If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$ I am a beginner at trigonometry, I want to know the answer to this question.

If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$.

These are the steps I have tried and got stuck in the middle.
$$m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{7\pi}{15}}\cos{\frac{\pi}{15}} = n^2 $$
$$m^2(\frac{1}{4})(2\cos{\frac{2\pi}{15}}\cos{\frac{\pi}{15}})(2\cos{\frac{4\pi}{15}}\cos{\frac{7\pi}{15}}) = n^2$$
$$m^2(\frac{1}{4})(\cos{\frac{3\pi}{15}} + \cos{\frac{\pi}{15}})(\cos{\frac{11\pi}{15}} + \cos{\frac{3\pi}{15}}) = n^2$$
Fro here onwards I could not continue. These steps may be wrong so please check if my method of solving is correct and help me solve the question. Thanks:)
 A: Hint:
$\cos\dfrac{14\pi}{15}=\cos\left(\pi-\dfrac\pi{15}\right)=?$
Use $\sin2x=2\sin x\cos x\iff\cos x=\dfrac{\sin2x}{2\sin x}$ repeatedly
Finally here $\sin\dfrac{16\pi}{15}=\sin\left(\pi+\dfrac\pi{15}\right)=?$
A: You can use Morrie's law:


*

*$2^n\prod_{k=0}^{n-1}\cos(2^k a) = \frac{\sin(2^na)}{\sin a}$
To do so, note that


*

*$\cos\left(\frac{14}{15}\pi \right) = -\cos\left(\frac{1}{15}\pi\right)$ and

*$\sin\left(\pi + \frac{1}{15}\pi \right) = -\sin\left(\frac{1}{15}\pi \right) $
It follows
$$\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = -2^4\prod_{k=0}^{3}\cos\left(2^k \frac{1}{15}\pi\right)$$ $$ = \frac{\sin\left(2^4 \frac{1}{15}\pi\right)}{\sin \frac{1}{15}\pi}= \frac{\sin\left(\pi +\frac{1}{15}\pi\right)}{\sin \frac{1}{15}\pi} = -1$$
Hence
$$\boxed{p :=\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = \frac{1}{16}}$$
Now, remember that


*

*$m^2p = n^2 \Rightarrow \color{blue}{\frac{m^2-n^2}{n^2}}= \frac{m^2}{n^2}-1 = \frac{1}{p}-1 \color{blue}{= 15}$
