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Prove that $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.

What I've tried doing : If $\theta=\frac{\pi}{7}:$ $$ 3\theta+4\theta=\pi $$ This allowed me to prove that : $$ \tan^2\frac{\pi}{7}+\tan^2\frac{2\pi}{7}+\tan^2\frac{3\pi}{7}=21 \\ \cot^2\frac{\pi}{7}+\cot^2\frac{2\pi}{7}+\cot^2\frac{3\pi}{7}=5 $$ Is my reasoning wrong or is this entirely the wrong way to approach this question ?

Vishnu V.S
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  • What is your actual question? – lab bhattacharjee Mar 17 '17 at 15:36
  • @labbhattacharjee I believe I put it in the title. The other two formulae, I've already managed to prove. I just mentioned them, thinking they would help. – Vishnu V.S Mar 17 '17 at 15:38
  • http://math.stackexchange.com/questions/818749/trigo-problem-find-the-value-of-sin-frac2-pi7-sin-frac4-pi7-sin-fr/937553#937553 – lab bhattacharjee Mar 17 '17 at 15:39
  • Are you adding the sines or multiplying? And where did you get tan and cot from? – browngreen Mar 17 '17 at 15:39
  • @browngreen Multiplying. I just... uh... found them along the way. I'll just delete that part of the question so it doesn't cause any more confusion. SOrry. – Vishnu V.S Mar 17 '17 at 15:41
  • Through $\sin\theta=\frac{e^{i\theta}-e^{i\theta}}{2i}$ that identity follows from a Gauss sum: https://en.wikipedia.org/wiki/Gauss_sum – Jack D'Aurizio Mar 17 '17 at 15:41
  • See also : http://math.stackexchange.com/questions/265229/prove-that-cot2-pi-7-cot22-pi-7-cot23-pi-7-5?noredirect=1&lq=1 and http://math.stackexchange.com/questions/175736/evaluate-tan-220-circ-tan-240-circ-tan-280-circ – lab bhattacharjee Mar 17 '17 at 15:43

2 Answers2

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You can show that $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ are the three roots of $64x^3-112x^2+56x-7 = 0$, as follows:

Take $\zeta_7$ a $7$'th root of unity (a root of $(x^7-1)/(x-1) = x^6+x^5+x^4+x^3+x^2+x+1 = 0$), then $2 \cos(k 2\pi/7) = \zeta_7^k + \zeta_7^{-k}$, and with $\sin^2+\cos^2=1$ you can now express $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ in terms of $\zeta_7$. After that you can verify that they are roots of $64x^3-112x^2+56x-7 = 0$.

Mark
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Since $\sin\frac{k\pi}{7}=\sin\frac{(7-k)\pi}{7}$ for k=1,2,3 we get:
$$(\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7})^2= \prod_{k=1}^6 {\sin\frac{k\pi}{7}}=\frac{7}{2^6}$$ where the last equality is the case n=7 in this question: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
So we get the desired result: $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.

Community
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Isberg
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