Show that the following product equals 1 (involves trig)

Question

How can I show that: $$\prod_{k=1}^{n}\left ( 1+2\cos\frac{2\pi .3^{k}}{3^{n}+1} \right )=1$$ Could you please explain to me how to approach this problem? Thank you.

Answer 1

Hint:

$$1+2 \cos{\theta} = \frac{\sin{(3 \theta/2)}}{\sin{(\theta/2})}$$

I get as a result for the product:

$$\frac{\sin{\left ( \frac{\pi \cdot 3^{n}}{3^n+1}\right)}}{\sin{\left( \frac{\pi}{3^n+1}\right)}} = 1$$

The way to get this result is to let $\theta_0 = 2 \pi/(3^n+1)$. The product is then

$$\frac{\sin{(3 \theta_0/2)}}{\sin{(\theta_0/2)}} \frac{\sin{(3^2 \theta_0/2)}}{\sin{(3 \theta_0/2)}} \cdots \frac{\sin{(3^n \theta_0/2)}}{\sin{(3^{n-1} \theta_0/2)}}$$

The product telescopes, so we are left with just the last term in the numerator and the first term in the denominator. The result follows.