On the left you have an expression of the form
$$
\frac{z^n-z^{-n}}{z-z^{-1}}=z^{n-1}+z^{n-3}+z^{n-5}+...+z^{1-n}
$$
which divides perfectly and has the unit roots of degree 2n except $\pm 1$ as its roots. Because of symmetry, for every root $λ$ one has also $\bar λ$, $-λ$, $-\bar λ$ as roots. Because of $n$ odd the special case $λ=i$ does not occur, so that all roots can be grouped into these quadrupels.
Instead of pairing conjugate roots one can here also pair mirror roots $λ$ and $-\bar λ$ relative to the imaginary axis resulting in
\begin{align}
\frac{z^n-z^{-n}}{z-z^{-1}}
&=\prod_{k=1}^{(n-1)/2}(z-e^{i\pi \,k/n})(1+z^{-1}e^{-i\pi\, k/n})\;(z-e^{-i\pi \,k/n})(1+z^{-1}e^{i\pi\, k/n})\\
&=\prod_{k=1}^{(n-1)/2}(z-z^{-1}-e^{i\pi \,k/n}+e^{-i\pi\, k/n})(z-z^{-1}+e^{i\pi \,k/n}-e^{-i\pi\, k/n})\\
&=\prod_{k=1}^{(n-1)/2}(z-z^{-1}-2i\sin(\pi \,k/n))(z-z^{-1}+2i\sin(\pi \,k/n))
\end{align}
Replacing $z=e^{ix}$, collecting exponentials into sine terms and applying a binomial formula gives a formula similar to the stated result.
Check that also using $λ=e^{i2\pi\,k/n}$ gives one representant per root quadruple, here it is also important that $n$ is odd. This will finally result in the stated formula.
