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I want to prove that Using $z^9=1$ and the fact that $1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8=0$ where $w=cis(\frac{2π}{9})$, that $$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$$

I am able to do this by using $w^n+w^{-n}=2\cos \frac{2nπ}{9}$ and hence expanding the LHS but it is laborious and, so, I was wondering is there is a more efficient or elegant method of proving this this identity using the ninth root of unity.

Thanks very much

Quanto
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Dfernando
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2 Answers2

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Note

$$\frac{1-z^9}{1-z}= \prod_{k=1}^8(w^k-z) $$ Set $z=-1$ and use $\prod_{k=1}^8w^{k}=1$

\begin{align} 1&=\prod_{k=1}^8(w^k+1)= (\prod_{k=1}^8w^{k})^{1/2}\prod_{k=1}^8(w^{k/2}+ w^{-k/2}) = 2^8 \prod_{k=1}^8\cos\frac{k\pi}9 =\left( 2^4 \prod_{k=1}^4\cos\frac{k\pi}9 \right)^2 \end{align}

Then, substitute $\cos\frac{3\pi}9=\frac12$ to obtain

$$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac18$$

Quanto
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Start out as by noticing that

$$ \omega = cis(\frac{2\pi}{9}),\quad \omega^{8}=cis(\frac{-2\pi}{9})\quad \\ \omega^2 = cis(\frac{4\pi}{9}),\quad\omega^7=cis(\frac{-4\pi}{9}) \\ etc. $$

Now see, $$ \omega\ + \omega^8 = 2\ cos(\frac{2\pi}{9}) \\ \omega^2\ + \omega^7= 2\ cos(\frac{4\pi}{9})\\ \omega^3 + \omega^6 = 2\ cos(\frac{6\pi}{9})\\ \omega^4\ + \omega^5= 2\ cos(\frac{8\pi}{9})\\ $$

If we multiply these we get:

$$ [2\ cos(\frac{2\pi}{9})][2\ cos(\frac{4\pi}{9})][2\ cos(\frac{6\pi}{9})][2\ cos(\frac{8\pi}{9})] = (\omega\ + \omega^8)(\omega^2\ + \omega^7)(\omega^3 + \omega^6)(\omega^4\ + \omega^5) $$

If we carefully pull out the $\omega$s, we can get the R.H.S to equal $ \ \omega^{10}(1\ + \omega)(1\ + \omega^3)(1\ + \omega^5)(1\ + \omega^7)$

We also know that $ \ \omega^{10} = \omega$, so we can simplify it to be $\ \omega(1\ + \omega)(1\ + \omega^3)(1\ + \omega^5)(1\ + \omega^7) $

Using the fact: $\ \omega\ + \omega^2\ + \omega^3\ + \omega^4\ + \omega^5\ + \omega^6\ + \omega^7\ + \omega^8\ = -1 $, the expression simplifies to -1.

Now simplifying the L.H.S. we get: $16[cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{6\pi}{9})\cos(\frac{8\pi}{9})]$. Knowing that $cos(\frac{2\pi}{3}) = -\frac{1}{2}$, we can finally get that:

$$ -8\ [cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{8\pi}{9})] = -1 \\ [cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{8\pi}{9})] = \frac{1}{8} \\ [cos(\frac{\pi}{9})\cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})] = \frac{1}{8} $$

Shaheer ziya
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  • Whoops. I think I didn't read your question quite carefully enough. Whelps, I mean this took me quite some while to write so I don't want to remove it, but hope this is somewhat useful even though it is exactly exactly what you asked not to do, sorry! – Shaheer ziya May 06 '20 at 12:30
  • Haha no problem that was still quite useful! Thanks very much – Dfernando May 06 '20 at 13:53