Question

If $\frac{\displaystyle\sum_{r=1}^{8} \sin \left((2 r-1) \frac{\pi}{36}\right)}{\displaystyle\prod_{k=0}^{6} \cos \left(2^{k} \cdot \frac{\pi}{36}\right)}=2^{n} \tan 140^{\circ}$ (where $n \in N$ ) then '$n$' is

Answer 1

$\frac{\displaystyle\sum_{r=1}^{8} \sin \left((2 r-1) \frac{\pi}{36}\right)}{\displaystyle\prod_{k=0}^{6} \cos \left(2^{k} \cdot \frac{\pi}{36}\right)}=\left(\frac{\sin \frac{8 \pi}{36} \cdot \sin \frac{8 \pi}{36}}{\sin \frac{\pi}{36}}\right)\left(\frac{2^{7} \sin \frac{\pi}{36}}{\sin 2^{7} \frac{\pi}{36}}\right)$
$\Rightarrow \frac{2^{7} \sin 40^{\circ} \sin 40^{\circ}}{\sin \left(640^{\circ}\right)}=\frac{2^{7} \sin 40^{\circ}}{-\cos 10^{\circ}}$
$=-2^{6}\left[\frac{1-\cos 80^{\circ}}{\left(\cos 5^{\circ}-\sin 5^{\circ}\right)\left(\cos 5^{\circ}+\sin 5^{\circ}\right)}\right]$
$=-2^{6}\left(\frac{\cos 5^{\circ}-\sin 5^{\circ}}{\cos 5^{\circ}+\sin 5^{\circ}}\right)=2^{6}\left(\frac{\cos 5^{\circ}-\sin 5^{\circ}}{\cos 5^{\circ}+\sin 5^{\circ}}\right)$
$=-2^{6} \tan 40^{\circ}=2^{6} \tan 140^{\circ}$