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 $\frac{m^2 - n^2}{n^2}$

Question

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:)

Answer 1

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 $\sin\dfrac{16\pi}{15}=\sin\left(\pi+\dfrac\pi{15}\right)=?$

Answer 2

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

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