Question

If $\alpha =\frac{\pi }{15}$, then $\cos 2\alpha \cos 4\alpha \cos 8\alpha \cos 14\alpha =\frac{1}{m}$.Find $m$

Answer 1

$\cos 2\alpha \cos 4\alpha \cos 8\alpha \cos 14\alpha =\frac{1}{m}$

$\alpha =\frac{\pi }{15}$

$\Rightarrow 14\alpha =\frac{14\alpha }{15}=(\pi -\frac{\pi }{15})=\pi -\alpha$

$\cos 2\alpha \cos 4\alpha \cos 8\alpha \cos 14\alpha =-\cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha$

$\Rightarrow \frac{1}{m}=\frac{-2\sin \alpha \cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha }{2\sin \alpha }$

$\Rightarrow \frac{1}{m}=\frac{-2\sin 2\alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha }{4\sin \alpha }$

$\Rightarrow \frac{1}{m}=\frac{-2\sin 4\alpha \cos 4\alpha \cos 8\alpha }{8\sin \alpha }$

$\Rightarrow \frac{1}{m}=\frac{-2\sin 8\alpha \cos 8\alpha }{16\sin \alpha }$

$\Rightarrow \frac{1}{m}=\frac{-\sin 16\alpha }{16\sin \alpha }$

$(16\alpha =\frac{16\pi }{15}=\pi +\frac{\pi }{15}=\pi +\alpha )$

$\Rightarrow \frac{1}{m}=\frac{-(-\sin \alpha )}{16\sin \alpha }$

$\Rightarrow \frac{1}{m}=\frac{1}{16}$

$\Rightarrow m=16$