A trigonometric product

Question

I have to prove:

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

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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!

Answer 1

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.

Answer 2

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).