Prove and generalize $\cos\frac\pi9\cos\frac{2\pi}9\cos\frac{3\pi}9\cos\frac{4\pi}9=\frac1{16}$

Question

Is there a simple proof for the following? $$16\cos\frac\pi9\cos\frac{2\pi}9\cos\frac{3\pi}9\cos\frac{4\pi}9=1$$ Is this statement valid for any odd number (not only 9)?

Is $\cos x=\frac{e^{ix}+e^{-ix}}2$ useful in this case?

Answer 1

Multiplying the given product by $2^4 \sin \frac{\pi}{9}$ and using the double-angle formula $\sin (2x)=2\sin x\cos x$ reduces the product one by one yields

$$ \begin{aligned} \because \quad & 2^4 \sin \frac{\pi}{9} \cos \frac{\pi}{9} \cos \frac{2 \pi}{9} \cos \frac{3 \pi}{9} \cos \frac{4 \pi}{9} \\ = & 2^3 \sin \frac{2 \pi}{9} \cos \frac{2 \pi}{9} \cdot \frac{1}{2} \cdot \cos \frac{4 \pi}{9} \\ = & 2 \sin \frac{4 \pi}{9} \cos \frac{4 \pi}{9} \\ = & \sin \frac{8 \pi}{9} \\ = & \sin \frac{\pi}{9} \end{aligned} $$ $$ \therefore \quad \sin \frac{\pi}{9} \cos \frac{\pi}{9} \cos \frac{2 \pi}{9} \cos \frac{3 \pi}{9} \cos \frac{4 \pi}{9}=\frac{1}{2^4}=\frac{1}{16} $$