Simplification a trigonometric equation $$16 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
$$=4\times 2 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \times2 \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
I am intending in this way and then tried to apply the formula, $2\cos A \cos B$ but i think I might not get the answer. What to do now? the result will be 1. 
 A: use this well know identity
$$2\sin{x}\cos{x}=\sin{2x}$$
so
\begin{align*}\cos{x}\cos{2x}\cos{4x}\cos{8x}&=\dfrac{2\sin{x}\cos{x}\cos{2x}\cos{4x}\cos{8x}}{2\sin{x}}\\
&=\dfrac{\sin{2x}\cos{2x}\cos{4x}\cos{8x}}{2\sin{x}}\\
&=\dfrac{2\sin{2x}\cos{2x}\cos{4x}\cos{8x}}{4\sin{x}}\\
&=\dfrac{\sin{4x}\cos{4x}\cos{8x}}{4\sin{x}}\\
&=\dfrac{2\sin{4x}\cos{4x}\cos{8x}}{8\sin{x}}\\
&=\dfrac{\sin{8x}\cos{8x}}{8\sin{x}}\\
&=\dfrac{2\sin{8x}\cos{8x}}{16\sin{x}}\\
&=\dfrac{\sin{16x}}{16\sin{x}}
\end{align*}
let
$x=\dfrac{\pi}{15}$
since
$$\cos{\dfrac{14\pi}{15}}=-\cos{\dfrac{\pi}{15}},\sin{\dfrac{16\pi}{15}}=-\sin{\dfrac{\pi}{15}}$$
so\begin{align*}
&16\cos{\dfrac{2\pi}{15}}\cos{\dfrac{4\pi}{15}}\cos{\dfrac{8\pi}{15}}\cos{\dfrac{14\pi}{15}}\\
&=-16\cos{\dfrac{\pi}{15}}\cos{\dfrac{2\pi}{15}}\cos{\dfrac{4\pi}{15}}\cos{\dfrac{8\pi}{15}}\\
&=-\dfrac{\sin{\dfrac{16\pi}{15}}}{\sin{\dfrac{\pi}{15}}}\\
&=1
\end{align*}
A: Using the identity, $\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=\dfrac{\sin2^n\theta}{2^n\sin\theta}$ 
By putting $n=4$ and $\theta=\dfrac{\pi}{15}$ you will get,
$- \cos\dfrac{\pi}{15}\cos \dfrac{2 \pi}{15} \cos\dfrac{4 \pi}{15} \cos\dfrac{8 \pi}{15}=-\dfrac{\sin2^4\dfrac{\pi}{15}}{2^4\sin\dfrac{\pi}{15}} $
Multiply both sides by $16$ and you get the required identity ,
$\implies - 16\cos\dfrac{\pi}{15}\cos \dfrac{2 \pi}{15} \cos\dfrac{4 \pi}{15} \cos\dfrac{8 \pi}{15}=-16\dfrac{\sin16\dfrac{\pi}{15}}{16\sin\dfrac{\pi}{15}} $
$\implies -\dfrac{\sin\dfrac{16\pi}{15}}{\sin\dfrac{\pi}{15}}= \dfrac{\sin\dfrac{\pi}{15}}{\sin\dfrac{\pi}{15}}=1$
