\sum\limits_{i=1}^n \tan a_i=\tan b$ holds, then there are rational functions $A(x)$, $B(x)$ with integer coefficients such that $A(\tan 1)=B(\tan 1)$. But by the Lindemann-Weierstrass Theorem, $\tan 1$ is transcendental. It follows that $A(x)=B(x)$ identically.

The fact that $A(x)$ is identically equal to $B(x)$ implies that $\sum\limits_{i=1}^n \tan(a_it)=\tan(bt)$ for all $t$. We will show that is not possible except in the trivial case $n=1$, $a_1=b$.

For if $\sum\limits_{i=1}^n \tan(a_it)=\tan(bt)$, then $$0=\lim_{t\to 0^+}\frac{\sum_{i=1}^n \tan(a_i t)-\tan(bt)}{t}=\sum_{i=1}^n a_i -b.$$ But if $\sum\limits_{i=1}^n a_i=b$, and $n>1$, we cannot have $\sum_{i=1}^n \tan(a_it)=\tan(bt)$ when $t>0$ is such that the $a_it$, $bt$ are less than $\pi/2$ (in this interval, the tangent of a sum is greater than the sum of the tangents).

Added: For the more general $\sum\limits_{i=0}^n \tan a_i = \sum\limits_{j=0}^m \tan b_j$, without loss of generality no $a_i$ is a $b_j$. We argue as before that $\sum\limits_{i=0}^n \tan(a_i t) = \sum\limits_{j=0}^m \tan (b_j t)$ for all positive $t$. We can assume that $b_1 \ge b_j$ for all $j$, and that $a_i<b_1$ for all $i$. Let $t$ be such that $tb_1$ is very close to but below $\frac{\pi}{2}$. Then $\sum\limits_{j=1}^m \tan(b_j t)$ is very large positive, but $\sum\limits_{i=1}^n \tan(a_i t)$ is not, which gives the desired contradiction.

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Nice proof! Couldn't you simplify the last part by saying that the functions $t \mapsto \tan(n t)$ for $n \in \mathbb{N}^*$ are linearly independent because the have a singularities at different points ? –  Joel Cohen Nov 30 '11 at 18:25
    
@Joel Cohen: I am sure there are improvements possible. When I saw that one could insert a $t$, the first thing that jumped to mind was make the largest of the $a_i t$, bt$ equal to $\pi/2$. But in thinking about possible loose ends, the convexity argument seemed simpler. –  André Nicolas Nov 30 '11 at 18:33

Somewhat more generally, suppose $\sum_{j=1}^n c_j \tan(a_j) = 0$ where $c_j$ are nonzero algebraic numbers and $a_j$ are distinct positive rationals. We can write $a_j = k_j/N$ where $N$ and $k_j$ are positive integers. Since $\tan(t) = \frac{e^{it} - e^{-it}}{i (e^{it} + e^{-it})} = \frac{e^{2it} - 1}{i (e^{2it} + 1)}$, our equation can be rewritten as $$ \sum_{j=1}^n c_j \frac{z^{k_j} -1}{z^{k_j}+1} = 0 $$ where $z = e^{2 i/N}$. Since $e^{2i/N}$ is transcendental, the rational function $\sum_{j=1}^n c_j \frac{z^{k_j} -1}{z^{k_j}+1} $ must be identically 0. However, if $k_m$ is the maximum of the $k_j$, the term $c_m (z^{k_m} - 1)/(z^{k_m} + 1)$ has a pole at each primitive $k_m$'th root of $-1$, while the other terms do not, so this is impossible.

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