MHB Trigonometric product challenge

[lfdahl]

Prove, that

$$\prod_{j = 1}^{n}\left(1+2\cos \left(\frac{3^j}{3^n+1}2\pi\right)\right) = 1.$$

 

[lfdahl]

Suggested solution:

Spoiler

We have

\[1 + 2\cos x = \frac{\sin x+\sin 2x}{\sin x } = \frac{2\sin \frac{3x}{2}\cos \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \frac{\sin \frac{3x}{2}}{\sin \frac{x}{2}}\]

Hence, we can rewrite the product

\[\prod_{j=1}^{n}\left ( 1+2 \cos \left ( \frac{2\pi 3^j }{1+3^n}\right ) \right ) \]
\[=\prod_{j=1}^{n}\frac{\sin \left ( \frac{3\pi 3^{j}}{1+3^n} \right )}{\sin\left ( \frac{\pi 3^j }{1+3^n}\right )} \]
Recognizing a telescoping product:
\[ = \frac{\sin \left ( \frac{3\pi3^{n}}{1+3^n} \right )}{\sin\left ( \frac{3\pi }{1+3^n}\right )} \]

Rewriting the argument:
\[\frac{3\pi 3^n}{1+3^n} = 3\pi-\frac{3\pi}{1+3^n}\]
And making use of the identity: \[\sin (\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha \]
yields the desired result:
\[\frac{\sin \left ( \frac{3\pi3^{n}}{1+3^n} \right )}{\sin\left ( \frac{3\pi }{1+3^n}\right )}\]
\[= \frac{\sin 3\pi \cos \left ( \frac{3\pi}{1+3^n} \right)-\sin \left (\frac{3\pi}{1+3^n} \right )\cos 3\pi}{\sin \left (\frac{3\pi}{1+3^n} \right )} = 1.\]

 

[topsquark]

Good one!

-Dan