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I have to prove:

$$\prod_{i=1}^6 \left(2\cos\left(\frac{2^{i}\pi}{13}\right)-1\right)=1$$

I really have no idea about starting with this one. With the help of Wolfram Alpha, I noticed that:

$$\left(2\cos\left(\frac{2\pi}{13}\right)-1\right)\left(2\cos\left(\frac{8\pi}{13}\right)-1\right)\left(2\cos\left(\frac{32\pi}{13}\right)-1\right)=1$$

and

$$\left(2\cos\left(\frac{4\pi}{13}\right)-1\right)\left(2\cos\left(\frac{16\pi}{13}\right)-1\right)\left(2\cos\left(\frac{64\pi}{13}\right)-1\right)=1$$

Any help is appreciated. Thanks!

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    $\begingroup$ Why the downvote? $\endgroup$
    – Pranav Arora
    May 13, 2015 at 13:17

2 Answers 2

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Hint: There is an identity of the form $\cos 3 \alpha = \cos \alpha \cdot (2 \cos 2 \alpha - 1)$. Multiply this as $\alpha$ runs through $2^k \pi/13$ with $0 \leq k < 6$. Using standard properties of $\cos \alpha$, you will be able to cancel out the redundant terms.

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  • $\begingroup$ That's not true, $\sin(3\alpha)=\sin \alpha (2\cos(2\alpha)+1)$ $\endgroup$
    – Pranav Arora
    May 13, 2015 at 14:00
  • $\begingroup$ @PranavArora: Yes, sorry! Now it should be correct, though. $\endgroup$
    – Jakub Konieczny
    May 13, 2015 at 14:03
  • $\begingroup$ (At least now Wolfram says it's true) $\endgroup$
    – Jakub Konieczny
    May 13, 2015 at 14:03
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You have $\cos 3\alpha=\cos \alpha(2\cos 2\alpha-1)$

You have also

$\cos \frac{32\pi}{13}=\cos \frac{6\pi}{13}$

$\cos \frac{64\pi}{13}=\cos \frac{12\pi}{13}$

$\cos \frac{16\pi}{13}=\cos \frac{10\pi}{13}$

Thus at the end you have

$\prod_{i=1}^6 \left(2\cos\left(\frac{2^{i}\pi}{13}\right)-1\right)=\prod_{i=1}^6 \dfrac{\cos \frac{3i\pi}{13}}{\cos \frac{i\pi}{13}}$

And all terms cancel out smoothly (the first few very easily).

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