I want to prove this equality holds:
$$\tan\big(\frac{\pi}7\big) \tan\big(\frac{2\pi}7\big) \tan\big(\frac {3\pi}7\big)= \sqrt 7.$$
Please help me. Thanks.
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There's a helpful identity here: http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Identities_without_variables – SubSevn Mar 16 '15 at 11:19
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This already has an answer here – najayaz Mar 16 '15 at 11:31
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See also: http://math.stackexchange.com/questions/823819/if-alpha-frac2-pi7-then-the-find-the-value-of-tan-alpha-tan2-alpha/824178#824178 – lab bhattacharjee Mar 16 '15 at 12:04
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Note that \begin{equation} \tan \frac{\pi}{7} \tan \frac{2\pi}{7} \tan \frac {3\pi}{7} = \frac{\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac {3\pi}{7}}{{\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac {3\pi}{7}}} \end{equation} Now, it is trivial to see that \begin{eqnarray} \frac{3 \pi}{7}= \frac{\pi}{7} + \frac{2 \pi}{7} = \frac{\pi}{7}+ \frac{\pi}{7} + \frac{\pi}{7} \end{eqnarray} Thus, concentrating on the numerator and using the identity \begin{equation} \sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \end{equation} We find \begin{eqnarray} \sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac {3\pi}{7} &=& \sin \frac{\pi}{7}\left(\sin \frac{\pi}{7}\cos \frac{\pi}{7}+\cos \frac{\pi}{7}\sin \frac{\pi}{7} \right) \sin \frac {3\pi}{7}\\ &=& 2\sin^{2} \frac{\pi}{7} \cos \frac{\pi}{7} \sin \frac {3\pi}{7} \end{eqnarray} Now repeat this process for the term on $3 \pi/7$ and continue into the denominator and you should be ok.
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@user224013 I believe the links supplied sufficiently answer the question. – najayaz Mar 16 '15 at 15:34