Question

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

Answer 1

Answer: 6

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