Some years ago, I came across the following question:

Find the value of $\tan5°\tan55°\tan65°\tan75°$.

The value is 1 and I generalized the question as follows.

$\tan a°\tan b°\tan c°\tan d° =1 $ if $(a, b, c, d, )=(3p, 30+p, 90-p, 30-p)$

Here comes my question:

Are there $a, b, c, d \in \mathbb{N}$ such that $\tan a°\tan b°\tan c°\tan d° =2 $?

If yes,

What are the values of $n \in \mathbb{N}$ so that there are integer solutions for $\tan a°\tan b°\tan c°\tan d° =n $?

  • $\begingroup$ That second line is pretty awesome... do you have the proof and/or domain of $p$ available? Note this answer as well and recall the "sum of angles" formulas for $\sin$ and $\cos$ which will produce the values of $\tan$... $\endgroup$ – abiessu Nov 8 '13 at 22:33
  • $\begingroup$ @abiessu The line is equivalent to $\tan(30-x)\tan(30+x)\tan(3x)=\tan(x)$. Expand the LHS using the formulas for $\sin(\alpha+\beta)$, $\cos(\alpha+\beta)$ $\cos(3x)$ and $\sin(3x)$ (difference of squares make it doesn't go that ugly). $\endgroup$ – chubakueno Nov 9 '13 at 1:23

Let $\displaystyle \tan5x=\tan5A$

then $5x=r180^\circ+5A$ where $r$ is any inetger

$\displaystyle\implies x=r36^\circ+A$ where $r\equiv0,1,2,3,4\pmod 5\ \ \ \ (1)$

Again using this, $\displaystyle \tan5x=\frac{\tan^5x-10\tan^3x+5\tan x}{5\tan^4x-10\tan^2x+1}\ \ \ \ (2)$

$\displaystyle\implies \frac{\tan^5x-10\tan^3x+5\tan x}{5\tan^4x-10\tan^2x+1}=\tan5A $

$\displaystyle\implies \tan^5x-5\tan5A\tan^4x-10\tan^3x+10\tan5A\tan^2x+5\tan x-\tan5A=0$ which is a five degree equation in $\tan x$

Using Vieta's formulas, $\displaystyle \prod_{0\le r\le4}\tan\left(r36^\circ+A\right)=\tan5A\ \ \ \ (3)$

$\displaystyle\implies\prod_{0\le r\le4,r\ne m}\tan\left(r36^\circ+A\right)=\frac{\tan5A}{\tan\left(m36^\circ+A\right)}$ where $m$ should be so chosen that $\tan\left(m36^\circ+A\right)\ne0$

Now, $\displaystyle\prod_{0\le r\le4,r\ne m}\tan\left(r36^\circ+A\right)=n\ \ \ \ (4)$ if $\displaystyle\frac{\tan5A}{\tan\left(m36^\circ+A\right)}=n$

Let $\displaystyle m36^\circ+A=B\implies 5A=5(B-m36^\circ)=5B-m180^\circ\implies \tan5A=\tan5B$

$\displaystyle\implies \frac{\tan5B}{\tan B}=n$

$\displaystyle\implies n\tan B={\frac{\tan^5B-10\tan^3B+5\tan B}{5\tan^4B-10\tan^2B+1}}$

$\displaystyle\implies n={\frac{\tan^4B-10\tan^2B+5}{5\tan^4B-10\tan^2B+1}}$ as $\tan B\ne0$

On rearrangement we get,

$\displaystyle (5n-1)\tan^4B-10(n-1)\tan^2B+n-5=0\ \ \ \ (5)$ which is clearly a Quadratic Equation in $\tan^2B$

As $B$ is real, the discriminant must be $\ge0$(which is true for all real $n$) and $\tan^2B\ge0$

So, any real value of $n$ that satisfies the above condition can form relation $(4)$


$1:$ Let's start with $(3)$ and with known $A=3^\circ$

$\displaystyle\implies \tan3^\circ\tan39^\circ\tan75^\circ\tan111^\circ\tan147^\circ=\tan15^\circ\ \ \ \ (6A)$

Now, $\tan111^\circ=-\tan(180^\circ-111^\circ)=-\tan69^\circ$ and similarly, $\tan147^\circ=-\tan33^\circ$

$\displaystyle\implies \tan3^\circ\tan39^\circ\tan69^\circ\tan33^\circ=\tan15^\circ\cot75^\circ\ \ \ \ (6B)$

Now, $\cot75^\circ=\tan(90^\circ-75^\circ)=\tan15^\circ$ and $\tan15^\circ=\tan(60^\circ-45^\circ)=\cdots=2-\sqrt3$

Interestingly, we shall reach at $(6B)$ if we start with $A=39^\circ,69^\circ,33^\circ,75^\circ$

and we shall get the angles in complementary form if we start with $36^\circ m+15^\circ$ where $m$ is any integer

$2:$ Setting $A=6^\circ, \implies \tan6^\circ\tan42^\circ\tan78^\circ\tan114^\circ\tan150^\circ=\tan30^\circ\ \ \ \ (7A)$

Now, $\tan114^\circ=-\tan66^\circ$ and $\tan150^\circ=-\tan30^\circ$

$\displaystyle\implies \tan6^\circ\tan42^\circ\tan78^\circ\tan66^\circ=1\ \ \ \ (7B)$

Again, we shall reach at $(7B)$ if we start with $A=42^\circ,78^\circ,66^\circ$

We shall get $1$ in the Right Hand Side, if $5A=u180^\circ\pm(r36^\circ+A)$ where $u,r$ are any integers

Try with any multiple of $3$ as $84^\circ,63^\circ$ etc. for $A$

Finally don't miss to try some values of $n$ in $(5)$

  • $\begingroup$ For some reason I did not got ping'd with your comment.Thank you for showing the interesting results! I will see if your formulas help in proving $\tan87\tan51\tan15\tan9=1$ or $\tan75\tan69\tan33\tan9=1$. You do show they are equal to each other, but we can´t conclude they are both equal to $1$. $\endgroup$ – chubakueno Nov 10 '13 at 21:02
  • $\begingroup$ @labbhattacharjee Can have an example where $n=2$? $\endgroup$ – pipi Nov 12 '13 at 1:36
  • $\begingroup$ @pipi With integers value in sexagesimals, a direct verification by bruteforce says no. But there are infinitely many non-integer angles... $\endgroup$ – chubakueno Nov 12 '13 at 2:20
  • $\begingroup$ @pipi, can't you put $n=2$ in $(5)?$ and find $\tan B$ $\endgroup$ – lab bhattacharjee Nov 12 '13 at 4:28
  • $\begingroup$ @labbhattacharjee For non-integer angles to stastify $\tan a°\tan b°\tan c°\tan d° =2 $, I think we can let $b=c=d=45$ and search for $\tan a°=2$. $\endgroup$ – pipi Nov 12 '13 at 10:03

This answers the case $n\neq 1$. Without considering $\tan60\tan60\tan x\tan(90-x)=3$, the code

#include <stdio.h>
#include <math.h>
int main ()
    double result;
    double tantable[90];
    int a=0,b=0,c=0,d=0;
                    result = tantable[a]*tantable[b]*tantable[c]*tantable[d];
                    int appresult=round(result);
                    if(a+d==90&&b==60&&c==60) continue;//tan(x)tan60tan60tan(90-x)
                    if(c+d==90&&a==60&&b==60) continue;//tan60tan60tan(x)tan(90-x) [just a permutation]
                    if(appresult>1 && (appresult-result<0.000001) && (result-appresult<0.000001))
    return 0;

Produces the following log(commented by me):

89,36,34,14->7.000000//No. Error is beyond precision
80,60,40,20->3.000000//Yes! (1)
79,74,44,22->7.000000//No. Error is beyond precision
78,59,57,53->16.000000//No. Error is beyond precision
76,76,64,23->14.000000//No. Error is beyond precision
73,73,61,25->9.000000//No. Error is beyond precision

Verification of the identities: $(1)$: $$\tan(30+x)\tan(3x)\tan(30-x)\tan(90-x)=1$$ $$\implies \tan40\tan30\tan20\tan80=1$$ $$\implies \tan40\tan20\tan80=\tan60$$ $$\implies (1)=\tan60\tan60=3$$

$(2)$: We know(or if skeptic, we can verify by setting $\sin3x=\cos2x$) $$\tan18=\sqrt{\frac{5-2\sqrt{5}}{5}}$$ By double angle formula, we get $$\implies\tan36=\sqrt{5-2\sqrt{5}}$$ And by $\frac{1}{\tan{18}}=\tan72$ $$\implies\tan72=\sqrt{\frac{5}{5-2\sqrt{5}}}$$ $$\implies\tan72\tan36=\sqrt{5}\implies\tan72\tan72\tan36\tan36=5$$ And $(3)$ is trivial.

So, we see that the only other such identities are of the forms: $$\tan(x)\tan(60)\tan(60)\tan(90-x)=3$$ $$\tan(80)\tan(60)\tan(40)\tan(20)=3$$ $$\tan(72)\tan(72)\tan(36)\tan(36)=5$$ $$\tan(60)\tan(60)\tan(60)\tan(60)=9$$

Progress in the $n=1$ case(does not consider known cases):

#include <stdio.h>
#include <math.h>
int main ()
    double result;
    double tantable[90];
    int a=0,b=0,c=0,d=0;
                    result = tantable[a]*tantable[b]*tantable[c]*tantable[d];
                    if((1-result<0.000001) && (result-1<0.000001)&&(a+d!=90)&&(b+c!=90)){
                        if((d%3==0) && (b-d/3)==30 && (c+d/3)==30&&(a+d/3)==90) continue;
                        if(a+3*d==90 && b-d==60&&c+d==60) continue;
                        if(a-d==60 && b+d==60&&c+3*d==90) continue;
                        if(a-d==60 && c+d==60&&b+3*d==90) continue;
    return 0;

These are the remaining cases (equivalences commented by me, they are apparently true)

87,27,21,15->1.000000 //2

We can reduce the number of statements to be proved:

Assuming $(1)$ $$\tan87\tan51\tan15\tan9=1\implies\tan51\tan15\tan9=\tan3 \tag{a.1}$$ $$\tan(30-x)\tan(30+x)\tan3x=\tan x\implies\tan27\tan33\tan9=\tan3 \tag{a.2}$$ $$(\text{a}.1),(\text{a}.2)\implies\tan51\tan15=\tan27\tan33\iff (5)$$

Assuming $(4)$ $$\tan75\tan69\tan33\tan9=\implies\tan75\tan69\tan33=\tan81 \tag{b.1}$$ $$\tan(60+x)\tan(60-x)\tan x=\tan3x\implies\tan87\tan33\tan27=\tan81 \tag{b.2}$$ $$(\text{b}.1),(\text{b}.2)\implies\tan75\tan69=\tan87\tan27\iff (2)$$

So, we just need to prove or disprove $$\tan87\tan51\tan15\tan9=1\tag{1}$$ $$\tan75\tan69\tan33\tan9=1\tag{2}$$ $$\tan78\tan66\tan42\tan6=1\tag{3}$$

Verification of $(3)$ $$\tan(60+x)\tan(60-x)\tan(x)=\tan3x\implies\tan78\tan42\tan18=\tan54$$ $$\implies\tan78\tan42=\tan54\tan72 \tag{3.1}$$ $$\tan(60+x)\tan(60-x)\tan(x)=\tan3x\implies\tan66\tan54\tan6=\tan18$$ $$\implies\tan72\tan6\tan66\tan54=1 \tag{3.2}$$ $$(3.1),(3.2)\implies\tan6\tan66\tan78\tan42\iff (3)$$ I have not been able to prove/disprove the other identities until now.

  • 2
    $\begingroup$ In $\tt C$ and $\tt C++$, $\pi$ is defined in $\tt<math.h>$ and $\tt<cmath>$ as $\tt M\_PI$ ($\tt\mbox{double}$ type) and ($\tt M\_PIl$) ($\tt\mbox{long double}$ type). $\endgroup$ – Felix Marin Nov 9 '13 at 7:19
  • $\begingroup$ Thank you very much, @FelixMarin . When I followed your suggestion I revisited the code and I saw I had lost an identity! $\endgroup$ – chubakueno Nov 9 '13 at 15:55

To answer, the first question, from this, I shall utilize $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$

Put $x=5^\circ$ to get $$\tan5^\circ\tan55^\circ\tan65^\circ=\tan15^\circ$$

Now, $\displaystyle\tan15^\circ=\cot(90^\circ-15^\circ)=\cot75^\circ=\frac1{\tan75^\circ}$

  • $\begingroup$ This was known to the op, but he wanted a generalization $\endgroup$ – chubakueno Nov 9 '13 at 5:07
  • $\begingroup$ @chubakueno, what about the identity (with proof), I have used $\endgroup$ – lab bhattacharjee Nov 9 '13 at 5:08
  • $\begingroup$ @pipi, here it another : math.stackexchange.com/questions/508634/… $\endgroup$ – lab bhattacharjee Nov 9 '13 at 5:12
  • $\begingroup$ Yes, i think it completes the fermat-like second line of the post :) By the way, can you see any pattern in the last lines of my post?(I can't) $\endgroup$ – chubakueno Nov 9 '13 at 5:14
  • $\begingroup$ @chubakueno, not sure if I understood your first statement. But thanks a ton for your last statement. I think you will find this interesting : math.stackexchange.com/questions/537275/… $\endgroup$ – lab bhattacharjee Nov 9 '13 at 5:19

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