$\cos(x) \; \cdot \; \cos(2x) \; \cdot \; \cos(3x) \; \cdots \cos(nx)=$? Here is my attempt,
$2 \cos(x) = e^{ix}+e^{-ix}$
$2 \cos(kx) = e^{kix}+e^{-kix}$
$\displaystyle\implies\prod_{k=1}^{n} \cos(kx) = \frac{1}{2^n} \prod_{k=1}^{n} \left(e^{kix}+e^{-kix}\right)$
Any ideas for taking it from here? Thanks.
Please don't give away the entire solution :P
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\cos\pars{x}\cos\pars{2x}\cos\pars{3x}\ldots\cos\pars{nx} =
\prod_{k = 0}^{n - 1}\cos\pars{\bracks{k + 1}x} =
\prod_{k = 0}^{n - 1}
{\expo{\ic\pars{k + 1}x} + \expo{-\ic\pars{k + 1}x} \over 2}
\\[5mm] = &\
{1 \over 2^{n}}
\prod_{k = 0}^{n - 1}
\expo{\ic\pars{k + 1}x}\bracks{1 + \expo{-2\ic\pars{k + 1}x}} =
{1 \over 2^{n}}\prod_{k = 0}^{n - 1}
\pars{\expo{\ic x}}^{k + 1}\bracks{1 - \pars{-\expo{-2\ic x}}
\pars{\expo{-2\ic x}}^{k}}
\\[5mm] = &\
{1 \over 2^{n}}\,\expo{\ic n\pars{n + 1}x/2}
\prod_{k = 0}^{n - 1}\
\bracks{1 - \pars{-\expo{-2\ic x}}\pars{\expo{-2\ic x}}^{k}} =
\bbx{{\expo{\ic n\pars{n + 1}x/2} \over 2^{n}}\,
\pars{-\expo{-2\ic x};\expo{-2\ic x}}_{n}}
\end{align}

$\ds{\pars{a;b}_{m}}$ is the q-Pochhammer Symbol.

