Question

If $\frac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }=\tan k\theta$ then find value of $k$.

Answer 1

Answer: 6.00

$\frac{\sin 9\theta +\sin 7\theta +\sin 5\theta +\sin 3\theta }{\cos 9\theta +\cos 7\theta +\cos 5\theta +\cos 3\theta }$
$=\frac{2\sin 8\theta \cos \theta +2\sin 4\theta \cos \theta }{2\cos 8\theta \cos \theta +2\cos 4\theta \cos \theta }$
$=\frac{2\cos \theta (\sin 8\theta +\sin 4\theta )}{2\cos \theta (\cos 8\theta +\cos 4\theta )}$
$=\frac{2\sin 6\theta \cos 2\theta }{2\cos 6\theta \cos 2\theta }$
$=\tan 6\theta$