Prove that $\tan(6^\circ)\tan(42^\circ) = \tan(12^\circ) \tan(24^\circ)$.

I don't know how to approach this problem. One approach might be to note that $42-6= 24+12$, and then apply the identities for $\tan(A+B)$ and $\tan(A-B)$, but it just makes it more complex: $$\frac{\tan(42^{\circ}) - \tan(6^{\circ})}{1+\tan(42^{\circ})\tan(6^{\circ})}= \frac{\tan(24^{\circ}) + \tan(12^{\circ})}{1 - \tan(24^{\circ})\tan(12^{\circ})}.$$

Can anyone here please provide a hint?

  • $\begingroup$ Do you know the formula for $\tan(x+y)$? $\endgroup$ – Thomas Andrews Jul 29 '13 at 19:22
  • $\begingroup$ @ThomasAndrews $tan(x+y)= \frac{tan(x) + tan(y)}{1-tan(x)tan(y)}$ $\endgroup$ – HelloWorld Jul 29 '13 at 19:25
  • $\begingroup$ Okay, now use $\tan(-6^\circ)=-\tan(6^\circ)$, and you should be well on your way. $\endgroup$ – Thomas Andrews Jul 29 '13 at 19:26
  • $\begingroup$ Oh yes, it was so easy! Thanks for the hint. @ThomasAndrews :) $\endgroup$ – HelloWorld Jul 29 '13 at 19:27
  • $\begingroup$ My answer was wrong, please unselect it. Misread the problem. $\endgroup$ – Thomas Andrews Jul 29 '13 at 19:31

We can prove $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$ (Proof below)

Putting $x=6^\circ,$ $$\tan 6^\circ\tan 66^\circ\tan 54^\circ=\tan18^\circ$$

Putting $x=18^\circ,$ $$\tan 18^\circ\tan42^\circ\tan78^\circ=\tan54^\circ$$

On multiplication, $$\tan 6^\circ\tan 66^\circ\tan42^\circ\tan78^\circ=1$$

$$\tan 6^\circ\tan42^\circ=\cot66^\circ\cot78^\circ=\tan(90^\circ-66^\circ)\tan(90^\circ-78^\circ)=\tan24^\circ\tan12^\circ$$


Proof $1:$ $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan x\cdot\frac{\tan60^\circ-\tan x}{1+\tan60^\circ\tan x}\cdot\frac{\tan60^\circ+\tan x}{1-\tan60^\circ\tan x}$$

$$=\tan x\cdot \frac{\sqrt3-\tan x}{1+\sqrt3\tan x}\cdot \frac{\sqrt3+\tan x}{1-\sqrt3\tan x}$$

$$=\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3x$$

Proof $2:$

If $\tan3x=\tan 3y$

$$\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3y$$

$$\implies \tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$

If $\tan x_1,\tan x_2,\tan x_3$ are the three roots of above cubic equation,

$\implies \tan x_1\cdot\tan x_2\cdot\tan x_3=-\tan3y$

Again, as $\tan3x=\tan 3y$

$3x=180^\circ n+3y$ where $n$ is any integer $\implies x=60^\circ n+y$

So, the in-congruent values of $x$ are $y,60^\circ +y,60^\circ\cdot2+y$

So, the corresponding roots are $\tan y,\tan(60^\circ +y), \tan(60^\circ\cdot2+y)=\tan\{180^\circ-(60^\circ-y)\}=-\tan(60^\circ-y)$

$\implies -\tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=-\tan3y$

$\implies \tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=\tan3y$


  • $\begingroup$ Nice! How did you figure it out though? $\endgroup$ – Alraxite Aug 3 '13 at 0:16
  • $\begingroup$ @Alraxite, First of all, the problem is equivalent to $$\tan 6^\circ\tan 66^\circ\tan42^\circ\tan78^\circ=1$$ Now, $$\tan 6^\circ\tan 66^\circ\tan42^\circ\tan78^\circ=(\tan 6^\circ\tan54^\circ\tan 66^\circ)\tan42^\circ\tan78^\circ\cdot\cot54^\circ=(\tan18^\circ\tan42^\circ\tan78^\circ)\cdot\cot54^\circ=\tan54^\circ\cdot\cot54^\circ=1$$ $\endgroup$ – lab bhattacharjee Aug 3 '13 at 4:02

Here is a proof that I got after back calculation. Let $\theta=36^{\circ}$ Then, one can show that $\theta$ satisfies the following equation $$(\cos \theta +1)(4\cos^2\theta -2\cos\theta-1)=0$$since $\cos 36^{\circ}>0$, $\theta $ satisfies $$4\cos^2\theta-2\cos \theta-1=0\\ \Rightarrow 2\cos 2\theta-2\cos \theta=-1\ \\ \Rightarrow \cos 2\theta-\cos \theta=-\frac{1}{2}$$ Let $\phi=12^{\circ}$. Then $\theta=3\phi$ and $$\cos 6\phi-\cos 3\phi=-\frac{1}{2}\\ \Rightarrow \cos 6\phi+1=\cos 3\phi+\frac{1}{2}=\cos 3\phi+\cos 5\phi\quad (\because 5\phi=60^{\circ})\\ \Rightarrow 2\cos^2 3\phi=2\cos 4\phi \cos \phi\\ \Rightarrow \frac{\cos 36^{\circ}}{\cos 48^{\circ}}=\frac{\cos 12^{\circ}}{\cos 36^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}+\sin 6^{\circ}\sin 42^{\circ}}{\cos 6^{\circ}\cos 42^{\circ}-\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}+\sin 24^{\circ}\sin 12^{\circ}}{\cos 24^{\circ}\cos 12^{\circ}-\sin 24^{\circ}\sin 12^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}}{\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}}{\sin 24^{\circ}\sin 12^{\circ}}\quad (\mbox{by Componendo-Dividendo})\\ \Rightarrow \tan 6^{\circ}\tan 42^{\circ}=\tan 24^{\circ}\tan 12^{\circ}\hspace{6cm}\Box$$

  • $\begingroup$ I suspected as I was working out an answer (which I finally deleted) that the equation being sought would not come from general trig identities, but from some specific relation among multiples of 6º or 12º. I had reached what you've shown to be the key relation $ \ \cos^2 36º \ = \ \cos 48º \cdot \cos 12º \ , $ but it was not clear how that would follow from more general identities... [I'll mention that this equation can also be obtained from the product-to-sum formula for tangent.] $\endgroup$ – colormegone Jul 29 '13 at 21:49
  • $\begingroup$ @RecklessReckoner You were almost there... $\cos 36 = {-1 + \sqrt{5} \over 4}$ and then using $\cos a \cos b = {1 \over 2}\cos(a - b) - {1 \over 2}\cos (a+b)$ you're done. $\endgroup$ – Zarrax Jul 29 '13 at 22:01
  • $\begingroup$ Yes, of course @RecklessReckoner there maybe several different ways to find out these equations, but the point is they all will use some properties specific to $\cos 36^{\circ}$. $\endgroup$ – Samrat Mukhopadhyay Jul 30 '13 at 8:43

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