5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $\tan y,\tan(60^\circ +y), \tan(60^\circ\cdot2+y)=\tan\{180^\circ-(60^\circ-y)\}=-\tan(60^\circ-y)$

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

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

]

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $$\cos 6\phi-\cos 3\phi=-\frac{1}{2}\\ \Rightarrow \cos 6\phi+1=\cos 3\phi+\frac{1}{2}=\cos 3\phi+\cos 5\phi\quad (\because 5\phi=60^{\circ})\\ \Rightarrow 2\cos^2 3\phi=2\cos 4\phi \cos \phi\\ \Rightarrow \frac{\cos 36^{\circ}}{\cos 48^{\circ}}=\frac{\cos 12^{\circ}}{\cos 36^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}+\sin 6^{\circ}\sin 42^{\circ}}{\cos 6^{\circ}\cos 42^{\circ}-\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}+\sin 24^{\circ}\sin 12^{\circ}}{\cos 24^{\circ}\cos 12^{\circ}-\sin 24^{\circ}\sin 12^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}}{\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}}{\sin 24^{\circ}\sin 12^{\circ}}\quad (\mbox{by Componendo-Dividendo})\\ \Rightarrow \tan 6^{\circ}\tan 42^{\circ}=\tan 24^{\circ}\tan 12^{\circ}\hspace{6cm}\Box$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.