Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $ 
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$

I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help will be appreciated.
Thanks.

Note: Here's another variation inspired by an answer to this question.

We consider the roots of unity $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ of the polynomial
$$p(z)=z^{15}-1=\prod_{k=0}^{14}(z-e^{\frac{2\pi i k}{15}})$$
We obtain 
\begin{align*}
-p(-z)=z^{15}+1&=\prod_{k=0}^{14}(z+e^{\frac{2\pi i k}{15}})\\
&=(z+1)\prod_{k=1}^{7}\left[(z+e^{\frac{2\pi i k}{15}})(z+e^{-\frac{2\pi i k}{15}})\right]\\
\end{align*}
Evaluating the polynomial $-p(-z)$ at $z=1$ gives
\begin{align*}
1&=\prod_{k=1}^{7}\left[(1+e^{\frac{2\pi i k}{15}})(1+e^{-\frac{2\pi i k}{15}})\right]\\
&=\prod_{k=1}^{7}\left[(e^{-\frac{\pi i k}{15}}+e^{\frac{\pi i k}{15}})e^{\frac{\pi i k}{15}}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})e^{-\frac{\pi i k}{15}}\right]\\
&=\prod_{k=1}^{7}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})^2\tag{1}\\
&=\prod_{k=1}^{7}\left(2\cos\left(\frac{k \pi}{15}\right)\right)^2\tag{2}\\
\end{align*}

In (1) we use the formula $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

We conclude from (2)
  \begin{align*}
\prod_{k=1}^{7}\cos\left(\frac{k\pi}{15}\right)=\frac{1}{2^7}
\end{align*}

Note: Writing $-p(-z)$ as
\begin{align*}
-p(-z)=\prod_{k=1}^{7}\left[z^2+\left(e^{\frac{2\pi i k}{15}}+e^{-\frac{2\pi i k}{15}}\right)z+1\right]\\
\end{align*}
and evaluating the polynomial $-p(-z)$ at $z=i$ we obtain the related formula

\begin{align*}
\prod_{k=1}^{7}\cos\left(\frac{2k\pi}{15}\right)=\frac{1}{2^7}
\end{align*}

Doubling the argument does not change the value of the product.
