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Minimize $\sum_{1≤ i≤ n} \tan(A_i)$ where $A_i>0$ and $\sum_{1≤ i≤ n} A_i=C$ where $\frac{\pi}{2}>C>0$ is some constant.

I tried to use $\tan(\sum_{1≤ i≤ n} A_i)=\frac{p_1-p_3+p_5-...}{1-p_2+p_4-...}$ where $p_r=\sum \tan A_1\tan A_2...\tan A_r$ where $r=1,2,...,n$

So, $\frac{p_1-p_3+p_5-...}{1-p_2+p_4-...}=d$ where $d=\tan(\sum_{1≤ r≤ n} A_i)$ some constant as $\sum_{1≤ i≤ n} A_i$ is constant.


I find $\frac{\partial{p_j}}{\partial{A_i}}=p_{j-1}(sec^2A_i)$ taking $p_0=1$ and $j>0$.

Now taking partial derivative w.r.t. $A_i$,


$=>d(0-p_1+p_3-p_5)=(1-p_2+p_4-...)$ as $sec^2A_i≠0$

$=>\tan(\sum_{1≤ r≤ n, r≠i} A_r)=-\frac{1}{d}$

$=>\tan(C-A_i)=-\frac{1}{d}$ for all $1≤ i≤ n$,

So,$\tan(C- A_i)=\tan(C-A_j)$ where $1≤ j≤ n$ and $i≠j$

One of the solution is $A_i=A_j$

I could not proceed further to show that these equalities($=>A_i=\frac{C}{n}$) lead to the minimum value of $\sum_{1≤ i≤ n} \tan(A_i)$?

Please rectify me if there is any mistake.

Another confusion is

$d>0$ as $\frac{\pi}{2}>C>0$

But $\tan(C-A_i)$ is also $>0$ unless $A_i>C$ which is impossible $=>\tan(C-A_i)=-\frac{1}{d}$ is also untenable.

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Is there a typo somewhere in the beginning? Did you mean $\tan(\sum A_i)$ instead of $\sum\tan(A_i)$? – Alex R. Aug 28 '12 at 18:40
Which line you are talking about? If it's the 2nd line, then it's not a typo. – lab bhattacharjee Aug 28 '12 at 18:47
"Maximize..." followed by "I tried to use..." Otherwise it's not entirely clear to me why you are considering the tangent of the sum rather than the sum of tangents. – Alex R. Aug 28 '12 at 18:48
As the sum of angles(S,say) is constant, so will be $\tan S$, then I used the expansion formula of $\tan(A+B+C+...)$ – lab bhattacharjee Aug 28 '12 at 18:55
up vote 2 down vote accepted

Just check if this works:

I'm assuming $A_i > 0 \ \forall \ i$. Let $f(x) = \tan x$. Now, $f'(x) = \sec^2x, f''(x) = 2 \sec^2 x \tan x > 0$ for $\pi/2 >x>0$. Hence, $\tan x$ is convex.

Now, by Weierstrass Theorem, a convex function is maximized at its endpoints. So, the function is maximized when all but one variable is equal to $C$.

Edit: Now that the OP has been edited to find instead the minimum of the function, it is even simpler by convexity. Just employ the Jensen's Inequality to get that $$\sum_{i=1}^n \tan A_i \ge n \cdot \tan \left( \frac{A_1 + A_2 + \cdots+ A_n}{n} \right) = n \tan \frac C n$$

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The variables add up to $C$, so I think you want one variable to be $C-\epsilon$ and the others to be $\epsilon/(n-1)$ to approach the unattainable maximum arbitrarily closely. – Gerry Myerson Aug 29 '12 at 2:45
@Rijul Saini, thanks for your answer. I think, the problem should be to minimize instead of to maximize. If $A+B=\frac{\pi}{2},$ then if $A=B=\frac{\pi}{4}=>\tan A+\tan B=1+1=2$, if $A=\frac{\pi}{3},B=\frac{\pi}{6}=>\tan A+\tan B=\sqrt3+\frac{1}{\sqrt3}=\frac{4}{\sqrt3}>2$ – lab bhattacharjee Aug 29 '12 at 4:14
If you want to minimize, then convexity would seem to imply that you want to take $A_i=C/n$ for all $i$. But you really should figure out what you want before you post the question. – Gerry Myerson Aug 29 '12 at 6:05
Sorry, there was probably some typo in the source itself, which I could validate by the test mentioned above.Could anybody please share some link/idea on the minima/maxima of a convex function. – lab bhattacharjee Aug 29 '12 at 6:53
@labbhattacharjee I have now edited to accommodate the minimum of the function too. – Rijul Saini Aug 29 '12 at 10:54

Since the sum of the angles is constant, the variation of the sum of the angles is $0$. $$ \delta\,\sum_iA_i=\sum_i1\,\delta A_i=0\tag{1} $$ To maximize/minimize the sum of the tangents we must have for any $\delta A_i$ that satisfy $(1)$ $$ \delta\,\sum_i\tan(A_i)=\sum_i\sec^2(A_i)\,\delta A_i=0\tag{2} $$ Useful here is a standard orthogonality result.

Lemma (Orthogonality): If $\langle h,g\rangle=0$ for all $h$ such that $\langle h,f_k\rangle=0$ for all $k$, then $g$ is a linear combination of the $f_k$.

Proof: If all the $f_k$ are $0$, then $\langle f_k,g\rangle=0$ for all $k$; thus, by hypothesis, $\langle g,g\rangle=0$. Therefore, $g=0$ and $g$ is a linear combination of the $f_k$.

Assume that not all the $f_k$ are $0$. If the $f_k$ are not linearly independent, pick a maximal linearly independent subset of the $f_k$. This insures that the matrix $\langle f_j,f_k\rangle$ is invertible. Thus, for some vector $\{a_j\}$ we have $$ \langle g,f_k\rangle=\sum_ja_j\langle f_j,f_k\rangle\tag{O1} $$ Let $$ h=g-\sum_ja_j f_j\tag{O2} $$ By $\mathrm{(O1)}$, $\langle h,f_k\rangle=0$ for all $k$. By hypothesis, $\langle h,g\rangle=0$. Thus, by $\mathrm{(O2)}$ $\langle h,h\rangle=0$.

Therefore, $h = 0$; that is, $$ g=\sum_ja_jf_j\tag{O3} $$ QED

Applying this lemma to $(1)$ and $(2)$, we get that for the $A_i$ that are not on the boundary ($A_i>0$), we must have that $\sec^2(A_i)$ must be a linear combination of $1$; that is, the $A_i$ must be equal.

All that is left now is to figure out how many of the $A_i>0$ (the rest being $0$) gives the minimum.

Suppose that $m$ of the $A_i$ are not $0$, then we get that the sum of the tangents to be $$ m\tan(C/m)\tag{3} $$ Since $\tan(x)$ is a convex function on $[0,\pi/2]$, $(3)$ is a decreasing function of $m$. Thus, to minimize the sum of the tangents, set $m=n$. Therefore, we get that the minimum of the sum of the tangents is $$ n\tan(C/n)\tag{4} $$

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(I assume that $n\geq2$ is given.)

The set $S:=\bigl\{A=(A_1,\ldots,A_n)\bigm| A_i\geq0 \ (1\leq i\leq n), \ \sum_{i=1}^n A_i=C\bigr\}$ is compact; therefore the function $f(A):=\sum_{i=1}^n \tan A_i$ assumes on $S$ a global minimum.

By looking at the graph of $\tan$ in the interval $I:=\bigl[0,{\pi\over2}\bigr[\ $ we see that for different $A_1$, $A_2\in I$ one has $$\tan{A_1+A_2\over2}<{1\over2}(\tan A_1+\tan A_2)\ ;$$ therefore $f\bigl({A_1+A_2\over2},{A_1+A_2\over2}, A_3,\ldots, A_n)<f(A_1,A_2,A_3,\ldots,A_n)$.

It follows that the point $A_*\in S$ where $f$ attains its minimum has all $A_i$ equal. Therefore $A_*={C\over n}(1,1,\ldots, 1)$, and the minimum value of $f$ is $f(A_*)=n\tan{C\over n}$.

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