Show that $\tan{(\pi/7)} \tan{(2\pi/7)}\tan{(3\pi/7)}=\sqrt{7}$

Question

I tried in this way.$\tan(a+b)=\frac{(\tan a + \tan b)}{1 - \tan a \tan b }$value of $\tan \frac{\pi}{7}$ is coming in decimal.what to do

Answer 1

If $\theta = \frac{k\pi}{7}$ where $k = 1, 2, 3$, then $7\theta = k\pi$ and hence $4\theta = k\pi - 3\theta$. Thus $\tan(4\theta) = -\tan(3\theta)$. Expanding, and writing $t = \tan\theta$, we get $$ \frac{4t-4t^3}{1-4t^2 + t^4} = -\frac{3t-t^3}{1-3t^2} $$ Simplifying we get $$t^6 - 7t^4 + \cdots -7 = 0$$ The roots are $\tan \pi/7, \tan 2\pi/7, \ldots, \tan 6\pi/7$. Noting that $\tan 2\pi/7 = - \tan 5\pi/7, \tan 4\pi/7 = -\tan 3\pi/7$ etc, the roots are $\pm \tan \pi/7, \pm \tan 2\pi/7, \pm \tan 3\pi/7$. Thus the product of the roots is $$-\tan^2\pi/7 \tan^2 2\pi/7 \tan^2 3\pi/7 = -7$$ and since $\tan k\pi/7$ for $k = 1, 2, 3$ are all positive, we get $$ \tan \pi/7 \tan 2\pi/7 \tan 3\pi/7 = \sqrt{7}$$

Answer 2

Here is another possible approach. Using Euler's formula, we know: $$\sin \theta=\frac {e^{i\theta}-e^{-i\theta}}{2i}$$ And $$cos\theta=\frac {e^{i\theta}+e^{-i\theta}}{2}$$ Since $\tan \theta=\frac {\sin \theta}{\cos \theta}$ and $\tan \frac {2\pi}{7}=\tan \frac {5\pi}{7}$, we have the product: $$P=-i \frac {(1-\alpha_1)(1-\alpha_3)(1-\alpha_5)}{(1+\alpha_1)(1+\alpha_3)(1+\alpha_5)}$$ Here, $\alpha_k=e^{\frac {2ki\pi}{7}}$ for $0\leq k \leq 6$, which are the $7^{th}$ roots of unity. Note that: $$\overline{\alpha_k}=\alpha_{7-k}$$ where $\overline{z}$ represents the conjugate of $z$. Thus, taking conjugate on $P$ and multiplying with above equation, using $P\overline{P}=|P|^2$, we get: $$|P|^2=\frac {\prod_{k=1}^6 (1-\alpha_k)}{\prod_{k=1}^6 (1+\alpha_k)}$$ Now, since $\alpha_k$ represent the $7^{th}$ roots of unity, they satisfy: $$1+z+z^2+z^3+...+z^6=\prod_{k=1}^6 (z-\alpha_k)$$ Putting $z=1$ gives numerator to be equal to $7$, while putting $z=-1$ gives denominator equal to $1$. Thus, $$|P|^2=7$$ and hence $P=\sqrt 7$ since $P$ must be real and positive as all angles in $\tan$ are less than $90°$. Hence, we obtain our answer.

PS: The origin of this problem lies in the so-called 'Coffin problems' of Soviet Russia, which were considered extremely difficult and asked specifically to Jewish students so that they would have much higher chances of rejection. This particular problem was one of those.