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

Answer 1

$m^2\cos \left(\frac{2\pi }{15}\right)\cos \left(\frac{4\pi }{15}\right)\cos \left(\frac{8\pi }{15}\right)\cos \left(\frac{14\pi }{15}\right)=n^2$

$\Rightarrow -m^2\cos \left(\frac{\pi }{15}\right)\cos \left(\frac{2\pi }{15}\right)\cos \left(\frac{4\pi }{15}\right)\cos \left(\frac{8\pi }{15}\right)=n^2$

$\Rightarrow (-m^2)2\sin \left(\frac{\pi }{15}\right)\cos \left(\frac{\pi }{15}\right)\cos \left(\frac{2\pi }{15}\right)\cos \left(\frac{4\pi }{15}\right)\cos \left(\frac{8\pi }{15}\right)=n^2\left(2\sin \frac{\pi }{15}\right)$

$\Rightarrow (-m^2)2\sin \left(\frac{2\pi }{15}\right)\cos \left(\frac{2\pi }{15}\right)\cos \left(\frac{4\pi }{15}\right)\cos \left(\frac{8\pi }{15}\right)=2n^2\sin \frac{\pi }{15}$

$\Rightarrow \left(\frac{-m^2}{2}\right)\sin \left(\frac{4\pi }{15}\right)\cos \left(\frac{4\pi }{15}\right)\cos \left(\frac{8\pi }{15}\right)=2n^2\left(\sin \frac{\pi }{15}\right)$

$\Rightarrow \left(\frac{-m^2}{8}\right)\sin \left(\frac{16\pi }{15}\right)=2n^2\left(\sin \frac{\pi }{15}\right)$

$\Rightarrow \left(\frac{-m^2}{8}\right)(-\sin (\frac{11\pi }{15}\left)\right)=2n^2\left(\sin \frac{\pi }{15}\right)$

$\Rightarrow m^2=16n^2$

$\Rightarrow \frac{m^2-n^2}{n^2}=\frac{15n^2}{n^2}=15$