Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$ 
Evaluate$$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$

My attempt: 
$$1+2\cos 2\theta= 1+2(1-2\sin^2\theta)=3-4\sin^2\theta$$
$$=\frac{3\sin \theta-4\sin^3\theta}{\sin \theta}=\frac{\sin 3\theta}{\sin \theta}$$
I did not understand how to solve after that. Help required.
 A: It's unclear if you are asking about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$ or about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi k\cdot 3^k}{3^{100}+1}\right]$$
I will assume it is the former. At the moment, that is in the body of your question, while the latter is in the title. If it's the former, then using the trig identity you found, you have a telescoping product which then further simplifies nicely: $$
\begin{align}
\prod^{100}_{k=1}\frac{\sin\left(\pi\frac{3^{k+1}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{k}}{3^{100}+1}\right)}&=\frac{\sin\left(\pi\frac{3^{2}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{1}}{3^{100}+1}\right)}\frac{\sin\left(\pi\frac{3^{3}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{2}}{3^{100}+1}\right)}\cdots\frac{\sin\left(\pi\frac{3^{101}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{100}}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3^{101}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{1}}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3\cdot3^{100}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3\left(\cdot3^{100}+1\right)-3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(3\pi-\pi\frac{3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=1
\end{align}$$ 
Note near the end that $\sin(3\pi-X)=\sin(X)$. 
A: Let $$z=\cos\bigg(\frac{2\pi}{3^n+1}\bigg)+i\sin\bigg(\frac{2\pi}{3^n+1}\bigg)$$
Then $z^{3^n+1}=1$ and also $\displaystyle 2\cos \bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)=z^{3^k}+\frac{1}{z^{3^k}}$
Write $$\prod^{n}_{k=1}\bigg[1+2\cos\bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)\bigg]$$
$$=\bigg(1+z^3+\frac{1}{z^3}\bigg)\bigg(1+z^9+\frac{1}{z^9}\bigg)\cdots \cdots \bigg(1+z^{3n}+\frac{1}{z^{3n}}\bigg)$$
$$=\frac{(1+z^3+z^6)(1+z^9+z^{18})\cdots \cdots (1+z^{3^n}+(z^{3^n})^2)}{z^{3+9+\cdots \cdots +3^n}}$$
Multiply both Nr and Dr by $(1-z^3)$
$$=\frac{1-z^{3^{n+1}}}{(1-z^3)\cdot z^{3\frac{(3^n-1)}{2}}}= \frac{1-z^{-3}}{-(1-z^3)\cdot z^{-3}}=1.$$
$\text{Simplification}:\;\; $ From $z^{3n+1}=1\Rightarrow z^{3n}=z^{-1}$
And $$z^{3\frac{(3^n-1)}{2}}=z^{-3}\cdot z^{3\frac{(3^+1)}{2}}=z^{-3}\cdot \bigg(z^{\frac{(3^+1)}{2}}\bigg)^3=-z^3$$
