Question

If $\frac{\sin \theta \sin 2 \theta+\sin 3 \theta \sin 6 \theta+\sin 4 \theta \sin 13 \theta}{\sin \theta \cos 2 \theta+\sin 3 \theta \cos 6 \theta+\sin 4 \theta \cos 13 \theta} $ $=\tan ( k \theta)$ then find $k$.

Answer 1

$E =\frac{\sin \theta+\sin 2 \theta+\sin 3 \theta \sin 6 \theta+\sin 4 \theta \sin 13 \theta}{\sin \theta \cos 2 \theta+\sin 3 \theta \cos 6 \theta+\sin 4 \theta \cos 13 \theta}$
$=\frac{2 \sin \theta \sin 2 \theta+2 \sin 3 \theta \sin 6 \theta+2 \sin 4 \theta \sin 13 \theta}{2 \sin \theta \cos 2 \theta+2 \sin 3 \theta \cos 6 \theta+2 \sin 4 \theta \cos 13 \theta}$
$=\frac{(\cos \theta-\cos 3 \theta)+(\cos 3 \theta-\cos 9 \theta)+(\cos 9 \theta-\cos 17 \theta)}{(\sin 3 \theta-\sin \theta)+(\sin 9 \theta-\sin 3 \theta)+(\sin 17 \theta-\sin 9 \theta)}$
$=\frac{\cos \theta-\cos 17 \theta}{\sin 17 \theta-\sin \theta}=\frac{2 \sin 9 \theta \sin 8 \theta}{2 \cos 9 \theta \sin 8 \theta}=\tan (9 \theta)$
$=\tan ( k \theta)$
Hence, $k =9$